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**Unformatted text preview: **1 UNITS, PHYSICAL QUANTITIES AND VECTORS 1.1. 1.2. IDENTIFY: Convert units from mi to km and from km to ft.
SET UP: 1 in. = 2.54 cm , 1 km = 1000 m , 12 in. = 1 ft , 1 mi = 5280 ft .
⎛ 5280 ft ⎞⎛ 12 in. ⎞⎛ 2.54 cm ⎞⎛ 1 m ⎞⎛ 1 km ⎞
EXECUTE: (a) 1.00 mi = (1.00 mi) ⎜
⎟⎜
⎟⎜
⎟⎜ 2
⎟⎜ 3 ⎟ = 1.61 km
⎝ 1 mi ⎠⎝ 1 ft ⎠⎝ 1 in. ⎠⎝ 10 cm ⎠⎝ 10 m ⎠ ⎛ 103 m ⎞⎛ 102 cm ⎞ ⎛ 1 in. ⎞⎛ 1 ft ⎞
3
(b) 1.00 km = (1.00 km) ⎜
⎟⎜
⎟⎜
⎟⎜
⎟ = 3.28 × 10 ft
1 km ⎠⎝ 1 m ⎠ ⎝ 2.54 cm ⎠⎝ 12 in. ⎠
⎝
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
IDENTIFY: Convert volume units from L to in.3 .
SET UP: 1 L = 1000 cm3 . 1 in. = 2.54 cm
3 ⎛ 1000 cm3 ⎞ ⎛ 1 in. ⎞
3
0.473 L × ⎜
⎟×⎜
⎟ = 28.9 in. .
⎝ 1 L ⎠ ⎝ 2.54 cm ⎠
EVALUATE: 1 in.3 is greater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in cm3 ,
which is 473 cm3 .
IDENTIFY: We know the speed of light in m/s. t = d / v . Convert 1.00 ft to m and t from s to ns.
SET UP: The speed of light is v = 3.00 × 108 m/s . 1 ft = 0.3048 m . 1 s = 109 ns .
0.3048 m
EXECUTE: t =
= 1.02 × 10−9 s = 1.02 ns
3.00 × 108 m/s
EVALUATE: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km = 1.86 × 105 mi .
IDENTIFY: Convert the units from g to kg and from cm3 to m3 .
SET UP: 1 kg = 1000 g . 1 m = 1000 cm .
EXECUTE: 1.3. 1.4. 3 EXECUTE: 1.5. EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3 .
IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L = 1000 cm3 . 1 in. = 2.54 cm .
EXECUTE: 1.6. ( 327 in. ) × ( 2.54 cm in.) × (1 L 1000 cm ) = 5.36 L
3 3 3 EVALUATE: The volume is 5360 cm3 . 1 cm3 is less than 1 in.3 , so the volume in cm3 is a larger number than the
volume in in.3 .
IDENTIFY: Convert ft 2 to m 2 and then to hectares.
SET UP: 1.00 hectare = 1.00 × 104 m 2 . 1 ft = 0.3048 m .
EXECUTE: 1.7. g ⎛ 1 kg ⎞ ⎛ 100 cm ⎞
4 kg
×
11.3
⎟×⎜
⎟ = 1.13 × 10
3⎜
cm ⎝ 1000 g ⎠ ⎝ 1 m ⎠
m3 ⎛ 43,600 ft 2 ⎞ ⎛ 0.3048 m ⎞
The area is (12.0 acres) ⎜
⎟⎜
⎟
⎝ 1 acre ⎠ ⎝ 1.00 ft ⎠ 2 ⎛ 1.00 hectare ⎞
= 4.86 hectares .
⎜
4
2⎟
⎝ 1.00 × 10 m ⎠ EVALUATE: Since 1 ft = 0.3048 m , 1 ft 2 = (0.3048) 2 m 2 .
IDENTIFY: Convert seconds to years.
SET UP: 1 billion seconds = 1 × 109 s . 1 day = 24 h . 1 h = 3600 s .
EXECUTE: ⎛ 1 h ⎞⎛ 1 day ⎞ ⎛ 1 y ⎞
1.00 billion seconds = (1.00 × 109 s ) ⎜
⎟ = 31.7 y .
⎟⎜
⎟⎜
⎝ 3600 s ⎠⎝ 24 h ⎠ ⎝ 365 days ⎠
1-1 1-2 1.8. Chapter 1 EVALUATE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d , which is the average for one extra day
every four years, in leap years. The problem says instead to assume a 365-day year.
IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h. ⎛ 0.125 mi ⎞⎛ 1 fortnight ⎞⎛ 1 day ⎞
furlongs fortnight ) ⎜
⎟ = 67 mi/h
⎟⎜
⎟⎜
⎝ 1 furlong ⎠⎝ 14 days ⎠ ⎝ 24 h ⎠
EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much
smaller number.
IDENTIFY: Convert miles/gallon to km/L.
SET UP: 1 mi = 1.609 km . 1 gallon = 3.788 L.
EXECUTE: 1.9. (180,000 ⎛ 1.609 km ⎞⎛ 1 gallon ⎞
(a) 55.0 miles/gallon = (55.0 miles/gallon) ⎜
⎟⎜
⎟ = 23.4 km/L .
⎝ 1 mi ⎠⎝ 3.788 L ⎠
1500 km
64.1 L
(b) The volume of gas required is
= 64.1 L .
= 1.4 tanks .
23.4 km/L
45 L/tank
EVALUATE: 1 mi/gal = 0.425 km/L . A km is very roughly half a mile and there are roughly 4 liters in a gallon,
EXECUTE: 1.10. so 1 mi/gal ∼ 2 km/L , which is roughly our result.
4
IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg .
EXECUTE: ft
⎛ mi ⎞ ⎛ 1h ⎞ ⎛ 5280 ft ⎞
(a) ⎜ 60 ⎟ ⎜
⎟⎜
⎟ = 88
h ⎠ ⎝ 3600s ⎠ ⎝ 1mi ⎠
s
⎝ ⎛ ft ⎞ ⎛ 30.48cm ⎞
(b) ⎜ 32 2 ⎟ ⎜
⎟
⎝ s ⎠ ⎝ 1ft ⎠ m
⎛ 1m
⎜
⎟ = 9.8 2
100 cm ⎠
s
⎝
3 g ⎞ ⎛ 100 cm ⎞ ⎛ 1 kg ⎞
⎛
3 kg
(c) ⎜1.0 3 ⎟ ⎜
⎟ = 10 3
⎟⎜
cm ⎠ ⎝ 1 m ⎠ ⎝ 1000 g ⎠
m
⎝ 1.11. EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation 32 ft/s 2 = 9.8 m/s 2 is
accurate to only two significant figures.
IDENTIFY: We know the density and mass; thus we can find the volume using the relation
density = mass/volume = m / V . The radius is then found from the volume equation for a sphere and the result for
the volume.
4
SET UP: Density = 19.5 g/cm3 and mcritical = 60.0 kg. For a sphere V = 3 π r 3 .
EXECUTE: ⎛ 60.0 kg ⎞⎛ 1000 g ⎞
V = mcritical / density = ⎜
= 3080 cm 3 .
19.5 g/cm3 ⎟⎜ 1.0 kg ⎟
⎝
⎠⎝
⎠
r=3 1.12. EVALUATE: The density is very large, so the 130 pound sphere is small in size.
IDENTIFY: Use your calculator to display π × 107 . Compare that number to the number of seconds in a year.
SET UP: 1 yr = 365.24 days, 1 day = 24 h, and 1 h = 3600 s. ⎛ 24 h ⎞ ⎛ 3600 s ⎞
7
7
7
(365.24 days/1 yr) ⎜
⎟⎜
⎟ = 3.15567... × 10 s ; π × 10 s = 3.14159... × 10 s
1 day ⎠ ⎝ 1 h ⎠
⎝
The approximate expression is accurate to two significant figures.
EVALUATE: The close agreement is a numerical accident.
IDENTIFY: The percent error is the error divided by the quantity.
SET UP: The distance from Berlin to Paris is given to the nearest 10 km.
10 m
EXECUTE: (a)
= 1.1 × 10−3%.
890 × 103 m
(b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total
distance to only three significant figures.
EVALUATE: In this case a very small percentage error has disastrous consequences.
IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no
greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of
the decimal that matters.
EXECUTE: 1.13. 1.14. 3V 3 3 (
3080 cm3 ) = 9.0 cm .
=
4π
4π Units, Physical Quantities and Vectors 1-3 SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures.
EXECUTE: (a) (12 mm ) × ( 5.98 mm ) = 72 mm 2 (two significant figures) 5.98 mm
= 0.50 (also two significant figures)
12 mm
(c) 36 mm (to the nearest millimeter)
(d) 6 mm
(e) 2.0 (two significant figures)
EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are
known only to the nearest mm.
IDENTIFY and SET UP: In each case, estimate the precision of the measurement.
EXECUTE: (a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%.
(b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3 × 10−3%.
(c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the
error will be about 2.8 × 10−2%.
EVALUATE: The percent errors are those due only to the limit of precision of the measurement.
IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area.
SET UP: The length could be as large as 5.11 cm and the width could be as large as 1.91 cm.
0.07 cm 2
EXECUTE: The area is 9.69 ± 0.07 cm2. The fractional uncertainty in the area is
= 0.72%, and the
9.69 cm 2
0.01 cm
0.01 cm
= 0.20% and
= 0.53%. The sum of these
fractional uncertainties in the length and width are
5.10 cm
1.9 cm
fractional uncertainties is 0.20% + 0.53% = 0.73% , in agreement with the fractional uncertainty in the area.
EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of
the individual numbers.
IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities.
SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target variables
and from these we get the uncertainty.
EXECUTE: (a) The volume of a disk of diameter d and thickness t is V = π (d / 2) 2 t.
(b) 1.15. 1.16. 1.17. The average volume is V = π (8.50 cm/2) 2 (0.50 cm) = 2.837 cm3 . But t is given to only two significant figures so
the answer should be expressed to two significant figures: V = 2.8 cm3 .
We can find the uncertainty in the volume as follows. The volume could be as large as
V = π (8.52 cm/2) 2 (0.055 cm) = 3.1 cm 3 , which is 0.3 cm3 larger than the average value. The volume could be as
small as V = π (8.52 cm/2) 2 (0.045 cm) = 2.5 cm3 , which is 0.3 cm3 smaller than the average value. The 1.18. uncertainty is ±0.3 cm3 , and we express the volume as V = 2.8 ± 0.3 cm3 .
(b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm = 170. By taking the largest
possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio:
8.52 cm/0.045 cm = 190. The smallest possible value of the ratio is 8.48 / 0.055 = 150. Thus the uncertainty is
±20 and we write the ratio as 170 ± 20.
EVALUATE: The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much less, so
the percentage uncertainty in the volume and in the ratio should be about 10%.
IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the
number of gallons.
SET UP: Estimate 3 × 108 people, so 2 × 108 cars.
EXECUTE: ( Number of cars × miles/car day ) / ( mi/gal ) = gallons/day ( 2 ×10 8 1.19. cars × 10000 mi/yr/car × 1 yr/365 days ) / ( 20 mi/gal ) = 3 × 108 gal/day EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S.
IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express
200 months in years.
SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs. 1 in. = 2.54 cm . 1 y = 12 months .
EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man.
⎛ 1 in. ⎞
3
(b) 200 m = (2.00 × 104 cm) ⎜
⎟ = 7.9 × 10 inches . This is much greater than the height of a person.
2.54 cm ⎠
⎝
(c) 200 cm = 2.00 m = 79 inches = 6.6 ft . Some people are this tall, but not an ordinary man. 1-4 1.20. Chapter 1 (d) 200 mm = 0.200 m = 7.9 inches . This is much too short.
⎛ 1y ⎞
(e) 200 months = (200 mon) ⎜
⎟ = 17 y . This is the age of a teenager; a middle-aged man is much older than this.
⎝ 12 mon ⎠
EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units
in which it is being expressed.
IDENTIFY: The number of kernels can be calculated as N = Vbottle / Vkernel .
SET UP: Based on an Internet search, Iowan corn farmers use a sieve having a hole size of 0.3125 in. ≅ 8 mm to
remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth
as 3 mm. We must also apply the conversion factors 1 L = 1000 cm3 and 1 cm = 10 mm.
EXECUTE: The volume of the kernel is: Vkernel = (10 mm ) ( 6 mm ) ( 3 mm ) = 180 mm3 . The bottle’s volume is:
3
3
Vbottle = ( 2.0 L ) ⎡(1000 cm3 ) (1.0 L ) ⎤ ⎡(10 mm ) (1.0 cm ) ⎤ = 2.0 × 106 mm3 . The number of kernels is then
⎣
⎦⎣
⎦ 1.21. 1.22. N kernels = Vbottle / Vkernels ≈ ( 2.0 × 106 mm3 ) (180 mm3 ) = 11,000 kernels .
EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since these
dimensions vary amongst the different available types of corn, acceptable answers could range from 6,500 to
20,000.
IDENTIFY: Estimate the number of pages and the number of words per page.
SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has
between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises
and problems).
EXECUTE: An estimate for the number of words is about 106 .
EVALUATE: We can expect that this estimate is accurate to within a factor of 10.
IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to find the
volume in m3 breathed in a year.
⎛ 24 h ⎞⎛ 60 min ⎞
5
2
SET UP: Assume 10 breaths/min . 1 y = (365 d) ⎜
⎟⎜
⎟ = 5.3 × 10 min . 10 cm = 1 m so
⎝ 1 d ⎠⎝ 1 h ⎠
4
106 cm3 = 1 m3 . The volume of a sphere is V = 3 π r 3 = 1 π d 3 , where r is the radius and d is the diameter. Don’t
6
forget to account for four astronauts.
⎛ 5.3 × 105 min ⎞
4
3
EXECUTE: (a) The volume is (4)(10 breaths/min)(500 × 10−6 m3 ) ⎜
⎟ = 1 × 10 m / yr .
1y
⎝
⎠
1/ 3 ⎛ 6V ⎞
(b) d = ⎜
⎟
⎝π ⎠ 1.23. 1.24. 1/ 3 ⎛ 6[1 × 104 m3 ] ⎞
=⎜
⎟
π
⎝
⎠ = 27 m EVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all the
oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be
required.
IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical lifetime in
years.
SET UP: Estimate that we blink 10 times per minute. 1 y = 365 days . 1 day = 24 h , 1 h = 60 min . Use 80 years
for the lifetime.
⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞
8
EXECUTE: The number of blinks is (10 per min) ⎜
⎟⎜
⎟ (80 y/lifetime) = 4 × 10
⎟⎜
⎝ 1 h ⎠ ⎝ 1 day ⎠⎝ 1 y ⎠
EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is
surely accurate to a power of 10.
IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped
during this interval is then the volume per beat multiplied by the total beats.
SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To
calculate the number of beats in a lifetime, use the current average lifespan of 80 years.
⎛ 60 min ⎞⎛ 24 h ⎞⎛ 365 days ⎞⎛ 80 yr ⎞
9
EXECUTE: N beats = ( 75 beats/min ) ⎜
⎟
⎟⎜ lifespan ⎟ = 3 × 10 beats/lifespan
yr
⎝ 1 h ⎠ ⎜ 1 day ⎟⎜
⎝
⎠⎝
⎠⎝
⎠
9
⎛ 1 L ⎞⎛ 1 gal ⎞ ⎛ 3 × 10 beats ⎞
7
Vblood = ( 50 cm3 /beat ) ⎜
⎟⎜
⎟ = 4 × 10 gal/lifespan
3 ⎟⎜
⎝ 1000 cm ⎠⎝ 3.788 L ⎠ ⎝ lifespan ⎠
EVALUATE: This is a very large volume. Units, Physical Quantities and Vectors 1.25. 1-5 IDENTIFY: Estimation problem
SET UP: Estimate that the pile is 18 in. × 18 in. × 5 ft 8 in.. Use the density of gold to calculate the mass of gold in
the pile and from this calculate the dollar value.
EXECUTE: The volume of gold in the pile is V = 18 in. × 18 in. × 68 in. = 22,000 in.3 . Convert to cm3 : V = 22,000 in.3 (1000 cm3 / 61.02 in.3 ) = 3.6 × 105 cm3 .
The density of gold is 19.3 g/cm3 , so the mass of this volume of gold is m = (19.3 g/cm3 )(3.6 × 105 cm3 ) = 7 × 106 g.
The monetary value of one gram is $10, so the gold has a value of ($10 / gram)(7 × 106 grams) = $7 × 107 , or about 1.26. $100 × 106 (one hundred million dollars).
EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable.
IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3 . Convert m3 to L.
4
SET UP: Estimate the diameter of a drop to be d = 2 mm . The volume of a spherical drop is V = 3 π r 3 = 1 π d 3 .
6
103 cm3 = 1 L .
1000 cm3
= 2 × 105
4 × 10−3 cm3
EVALUATE: Since V ∼ d 3 , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the
number of drops is off by a factor of 8.
IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year.
SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate)
EXECUTE: They eat a total of 104 pizzas.
EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each.
IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill.
SET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the
measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance
from the earth to the moon is 3.8 × 108 m.
EXECUTE: 1.27. 1.28. V = 1 π (0.2 cm)3 = 4 × 10−3 cm3 . The number of drops in 1.0 L is
6 ⎛ 3.8 × 108 m ⎞⎛ 103 mm ⎞
12
12
N bills = ⎜
⎟ = 3.8 × 10 bills ≈ 4 × 10 bills
⎟⎜
⎝ 0.1 mm/bill ⎠ ⎝ 1 m ⎠
EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is
significantly less – roughly 1 billion dollars.
IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the
surface area of a single dollar bill.
SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or
3,380,000 mi 2 . This estimate is within 10 percent of the actual area, 3,794,083 mi 2 . The population is roughly
5
3.0 × 108 while the area of a dollar bill, as measured with a ruler, is approximately 6 1 in. by 2 8 in.
8
EXECUTE: 1.29. AU.S. = ( 3,380,000 mi 2 ) [( 5280 ft ) / (1 mi )] ⎡(12 in.) (1 ft )⎤ = 1.4 × 1016 in.2
⎣
⎦
2
= ( 6.125 in.)( 2.625 in.) = 16.1 in. EXECUTE: Abill 2 2 Total cost = N bills = AU.S. Abill = (1.4 × 1016 in.2 ) (16.1 in.2 / bill ) = 9 × 1014 bills 1.30. Cost per person = (9 × 1014 dollars) /(3.0 × 108 persons) = 3 × 106 dollars/person
EVALUATE: The actual cost would be somewhat larger, because the land isn’t flat.
IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative
orientation of the two displacements.
SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the
smallest magnitude is when the two displacements are antiparallel.
EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30.
EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between
0.6 m and 4.2 m. Figure 1.30 1-6 1.31. Chapter 1 IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant
displacement is the single vector that points from the starting point to the stopping point.
""
"
"
""""
SET UP: Call the three displacements A , B , and C . The resultant displacement R is given by R = A + B + C .
"
EXECUTE: The vector addition diagram is given in Figure 1.31. Careful measurement gives that R is
7.8 km, 38# north of east .
EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the
individual displacements, 2.6 km + 4.0 km + 3.1 km . Figure 1.31
1.32. IDENTIFY: Draw the vector addition diagram, so scale.
"
"
SET UP: The two vectors A and B are specified in the figure that accompanies the problem.
"""
"
EXECUTE: (a) The diagram for C = A + B is given in Figure 1.32a. Measuring the length and angle of C gives
C = 9.0 m and an angle of θ = 34° .
"
"""
(b) The diagram for D = A − B is given in Figure 1.32b. Measuring the length and angle of D gives D = 22 m and
an angle of θ = 250° .
""
""
""
(c) − A − B = −( A + B ) , so − A − B has a magnitude of 9.0 m (the same as A + B ) and an angle with the + x axis
""
of 214° (opposite to the direction of A + B ).
""
""
""
(d) B − A = −( A − B ) , so B − A has a magnitude of 22 m and an angle with the + x axis of 70° (opposite to the
""
direction of A − B ).
"
"
EVALUATE: The vector − A is equal in magnitude and opposite in direction to the vector A . Figure 1.32
1.33. IDENTIFY: Since she returns to the starting point, the vectors sum of the four displacements must be zero.
""
"
"
SET UP: Call the three given displacements A , B , and C , and call the fourth displacement D .
""""
A+ B + C + D = 0.
"
EXECUTE: The vector addition diagram is sketched in Figure 1.33. Careful measurement gives that D
is144 m, 41# south of west. Units, Physical Quantities and Vectors 1-7 """
"
D is equal in magnitude and opposite in direction to the sum A + B + C . EVALUATE: Figure 1.33
1.34. 1.35. IDENTIFY and SET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding
horizontal and vertical components.
EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m.
(b) Figure 1.34 gives components −15.6 km, 15.6 km .
(c) Figure 1.34 gives components 3.82 cm, − 5.07 cm .
EVALUATE: The signs of the components depend on the quadrant in which the vector lies. Figure 1.34
"
"
IDENTIFY: For each vector V , use that Vx = V cosθ and Vy = V sin θ , when θ is the angle V makes with the + x axis, measured counterclockwise from the axis.
"
"
"
"
SET UP: For A , θ = 270.0° . For B , θ = 60.0° . For C , θ = 205.0° . For D , θ = 143.0° .
EXECUTE: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m . Dx = −7.99 m , Dy = 6.02 m .
The signs of the components correspond to the quadrant in which the vector lies.
A
IDENTIFY: tan θ = y , for θ measured counterclockwise from the + x -axis.
Ax
"
"
SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies.
EVALUATE:
1.36. EXECUTE:
(a) tan θ =
(b) tan θ =
(c) tan θ =
(d) tan θ = 1.37. Ay
AX
Ay
Ax
Ay
Ax
Ay
Ax = −1.00 m
= −0.500 . θ = tan −1 ( −0.500 ) = 360° − 26.6° = 333° .
2.00 m = 1.00 m
= 0.500 . θ = tan −1 ( 0.500 ) = 26.6° .
2.00 m = 1.00 m
= −0.500 . θ = tan −1 ( −0.500 ) = 180° − 26.6° = 153° .
−2.00 m = −1.00 m
= 0.500 . θ = tan −1 ( 0.500 ) = 180° + 26.6° = 207°
−2.00 m EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value
of θ .
IDENTIFY: Find the vector sum of the two forces.
SET UP: Use components to add the two forces. Take the + x -direction to be forward and the + y -direction to be
upward. 1-8 Chapter 1 The second force has components F2 x = F2 cos32.4° = 433 N and F2 y = F2 sin 32.4° = 275 N. The EXECUTE: first force has components F1x = 725 N and F1 y = 0. Fx = F1x + F2 x = 1158 N and Fy = F1 y + F2 y = 275 N 1.38. The resultant force is 1190 N in the direction 13.4° above the forward direction.
EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less than the
sum of their magnitudes.
IDENTIFY: Find the vector sum of the three given displacements.
SET UP: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are:
$
$
$
A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east . Rx = Ax + Bx + C x = 0 + 4.0 km + ( 3.1 km ) cos ( 45# ) = 6.2 km ; Ry = Ay + By + C y = EXECUTE: 2
2.6 km + 0 + (3.1 km) ( sin45# ) = 4.8 km ; R = Rx2 + Ry = 7.8 km ; θ = tan −1 ⎡( 4.8 km ) ( 6.2 km )⎤ = 38# ;
⎣
⎦
$
#
R = 7.8 km, 38 north of east. This result is confirmed by the sketch in Figure 1.38.
"
EVALUATE: Both Rx and Ry are positive and R is in the first quadrant. 1.39. Figure 1.38
"""
IDENTIFY: If C = A + B , then Cx = Ax + Bx and C y = Ay + By . Use C x and C y to find the magnitude and
"
direction of C .
SET UP: From Figure 1.34 in the textbook, Ax = 0 , Ay = −8.00 m and Bx = + B sin 30.0° = 7.50 m , By = + B cos30.0° = 13.0 m .
"""
EXECUTE: (a) C = A + B so Cx = Ax + Bx = 7.50 m and C y = Ay + By = +5.00 m . C = 9.01 m . 1.40. Cy 5.00 m
and θ = 33.7° .
Cx 7.50 m
""""
""
(b) B + A = A + B , so B + A has magnitude 9.01 m and direction specified by 33.7° .
"""
D
−21.0 m
(c) D = A − B so Dx = Ax − Bx = −7.50 m and Dy = Ay − By = −21.0 m . D = 22.3 m . tan φ = y =
and
Dx −7.50 m
"
φ = 70.3° . D is in the 3rd quadrant and the angle θ counterclockwise from the + x axis is 180° + 70.3° = 250.3° .
""
""
""
(d) B − A = −( A − B ) , so B − A has magnitude 22.3 m and direction specified by θ = 70.3° .
EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.32.
IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors.
"
"
SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies.
tan θ = EXECUTE:
(b) = (a) ⎛ 5.20 ⎞
(−8.60 cm) 2 + (5.20 cm) 2 = 10.0 cm, arctan ⎜
⎟ = 148.8° (which is 180° − 31.2° ).
⎝ −8.60 ⎠ ⎛ −2.45 ⎞
(−9.7 m) 2 + (−2.45 m) 2 = 10.0 m, arctan ⎜
⎟ = 14° + 180° = 194°.
⎝ −9.7 ⎠ ⎛ −2.7 ⎞
(7.75 km)2 + (−2.70 km) 2 = 8.21 km, arctan ⎜
⎟ = 340.8° (which is 360° − 19.2° ).
⎝ 7.75 ⎠
EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agree with
our sketches.
(c) Units, Physical Quantities and Vectors 1.41. 1-9 IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to
find their sum.
SET UP: A = 3.25 km
B = 4.75 km
C = 1.50 km Figure 1.41a ""
"
Select a coordinate system where + x is east and + y is north. Let A, B and C be the three displacements of the
"
""""
professor. Then the resultant displacement R is given by R = A + B + C . By the method of components,
Rx = Ax + Bx + C x and Ry = Ay + By + C y . Find the x and y components of each vector; add them to find the
components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y
components that we have calculated. As always it is essential to draw a sketch.
EXECUTE:
Ax = 0, Ay = +3.25 km
Bx = −4.75 km, By = 0
Cx = 0, C y = −1.50 km Rx = Ax + Bx + C x
Rx = 0 − 4.75 km + 0 = −4.75 km
Ry = Ay + By + C y
Ry = 3.25 km + 0 − 1.50 km = 1.75 km
Figure 1.41b 2
R = Rx2 + Ry = ( −4.75 km) 2 + (1.75 km) 2 R = 5.06 km
R
1.75 km
= −0.3684
tan θ = y =
Rx −4.75 km
θ = 159.8°
Figure 1.41c 1.42. The angle θ measured counterclockwise from the + x -axis. In terms of compass directions, the resultant
displacement is 20.2° N of W.
"
EVALUATE: Rx < 0 and Ry > 0, so R is in 2nd quadrant. This agrees with the vector addition diagram.
"
"""
IDENTIFY: Add the vectors using components. B − A = B + (− A) .
"""
"""
SET UP: If C = A + B then Cx = Ax + Bx and C y = Ay + By . If D = B − A then Dx = Bx − Ax and Dy = By − Ay .
EXECUTE: (a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm,
2.25 cm + ( −3.75 cm) = −1.50 cm.
(b) Using Equations (1.7) and (1.8), ⎛ −1.50 ⎞
(5.40cm) 2 (−1.50 cm) 2 = 5.60 cm, arctan ⎜
⎟ = 344.5° ccw.
⎝ +5.40 ⎠ 1-10 Chapter 1 (c) Similarly, 4.10 cm − (1.30 cm ) = 2.80 cm, 23.75 cm − ( 2.25 cm ) = 26.00 cm. ⎛ −6.00 ⎞
(2.80cm) 2 + (−6.00cm) 2 = 6.62 cm, arctan ⎜
⎟ = 295° (which is 360° − 65° ).
⎝ 2.80 ⎠
EVALUATE: We can draw the vector addition diagram in each case and verify that our results are qualitatively
correct.
"""
"
IDENTIFY: Vector addition problem. A − B = A + (− B ).
"
"
SET UP: Find the x- and y-components of A and B. Then the x- and y-components of the vector sum are
"
"
calculated from the x- and y-components of A and B.
EXECUTE:
Ax = A cos(60.0°)
(d) 1.43. Ax = ( 2.80 cm)cos(60.0°) = +1.40 cm
Ay = A sin(60.0°)
Ay = (2.80 cm)sin(60.0°) = +2.425 cm
Bx = B cos( −60.0°)
Bx = (1.90 cm)cos(−60.0°) = +0.95 cm
By = B sin(−60.0°)
By = (1.90 cm)sin(−60.0°) = −1.645 cm
Note that the signs of the components
correspond to the directions of the component
vectors.
Figure 1.43a
"""
(a) Now let R = A + B.
Rx = Ax + Bx = +1.40 cm + 0.95 cm = +2.35 cm. Ry = Ay + By = +2.425 cm − 1.645 cm = +0.78 cm.
2
R = Rx2 + Ry = (2.35 cm) 2 + (0.78 cm) 2 R = 2.48 cm
R
+0.78 cm
tan θ = y =
= +0.3319
Rx +2.35 cm
θ = 18.4°
Figure 1.43b
EVALUATE: """
The vector addition diagram for R = A + B is
"
R is in the 1st quadrant, with
Ry < Rx , in agreement with
our calculation. Figure 1.43c Units, Physical Quantities and Vectors """
(b) EXECUTE: Now let R = A − B.
Rx = Ax − Bx = +1.40 cm − 0.95 cm = +0.45 cm.
Ry = Ay − By = +2.425 cm + 1.645 cm = +4.070 cm. 2
R = Rx2 + Ry = (0.45 cm) 2 + (4.070 cm) 2 R = 4.09 cm
R
4.070 cm
= +9.044
tan θ = y =
Rx
0.45 cm
θ = 83.7° Figure 1.43d
""
"
EVALUATE: The vector addition diagram for R = A + − B is () "
R is in the 1st quadrant,
with Rx < Ry , in
agreement with our
calculation. Figure 1.43e
(c) EXECUTE: ""
""
B − A= − A−B
""
""
B − A and A − B are
equal in magnitude and
opposite in direction.
R = 4.09 cm and
θ = 83.7° + 180° = 264° ( Figure 1.43f ) 1-11 1-12 Chapter 1 "
""
EVALUATE: The vector addition diagram for R = B + − A is () "
R is in the 3rd quadrant,
with Rx < Ry , in
agreement with our
calculation. Figure 1.43g
1.44. "
"
IDENTIFY: The velocity of the boat relative to the earth, vB/E , the velocity of the water relative to the earth, vW/E ,
"
"
"
"
and the velocity of the boat relative to the water, vB/W , are related by vB/E = vB/W + v W/E .
"
"
SET UP: vW/E = 5.0 km/h , north and vB/W = 7.0 km/h , west. The vector addition diagram is sketched in
Figure 1.44.
v
5.0 km/h
2
2
2
EXECUTE: vB/E = vW/E + vB/W and vB/E = (5.0 km/h) 2 + (7.0 km/h) 2 = 8.6 km/h . tan φ = W/E =
and
vB/W 7.0 km/h φ = 36° , north of west.
EVALUATE: Since the two vectors we are adding are perpendicular we can use the Pythagorean theorem directly
to find the magnitude of their vector sum. Figure 1.44
1.45. "
"""
IDENTIFY: Let A = 625 N and B = 875 N . We are asked to find the vector C such that A + B = C = 0 .
SET UP: Ax = 0 , Ay = −625 N . Bx = (875 N)cos30° = 758 N , By = (875 N)sin 30° = 438 N .
EXECUTE: C x = −( Ax + Bx ) = −(0 + 758 N) = −758 N . C y = −( Ay + By ) = −( −625 N + 438 N) = +187 N . Vector "
C y 187 N
2
C and its components are sketched in Figure 1.45. C = Cx2 + C y = 781 N . tan φ =
and φ = 13.9° .
=
C x 758 N
"
C is at an angle of 13.9° above the − x -axis and therefore at an angle 180° − 13.9° = 166.1° counterclockwise from
the + x -axis .
"""
EVALUATE: A vector addition diagram for A + B + C verifies that their sum is zero. Figure 1.45 Units, Physical Quantities and Vectors 1.46. 1-13 IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of
components.
"
"
"
SET UP: The two vectors A and B and their resultant C are shown in Figure 1.46. Let + y be in the direction of
the resultant. A = B .
EXECUTE: C y = Ay + By . 372 N = 2 A cos 43.0° and A = 254 N .
EVALUATE: The sum of the magnitudes of the two forces exceeds the magnitude of the resultant force because
only a component of each force is upward. Figure 1.46
1.47. IDENTIFY: Find the components of each vector and then use Eq.(1.14).
SET UP: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m . Dx = −7.99 m , Dy = 6.02 m . 1.48. "
"
"
ˆ
ˆ
EXECUTE: A = (−8.00 m) ˆ ; B = (7.50 m)i + (13.0 m) ˆ ; C = (−10.9 m) i + (−5.07 m) ˆ ;
j
j
j
"
ˆ
D = (−7.99 m)i + (6.02 m) ˆ .
j
EVALUATE: All these vectors lie in the xy-plane and have no z-component.
"
ˆ
IDENTIFY: The general expression for a vector written in terms of components and unit vectors is A = Ax i + Ay ˆ
j
"
"
"
ˆ
SET UP: 5.0 B = 5.0(4i − 6 ˆ) = 20i − 30 j
j EXECUTE: (a) Ax = 5.0 , Ay = −6.3 (b) Ax = 11.2 , Ay = −9.91 (c) Ax = −15.0 , Ay = 22.4
(d) Ax = 20 , Ay = −30
1.49. EVALUATE: The components are signed scalars.
IDENTIFY: Use trig to find the components of each vector. Use Eq.(1.11) to find the components of the vector
sum. Eq.(1.14) expresses a vector in terms of its components.
SET UP: Use the coordinates in the figure that accompanies the problem.
"
ˆ
ˆ
EXECUTE: (a) A = ( 3.60 m ) cos70.0°i + ( 3.60 m ) sin 70.0° ˆ = (1.23 m ) i + ( 3.38 m ) ˆ
j
j
"
ˆ
ˆ
B = − ( 2.40 m ) cos 30.0°i − ( 2.40 m ) sin 30.0° ˆ = ( −2.08 m ) i + ( −1.20 m ) ˆ
j
j
"
"
"
ˆ
ˆ
(b) C = ( 3.00 ) A − ( 4.00 ) B = ( 3.00 )(1.23 m ) i + ( 3.00 )( 3.38 m ) ˆ − ( 4.00 )( −2.08 m ) i − ( 4.00 )( −1.20 m ) ˆ
j
j
ˆ
j
= (12.01 m)i + (14.94) ˆ
(c) From Equations (1.7) and (1.8),
2
2
⎛ 14.94 m ⎞
C = (12.01 m ) + (14.94 m ) = 19.17 m, arctan ⎜
⎟ = 51.2°
⎝ 12.01 m ⎠ EVALUATE:
1.50. C x and C y are both positive, so θ is in the first quadrant. IDENTIFY: Find A and B. Find the vector difference using components.
SET UP: Deduce the x- and y-components and use Eq.(1.8).
"
ˆ
EXECUTE: (a) A = 4.00i + 3.00 ˆ; Ax = +4.00; Ay = +3.00
j
2
A = Ax2 + Ay = (4.00) 2 + (3.00) 2 = 5.00 1-14 Chapter 1 "
ˆ
B = 5.00i − 2.00 ˆ; Bx = +5.00; By = −2.00
j
2
B = Bx2 + By = (5.00) 2 + (−2.00) 2 = 5.39 "
"
EVALUATE: Note that the magnitudes of A and B are each larger than either of their components.
""
ˆ
ˆ
ˆ
EXECUTE: (b) A − B = 4.00i + 3.00 ˆ − 5.00i − 2.00 ˆ = (4.00 − 5.00) i + (3.00 + 2.00) ˆ
j
j
j ( ) ""
ˆ
A − B = −1.00i + 5.00 ˆ
j
"""
ˆ
(c) Let R = A − B = −1.00i + 5.00 ˆ. Then Rx = −1.00, Ry = 5.00.
j 2
R = Rx2 + Ry R = ( −1.00) 2 + (5.00) 2 = 5.10.
tan θ = EVALUATE:
1.51. 1.52. 1.53. 5.00
= −5.00
Rx −1.00
θ = −78.7° + 180° = 101.3°.
= Figure 1.50
"
Rx < 0 and Ry > 0, so R is in the 2nd quadrant. IDENTIFY: A unit vector has magnitude equal to 1.
SET UP: The magnitude of a vector is given in terms of its components by Eq.(1.12).
ˆjˆ
EXECUTE: (a) i + ˆ + k = 12 + 12 + 12 = 3 ≠ 1 so it is not a unit vector. "
"
2
(b) A = Ax2 + Ay + Az2 . If any component is greater than +1 or less than −1, A > 1 , so it cannot be a unit
"
vector. A can have negative components since the minus sign goes away when the component is squared.
"
1
2
2
(c) A = 1 gives a 2 ( 3.0 ) + a 2 ( 4.0 ) = 1 and a 2 25 = 1 . a = ±
= ±0.20 .
5.0
EVALUATE: The magnitude of a vector is greater than the magnitude of any of its components.
"
""""
"
IDENTIFY: If vectors A and B commute for addition, A + B = B + A . If they commute for the scalar product,
"" ""
A⋅ B = B ⋅ A .
"
"
SET UP: Express the sum and scalar product in terms of the components of A and B .
"
""
"
ˆ
ˆ
ˆ
EXECUTE: (a) Let A = Ax i + Ay ˆ and B = Bx i + By ˆ . A + B = ( Ax + Bx ) i + ( Ay + By ) ˆ .
j
j
j
""""
""
ˆ
B + A = ( Bx + Ax ) i + ( By + Ay ) ˆ . Scalar addition is commutative, so A + B = B + A .
j
""
"" ""
""
A ⋅ B = Ax Bx + Ay By and B ⋅ A = Bx Ax + By Ay . Scalar multiplication is commutative, so A ⋅ B = B ⋅ A .
""
ˆ
ˆ
(b) A × B = ( Ay Bz − Az By ) i + ( Az Bx − Ax Bz ) ˆ + ( Ax By − Ay Bx ) k .
j
""
ˆ
ˆ
B × A = ( By Az − Bz Ay ) i + ( Bz Ax − Bx Az ) ˆ + ( Bx Ay − By Ax ) k . Comparison of each component in each vector
j product shows that one is the negative of the other.
""
""
EVALUATE: The result in part (b) means that A × B and B × A have the same magnitude and opposite direction.
""
IDENTIFY: A ⋅ B = AB cos φ
"
"
"
"
"
"
SET UP: For A and B , φ = 150.0° . For B and C , φ = 145.0° . For A and C , φ = 65.0° .
""
EXECUTE: (a) A ⋅ B = (8.00 m)(15.0 m)cos150.0° = −104 m 2
""
(b) B ⋅ C = (15.0 m)(12.0 m)cos145.0° = −148 m 2
""
(c) A ⋅ C = (8.00 m)(12.0 m)cos65.0° = 40.6 m 2
When φ < 90° the scalar product is positive and when φ > 90° the scalar product is negative.
""
IDENTIFY: Target variables are A ⋅ B and the angle φ between the two vectors.
"
"
SET UP: We are given A and B in unit vector form and can take the scalar product using Eq.(1.19). The angle
φ can then be found from Eq.(1.18).
EVALUATE: 1.54. Ry Units, Physical Quantities and Vectors 1.55. EXECUTE:
"
"
ˆ
ˆ
(a) A = 4.00i + 3.00 ˆ, B = 5.00i − 2.00 ˆ; A = 5.00, B = 5.39
j
j
""
ˆ + 3.00 ˆ) ⋅ (5.00i − 2.00 ˆ) = (4.00)(5.00) + (3.00)( −2.00) = 20.0 − 6.0 = +14.0.
ˆ
A ⋅ B = (4.00i
j
j
""
A⋅ B
14.0
=
= 0.519; φ = 58.7°.
(b) cos φ =
AB (5.00)(5.39)
"
"
"
EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and
the angle φ is less than 90°.
IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to
""
⎛ A⋅ B ⎞
⎛ Ax Bx + Ay By ⎞
give the angle φ as φ = arccos ⎜
⎟ = arccos ⎜
⎟.
AB
⎝
⎠
⎝ AB ⎠
SET UP: Eq.(1.14) shows how to obtain the components for a vector written in terms of unit vectors.
""
⎛ −22 ⎞
EXECUTE: (a) A ⋅ B = −22, A = 40, B = 13, and so φ = arccos ⎜
⎟ = 165° .
⎝ 40 13 ⎠ ""
(b) A ⋅ B = 60, A = 34, B = 136, 1.56. 1-15 ⎛ 60
⎞
⎟ = 28° .
⎝ 34 136 ⎠ φ = arccos ⎜ ""
(c) A ⋅ B = 0 and φ = 90° .
""
""
""
EVALUATE: If A ⋅ B > 0 , 0 ≤ φ < 90° . If A ⋅ B < 0 , 90° < φ ≤ 180° . If A ⋅ B = 0 , φ = 90° and the two vectors are
perpendicular.
"
""
""
"
IDENTIFY: A ⋅ B = AB cos φ and A × B = AB sin φ , where φ is the angle between A and B .
"
"
"
"
"
"
SET UP: Figure 1.56 shows A and B . The components A% of A along B and A⊥ of A perpendicular to B are
"
"
"
"
shown in Figure 1.56a. The components of B% of B along A and B⊥ of B perpendicular to A are shown in
Figure 1.56b.
""
EXECUTE: (a) From Figures 1.56a and b, A% = A cos φ and B% = B cos φ . A ⋅ B = AB cos φ = BA% = AB% .
""
(b) A⊥ = A sin φ and B⊥ = B sin φ . A × B = AB sin φ = BA⊥ = AB⊥ .
"
"
"
"
"
EVALUATE: When A and B are perpendicular, A has no component along B and B has no component along
"
"
""
"
"
"
"
A and A ⋅ B = 0 . When A and B are parallel, A has no component perpendicular to B and B has no component
"
""
perpendicular to A and A × B = 0 . 1.57. Figure 1.56
""
IDENTIFY: A × D has magnitude AD sin φ . Its direction is given by the right-hand rule.
SET UP: φ = 180° − 53° = 127°
""
""
EXECUTE: A × D = (8.00 m)(10.0 m)sin127° = 63.9 m 2 . The right-hand rule says A × D is in the − z -direction (into the page).
EVALUATE:
1.58. "
""
"
The component of D perpendicular to A is D⊥ = D sin 53.0° = 7.00 m . A × D = AD⊥ = 63.9 m 2 , which agrees with our previous result.
""
IDENTIFY: Target variable is the vector A × B , expressed in terms of unit vectors.
"
"
SET UP: We are given A and B in unit vector form and can take the vector product using Eq.(1.24).
"
"
ˆ
ˆ
EXECUTE: A = 4.00i + 3.00 ˆ, B = 5.00i − 2.00 ˆ
j
j 1-16 Chapter 1 ""
ˆ
ˆ
ˆˆ
ˆj
A × B = 4.00i + 3.00 ˆ × 5.00i − 2.00 ˆ = 20.0i × i − 8.00i × ˆ + 15.0 ˆ × i − 6.00 ˆ × ˆ
j
j
jˆ
jj
""
ˆ
ˆ
ˆ
ˆj ˆjˆ
ˆ
ˆˆjj
But i × i = ˆ × ˆ = 0 and i × ˆ = k , ˆ × i = − k , so A × B = −8.00k + 15.0 − k = −23.0k.
""
The magnitude of A × B is 23.0.
"
"
EVALUATE: Sketch the vectors A and B in a coordinate system where the xy-plane is in the plane of the paper
and the z-axis is directed out toward you. ( )( ) () 1.59. Figure 1.58
""
By the right-hand rule A × B is directed into the plane of the paper, in the − z -direction. This agrees with the
above calculation that used unit vectors.
IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude.
SET UP: φ = 120.0° .
""
EXECUTE: (a) The direction of A× B is into the page (the − z -direction ). The magnitude of the vector product is AB sin φ = ( 2.80 cm )(1.90 cm ) sin120# = 4.61 cm 2 . ""
(b) Rather than repeat the calculations, Eq. (1.23) may be used to see that B × A has magnitude 4.61 cm2 and is in
the + z -direction (out of the page).
EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is
Cz = Ax By − Ay Bx = ( 2.80 cm ) cos 60.0° ( −1.90 cm ) sin 60°
− ( 2.80 cm ) sin 60.0° (1.90 cm ) cos60.0° = −4.61 cm 2 . 1.60. This gives the same result.
IDENTIFY: Area is length times width. Do unit conversions.
SET UP: 1 mi = 5280 ft . 1 ft 3 = 7.477 gal .
EXECUTE: (a) The area of one acre is 1
8 1
mi × 80 mi = mi 2 , so there are 640 acres to a square mile. 1
640 2 1.61. ⎛ 1 mi 2 ⎞ ⎛ 5280 ft ⎞
2
(b) (1 acre ) × ⎜
⎟×⎜
⎟ = 43,560 ft
640 acre ⎠ ⎝ 1 mi ⎠
⎝
(all of the above conversions are exact).
⎛ 7.477 gal ⎞
5
(c) (1 acre-foot) = ( 43,560 ft 3 ) × ⎜
⎟ = 3.26 × 10 gal, which is rounded to three significant figures.
1 ft 3 ⎠
⎝
EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-foot is
much larger than a gallon.
IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from
this the radius.
SET UP: The earth has mass mE = 5.97 × 1024 kg and radius rE = 6.38 × 106 m . The volume of a sphere is
4
V = 3 π r 3 . ρ = 1.76 g/cm3 = 1760 km/m3 . EXECUTE: (a) The planet has mass m = 5.5mE = 3.28 × 1025 kg . V =
1/ 3 ⎛ 3V ⎞
r =⎜
⎟
⎝ 4π ⎠ 3 ρ = 3.28 × 1025 kg
= 1.86 × 1022 m3 .
1760 kg/m3 1/ 3 ⎛ 3[1.86 × 10 m ] ⎞
=⎜
⎟
4π
⎝
⎠
22 m = 1.64 × 107 m = 1.64 × 104 km (b) r = 2.57 rE
EVALUATE: Volume V is proportional to mass and radius r is proportional to V 1 / 3 , so r is proportional to m1 / 3 . If
the planet and earth had the same density its radius would be (5.5)1 / 3 rE = 1.8rE . The radius of the planet is greater
than this, so its density must be less than that of the earth. Units, Physical Quantities and Vectors 1.62. IDENTIFY and SET UP: (b) Unit conversion. (a) f = 1.420 × 109 cycles/s, so EXECUTE: 1-17 1
s = 7.04 × 10−10 s for one cycle.
1.420 × 109 3600 s/h
= 5.11 × 1012 cycles/h
7.04 × 10−10 s/cycle (c) Calculate the number of seconds in 4600 million years = 4.6 × 109 y and divide by the time for 1 cycle: (4.6 × 109 y)(3.156 × 107 s/y)
= 2.1 × 1026 cycles
7.04 × 10−10 s/cycle 1.63. ⎛ 4.60 × 109 ⎞
4
(d) The clock is off by 1 s in 100,000 y = 1 × 105 y, so in 4.60 × 109 y it is off by (1 s) ⎜
⎟ = 4.6 × 10 s
5
⎝ 1 × 10 ⎠
(about 13 h).
EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer.
IDENTIFY: The number of atoms is your mass divided by the mass of one atom.
SET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the mass of
one H2O molecule: 18.015 u × 1.661 × 10227 kg/u = 2.992 × 10226 kg/molecule. ( 70 kg ) / ( 2.992 × 10226 kg/molecule ) = 2.34 × 1027 EXECUTE: 1.64. molecules. Each H 2O molecule has 3 atoms, so there are about 6 × 1027 atoms.
EVALUATE: Assuming carbon to be the most common atom gives 3 × 1027 molecules, which is a result of the
same order of magnitude.
IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume.
4
SET UP: The volume of a sphere of radius r is V = 3 π r 3 . The volume of a cylinder of radius r and length l is V = π r 2l . The density of water is 1000 kg m3 .
(a) Estimate the volume as that of a sphere of diameter 10 cm: V = 5.2 × 10−4 m3 . EXECUTE: m = ( 0.98 ) (1000 kg m3 ) ( 5.2 × 10−4 m3 ) = 0.5 kg . (b) Approximate as a sphere of radius r = 0.25 μm (probably an over estimate): V = 6.5 × 10−20 m3 . m = ( 0.98 ) (1000 kg m3 ) ( 6.5 × 10−20 m3 ) = 6 × 10−17 kg = 6 × 10−14 g .
(c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V = π r 2l = 2.8 × 10−7 m3 . m = ( 0.98 ) (1000 kg m3 ) ( 2.8 × 10−7 m3 ) = 3 × 10−4 kg = 0.3 g . EVALUATE:
1.65. IDENTIFY:
SET UP:
EXECUTE: The mass is directly proportional to the volume.
Use the volume V and density ρ to calculate the mass M: ρ = M
M
, so V =
.
V
ρ 4
The volume of a cube with sides of length x is x3 . The volume of a sphere with radius R is 3 π R 3 . (a) x3 = 0.200 kg
= 2.54 × 10−5 m3 . x = 2.94 × 10−2 m = 2.94 cm .
7.86 × 103 kg/m3 43
π R = 2.54 × 10−5 m 3 . R = 1.82 × 10−2 m = 1.82 cm .
3
4
EVALUATE: 3 π = 4.2 , so a sphere with radius R has a greater volume than a cube whose sides have length R.
IDENTIFY: Estimate the volume of sand in all the beaches on the earth. The diameter of a grain of sand determines
its volume. From the volume of one grain and the total volume of sand we can calculate the number of grains.
SET UP: The volume of a sphere of diameter d is V = 1 π d 3 . Consulting an atlas, we estimate that the continents
6
(b) 1.66. have about 1.45 × 105 km of coastline. Add another 25% of this for rivers and lakes, giving 1.82 × 105 km of
coastline. Assume that a beach extends 50 m beyond the water and that the sand is 2 m deep. 1 billion = 1 × 109 .
EXECUTE: (a) The volume of sand is (1.82 × 108 m)(50 m)(2 m) = 2 × 1010 m3 . The volume of a grain is V = 1 π (0.2 × 10−3 m)3 = 4 × 10−12 m 3 . The number of grains is
6 2 × 1010 m3
= 5 × 1021 . The number of grains of sand
4 × 10−12 m3 is about 1022 .
(b) The number of stars is (100 × 109 )(100 × 109 ) = 1022 . The two estimates result in comparable numbers for these
two quantities. 1-18 1.67. Chapter 1 EVALUATE: Both numbers are crude estimates but are probably accurate to a few powers of 10.
IDENTIFY: The number of particles is the total mass divided by the mass of one particle.
SET UP: 1 mol = 6.0 × 1023 atoms . The mass of the earth is 6.0 × 1024 kg . The mass of the sun is 2.0 × 1030 kg .
4
The distance from the earth to the sun is 1.5 × 1011 m . The volume of a sphere of radius R is 3 π R 3 . Protons and neutrons each have a mass of 1.7 × 10−27 kg and the mass of an electron is much less.
⎛ 6.0 × 1023 atoms ⎞
50
mole
(a) (6.0 × 1024 kg) × ⎜
⎟ = 2.6 × 10 atoms.
kg
14 × 10−3 mole ⎠
⎝
(b) The number of neutrons is the mass of the neutron star divided by the mass of a neutron:
(2)(2.0 × 1030 kg)
= 2.4 × 1057 neutrons.
(1.7 × 10−27 kg neutron)
EXECUTE: 2
(c) The average mass of a particle is essentially 3 the mass of either the proton or the neutron, 1.7 × 10−27 kg. The
total number of particles is the total mass divided by this average, and the total mass is the volume times the
average density. Denoting the density by ρ , 1.68. 4
π R 3 ρ (2π )(1.5 × 1011 m)3 (1018 kg m3 )
M
3
=
=
= 1.2 × 1079.
2
mave
1.7 × 10−27 kg
mp
3
Note the conversion from g/cm3 to kg/m3.
EVALUATE: These numbers of particles are each very, very large but are still much less than a googol.
""""
"""
"
"
"
IDENTIFY: Let D be the fourth force. Find D such that A + B + C + D = 0 , so D = − A + B + C .
"
SET UP: Use components and solve for the components Dx and Dy of D . ( EXECUTE: ) Ax = + A cos30.0° = +86.6 N, Ay = + A cos30.0° = +50.00 N . Bx = − B sin 30.0° = −40.00 N, By = + B cos30.0° = +69.28 N .
Cx = +C cos53.0° = −24.07 N, C y = −C sin 53.0° = −31.90 N .
2
Then Dx = −22.53 N , Dy = −87.34 N and D = Dx2 + Dy = 90.2 N . tan α = Dy / Dx = 87.34 / 22.53 . α = 75.54° . φ = 180° + α = 256° , counterclockwise from the + x-axis.
EVALUATE: "
As shown in Figure 1.68, since Dx and Dy are both negative, D must lie in the third quadrant. Figure 1.68
1.69. IDENTIFY: We know the magnitude and direction of the sum of the two vector pulls and the direction of one pull.
We also know that one pull has twice the magnitude of the other. There are two unknowns, the magnitude of the
smaller pull and its direction. Ax + Bx = Cx and Ay + By = C y give two equations for these two unknowns.
"""
"
"
SET UP: Let the smaller pull be A and the larger pull be B . B = 2 A . C = A + B has magnitude 350.0 N and is
northward. Let + x be east and + y be north. Bx = − B sin 25.0° and By = B cos 25.0° . C x = 0 , C y = 350.0 N .
"
"
A must have an eastward component to cancel the westward component of B . There are then two possibilities, as
"
"
sketched in Figures 1.69 a and b. A can have a northward component or A can have a southward component.
EXECUTE: In either Figure 1.69 a or b, Ax + Bx = Cx and B = 2 A gives (2 A)sin 25.0° = A sin φ and φ = 57.7° . In Figure 1.69a, Ay + By = C y gives 2 A cos 25.0° + A cos57.7° = 350.0 N and A = 149 N . In Figure 1.69b,
2 A cos 25.0° − A cos57.7° = 350.0 N and A = 274 N . One solution is for the smaller pull to be 57.7° east of north.
In this case, the smaller pull is 149 N and the larger pull is 298 N. The other solution is for the smaller pull to be
57.7° east of south. In this case the smaller pull is 274 N and the larger pull is 548 N. Units, Physical Quantities and Vectors 1-19 "
EVALUATE: For the first solution, with A east of north, each worker has to exert less force to produce the given
resultant force and this is the sensible direction for the worker to pull. Figure 1.69
1.70. IDENTIFY: Find the vector sum of the two displacements.
"
""""
"
"
SET UP: Call the two displacements A and B , where A = 170 km and B = 230 km . A + B = R . A and B are
as shown in Figure 1.70.
EXECUTE: Rx = Ax + Bx = (170 km) sin 68° + (230 km) cos 48° = 311.5 km . Ry = Ay + By = (170 km) cos 68° − (230 km) sin 48° = −107.2 km .
2
2
R = Rx + Ry = ( 311.5 km ) + ( −107.2 km )
2 2 = 330 km . tanθR = Ry
Rx = 107.2 km
= 0.344 .
311.5 km θR = 19° south of east .
EVALUATE: 1.71. "
Our calculation using components agrees with R shown in the vector addition diagram, Figure 1.70. Figure 1.70
"
"""
"""
IDENTIFY: A + B = C (or B + A = C ). The target variable is vector A.
"
SET UP: Use components and Eq.(1.10) to solve for the components of A. Find the magnitude and direction of
"
A from its components.
EXECUTE: (a)
Cx = Ax + Bx , so Ax = Cx − Bx C y = Ay + By , so Ay = C y − By
Cx = C cos 22.0° = (6.40 cm)cos 22.0°
C x = +5.934 cm
C y = C sin 22.0° = (6.40 cm)sin 22.0°
C y = +2.397 cm
Bx = B cos(360° − 63.0°) = (6.40 cm)cos 297.0°
Bx = +2.906 cm
By = B sin 297.0° = (6.40 cm)sin 297.0°
By = −5.702 cm
Figure 1.71a
(b) Ax = Cx − Bx = +5.934 cm − 2.906 cm = +3.03 cm Ay = C y − By = +2.397 cm − ( −5.702) cm = +8.10 cm 1-20 Chapter 1 (c)
2
A = Ax2 + Ay A = (3.03 cm)2 + (8.10 cm) 2 = 8.65 cm
tan θ = Ay Ax
θ = 69.5° 1.72. = 8.10 cm
= 2.67
3.03 cm Figure 1.71b
"
"
EVALUATE: The A we calculated agrees qualitatively with vector A in the vector addition diagram in part (a).
IDENTIFY: Add the vectors using the method of components.
SET UP: Ax = 0 , Ay = −8.00 m . Bx = 7.50 m , By = 13.0 m . C x = −10.9 m , C y = −5.07 m .
EXECUTE: (a) Rx = Ax + Bx + C x = −3.4 m . Ry = Ay + By + C y = −0.07 m . R = 3.4 m . tan θ = −0.07 m
.
−3.4 m θ = 1.2° below the − x-axis .
(b) S x = Cx − Ax − Bx = −18.4 m . S y = C y − Ay − By = −10.1 m . S = 21.0 m . tan θ = Sy = −10.1 m
. θ = 28.8°
−18.4 m Sx
below the − x-axis .
"
"
EVALUATE: The magnitude and direction we calculated for R and S agree with our vector diagrams. Figure 1.72
1.73. IDENTIFY: Vector addition. Target variable is the 4th displacement.
SET UP: Use a coordinate system where east is in the + x -direction and north is in the + y -direction.
""
"
"
Let A, B, and C be the three displacements that are given and let D be the fourth unmeasured displacement.
"
"""""
Then the resultant displacement is R = A + B + C + D. And since she ends up back where she started, R = 0.
""""
"""
"
0 = A + B + C + D, so D = − A + B + C ( ) Dx = −( Ax + Bx + Cx ) and Dy = −( Ay + By + C y )
EXECUTE: Ax = −180 m, Ay = 0
Bx = B cos315° = (210 m)cos315° = +148.5 m
By = B sin 315° = (210 m)sin 315° = −148.5 m
Cx = C cos 60° = (280 m)cos60° = +140 m
C y = C sin 60° = (280 m)sin 60° = +242.5 m
Figure 1.73a Dx = −( Ax + Bx + Cx ) = −( −180 m + 148.5 m + 140 m) = −108.5 m Units, Physical Quantities and Vectors 1-21 Dy = −( Ay + By + C y ) = −(0 − 148.5 m + 242.5 m) = −94.0 m
2
D = Dx2 + Dy D = (−108.5 m) 2 + (−94.0 m) 2 = 144 m
−94.0 m
= 0.8664
Dx −108.5 m
θ = 180° + 40.9° = 220.9°
"
( D is in the third quadrant since both
Dx and Dy are negative.)
tan θ = Dy = Figure 1.73b
"
"
The direction of D can also be specified in terms of φ = θ − 180° = 40.9°; D is 41° south of west.
EVALUATE: The vector addition diagram, approximately to scale, is "
Vector D in this diagram
agrees qualitatively with
our calculation using
components. Figure 1.73c
1.74. IDENTIFY: Solve for one of the vectors in the vector sum. Use components.
SET UP: Use coordinates for which + x is east and + y is north. The vector displacements are:
$
$
$
A = 2.00 km, 0°of east; B = 3.50 m, 45° south of east; and R = 5.80 m, 0° east
EXECUTE: Cx = Rx − Ax − Bx = 5.80 km − ( 2.00 km ) − ( 3.50 km ) ( cos 45° ) = 1.33 km ; C y = Ry − Ay − By
2
2
= 0 km − 0 km − ( −3.50 km )( sin 45° ) = 2.47 km ; C = (1.33 km ) + ( 2.47 km ) = 2.81 km ;
θ = tan −1 ⎡( 2.47 km ) (1.33 km )⎤ = 61.7° north of east. The vector addition diagram in Figure 1.74 shows good
⎣
⎦ qualitative agreement with these values.
EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive. Figure 1.74
1.75. IDENTIFY: The sum of the vector forces on the beam sum to zero, so their x components and their y components
"
sum to zero. Solve for the components of F .
SET UP: The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The
"
100-N pull makes an angle of 30.0° + 40.0° = 70.0° with the horizontal. F and the 100-N pull have been replaced
by their x and y components.
EXECUTE: (a) The sum of the x-components is equal to zero gives Fx + (100 N)cos70.0° = 0 and Fx = −34.2 N .
"
The sum of the y-components is equal to zero gives Fy + (100 N)sin 70.0° − 124 N = 0 and Fy = +30.0 N . F and its components are sketched in Figure 1.75b. F = Fx2 + Fy2 = 45.5 N . tan φ = Fy
Fx = "
30.0 N
and φ = 41.3° . F is
34.2 N directed at 41.3° above the − x -axis in Figure 1.75a.
"
(b) The vector addition diagram is given in Figure 1.75c. F determined from the diagram agrees with
"
F calculated in part (a) using components. 1-22 Chapter 1 "
EVALUATE: The vertical component of the 100 N pull is less than the 124 N weight so F must have an upward
component if all three forces balance. Figure 1.75
1.76. """
""
"
IDENTIFY: The four displacements return her to her starting point, so D = − ( A + B + C ) , where A , B and
"
"
C are in the three given displacements and D is the displacement for her return.
START UP: Let + x be east and + y be north.
EXECUTE: (a) Dx = −[(147 km ) sin 85° + (106 km ) sin167° + (166 km ) sin 235°] = −34.3 km . Dy = −[(147 km ) cos85° + (106 km ) cos167° + (166 km ) cos 235°] = +185.7 km .
D = (−34.3 km) 2 + (185.7 km) 2 = 189 km . 1.77. "
⎛ 34.3 km ⎞
(b) The direction relative to north is φ = arctan ⎜
⎟ = 10.5° . Since Dx < 0 and Dy > 0 , the direction of D
⎝ 185.7 km ⎠
is 10.5° west of north.
EVALUATE: The four displacements add to zero.
"
IDENTIFY and SET UP: The vector A that connects points ( x1 , y1 ) and ( x2 , y2 ) has components Ax = x2 − x1 and
Ay = y2 − y1 .
⎛ 200 − 20 ⎞
(a) Angle of first line is θ = tan −1 ⎜
⎟ = 42°. Angle of second line is 42° + 30° = 72°.
⎝ 210 − 10 ⎠
Therefore X = 10 + 250 cos 72° = 87 , Y = 20 + 250 sin 72° = 258 for a final point of (87,258).
(b) The computer screen now looks something like Figure 1.77. The length of the bottom line is
258 − 200 ⎞
2
2
( 210 − 87 ) + ( 200 − 258) = 136 and its direction is tan −1 ⎛
⎜
⎟ = 25° below straight left.
⎝ 210 − 87 ⎠
EVALUATE: Figure 1.77 is a vector addition diagram. The vector first line plus the vector arrow gives the vector
for the second line.
EXECUTE: Figure 1.77 Units, Physical Quantities and Vectors 1.78. 1-23 ""
"
IDENTIFY: Let the three given displacements be A , B and C , where A = 40 steps , B = 80 steps and
"
"
""""
C = 50 steps . R = A + B + C . The displacement C that will return him to his hut is − R .
SET UP: Let the east direction be the + x -direction and the north direction be the + y -direction.
EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.78.
(b) Rx = ( 40 ) cos 45° − ( 80 ) cos 60° = −11.7 and Ry = ( 40 ) sin 45° + ( 80 ) sin 60° − 50 = 47.6.
The magnitude and direction of the resultant are ⎛ 47.6 ⎞
(−11.7) 2 + (47.6) 2 = 49, arctan ⎜
⎟ = 76° , north of west.
⎝ 11.7 ⎠ "
We know that R is in the second quadrant because Rx < 0 , Ry > 0 . To return to the hut, the explorer must take
49 steps in a direction 76° south of east, which is 14° east of south.
"
EVALUATE: It is useful to show Rx , Ry and R on a sketch, so we can specify what angle we are computing. Figure 1.78
1.79. IDENTIFY: Vector addition. One vector and the sum are given; find the second vector (magnitude and direction).
"
"
SET UP: Let + x be east and + y be north. Let A be the displacement 285 km at 40.0° north of west and let B
be the unknown displacement.
"""
"
A + B = R where R = 115 km, east
"""
B = R− A
Bx = Rx − Ax , By = Ry − Ay
EXECUTE: Ax = − A cos 40.0° = −218.3 km, Ay = + A sin 40.0° = +183.2 km Rx = 115 km, Ry = 0
2
Then Bx = 333.3 km, By = −183.2 km. B = Bx2 + By = 380 km; tan α = By / Bx = (183.2 km)/(333.3 km) α = 28.8°, south of east
Figure 1.79 1.80. "
"
EVALUATE: The southward component of B cancels the northward component of A. The eastward component
"
"
of B must be 115 km larger than the magnitude of the westward component of A.
IDENTIFY: Find the components of the weight force, using the specified coordinate directions.
SET UP: For parts (a) and (b), take + x direction along the hillside and the + y direction in the downward
direction and perpendicular to the hillside. For part (c), α = 35.0° and w = 550 N .
EXECUTE: (a) wx = w sin α
(b) wy = w cos α
(c) The maximum allowable weight is w = wx ( sin α ) = ( 550 N ) ( sin35.0° ) = 959 N .
EVALUATE: The component parallel to the hill increases as α increases and the component perpendicular to the
hill increases as α decreases. 1-24 1.81. Chapter 1 IDENTIFY: Vector addition. One force and the vector sum are given; find the second force.
SET UP: Use components. Let + y be upward. "
B is the force the biceps exerts. Figure 1.81a
"
"""
E is the force the elbow exerts. E + B = R, where R = 132.5 N and is upward.
Ex = Rx − Bx ,
E y = Ry − By
EXECUTE: Bx = − B sin 43° = −158.2 N, By = + B cos 43° = +169.7 N, Rx = 0, Ry = +132.5 N Then Ex = +158.2 N, E y = −37.2 N
2
E = Ex2 + E y = 160 N; tan α = E y / Ex = 37.2 /158.2 α = 13°, below horizontal
Figure 1.81b
"
"
EVALUATE: The x-component of E cancels the x-component of B. The resultant upward force is less than the
"
upward component of B, so E y must be downward.
1.82. IDENTIFY: Find the vector sum of the four displacements.
SET UP: Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction,
ˆ
ˆ
and the z-axis vertical. The first displacement is then (−30 m) k , the second is (−15 m) ˆ, the third is (200 m) i ,
j j
and the fourth is (100 m) ˆ.
EXECUTE: (a) Adding the four displacements gives
ˆ
ˆ
ˆ
ˆ
j
j
j
(−30 m) k + ( −15 m) ˆ + (200 m) i + (100 m) ˆ = (200 m) i + (85 m) ˆ − (30 m) k.
(b) The total distance traveled is the sum of the distances of the individual segments:
30 m + 15 m + 200 m + 100 m = 345 m. The magnitude of the total displacement is:
2
D = Dx2 + Dy + Dz2 = (200 m) 2 + (85 m) 2 + ( −30 m ) = 219 m.
2 1.83. EVALUATE: The magnitude of the displacement is much less than the distance traveled along the path.
IDENTIFY: The sum of the force displacements must be zero. Use components.
"""
"
"
SET UP: Call the displacements A , B , C and D , where D is the final unknown displacement for the return
""
"
""""
from the treasure to the oak tree. Vectors A , B , and C are sketched in Figure 1.83a. A + B + C + D = 0 says
Ax + Bx + Cx + Dx = 0 and Ay + By + C y + Dy = 0 . A = 825 m , B = 1250 m , and C = 1000 m . Let + x be eastward and + y be north.
EXECUTE: (a) Ax + Bx + Cx + Dx = 0 gives Dx = −( Ax + Bx + Cx ) = −(0 − [1250 m]sin30.0° +[1000 m]cos40.0°) = −141 m . Ay + By + C y + Dy = 0 gives Dy = −( Ay + By + C y ) = −( −825 m + [1250 m]cos30.0° + [1000 m]sin 40.0°) = −900 m .
"
2
The fourth displacement D and its components are sketched in Figure 1.83b. D = Dx2 + Dy = 911 m .
tan φ = Dx
Dy = 141 m
and φ = 8.9° . You should head 8.9° west of south and must walk 911 m.
900 m Units, Physical Quantities and Vectors 1-25 "
(b) The vector diagram is sketched in Figure 1.83c. The final displacement D from this diagram agrees with the
"
vector D calculated in part (a) using components.
""
"
"
EVALUATE: Note that D is the negative of the sum of A , B , and C . 1.84. Figure 1.83
"
"
IDENTIFY: If the vector from your tent to Joe’s is A and from your tent to Karl’s is B , then the vector from
""
Joe’s tent to Karl’s is B − A .
SET UP: Take your tent's position as the origin. Let + x be east and + y be north.
EXECUTE: The position vector for Joe’s tent is
j
j
([21.0 m]cos 23° ) iˆ − ([21.0 m]sin 23° ) ˆ = (19.33 m) iˆ − (8.205 m) ˆ. ˆ
ˆ
j
j
The position vector for Karl's tent is ([32.0 m]cos 37° ) i + ([32.0 m]sin 37° ) ˆ = (25.56 m) i + (19.26 m) ˆ.
The difference between the two positions is
j
j
(19.33 m − 25.56 m ) iˆ + ( −8.205 m − 19.25 m ) ˆ = −(6.23 m)iˆ − (27.46 m) ˆ. The magnitude of this vector is the
distance between the two tents: D = 1.85. ( −6.23 m ) 2 + ( −27.46 m ) = 28.2 m
2 EVALUATE: If both tents were due east of yours, the distance between them would be 32.0 m − 21.0 m = 17.0 m .
If Joe’s was due north of yours and Karl’s was due south of yours, then the distance between them would be
32.0 m + 21.0 m = 53.0 m . The actual distance between them lies between these limiting values.
"
"
IDENTIFY: In Eqs.(1.21) and (1.27) write the components of A and B in terms of A, B, θ A and θ B .
SET UP: From Appendix B, cos(a − b) = cos a cos b + sin a sin b and sin(a − b) = sin a cos b − cos a sin b .
EXECUTE: (a) With Az = Bz = 0 , Eq.(1.21) becomes Ax Bx + Ay By = ( A cos θ A )( B cos θB ) + ( A sin θ A ) ( B sin θB )
Ax Bx + Ay By = AB ( cos θ Acos θB + sin θ Asin θB ) = AB cos ( θ A − θB ) = AB cos φ , where the expression for the cosine
of the difference between two angles has been used.
"
ˆ
(b) With Az = Bz = 0 , C = Cz k and C = C z . From Eq.(1.27),
C = Ax By − Ay Bx = ( A cos θ A )( B sin θB ) − ( A sin θ A )( B cos θ A )
C = AB cos θ A sin θB − sin θ A cos θB = AB sin ( θB − θ A ) = AB sin φ , where the expression for the sine of the 1.86. difference between two angles has been used.
EVALUATE: Since they are equivalent, we may use either Eq.(1.18) or (1.21) for the scalar product and either
(1.22) or (1.27) for the vector product, depending on which is the more convenient in a given application.
IDENTIFY: Apply Eqs.(1.18) and (1.22).
SET UP: The angle between the vectors is 20° + 90° + 30° = 140°.
""
EXECUTE: (a) Eq. (1.18) gives A ⋅ B = ( 3.60 m ) ( 2.40 m ) cos 140° = −6.62 m 2 .
(b) From Eq.(1.22), the magnitude of the cross product is ( 3.60 m ) ( 2.40 m ) sin 140° = 5.55 m 2 and the direction, from the right-hand rule, is out of the page (the + z -direction ).
"
"
EVALUATE: We could also use Eqs.(1.21) and (1.27), with the components of A and B . 1-26 1.87. 1.88. Chapter 1 Compare the magnitude of the cross product, AB sin φ , to the area of the parallelogram.
"
"
SET UP: The two sides of the parallelogram have lengths A and B. φ is the angle between A and B .
EXECUTE: (a) The length of the base is B and the height of the parallelogram is A sin φ , so the area is AB sin φ .
This equals the magnitude of the cross product.
"
""
"
(b) The cross product A × B is perpendicular to the plane formed by A and B , so the angle is 90° .
EVALUATE: It is useful to consider the special cases φ = 0° , where the area is zero, and φ = 90° , where the
parallelogram becomes a rectangle and the area is AB.
IDENTIFY: Use Eq.(1.27) for the components of the vector product.
SET UP: Use coordinates with the + x -axis to the right, + y -axis toward the top of the page, and + z -axis out of
IDENTIFY: the page. Ax = 0 , Ay = 0 and Az = −3.50 cm . The page is 20 cm by 35 cm, so Bx = 20 cm and By = 35 cm .
""
""
""
EXECUTE:
A × B = 122 cm 2 , A × B = −70 cm 2 , A × B = 0. ( 1.89. ) ( x ) y ( ) z EVALUATE: From the components we calculated the magnitude of the vector product is 141 cm 2 .
B = 40.3 cm and φ = 90° , so AB sin φ = 141 cm 2 , which agrees.
"
""
"
IDENTIFY: A and B are given in unit vector form. Find A, B and the vector difference A − B.
"
"
"
""
"
"
"
SET UP: A = −2.00i + 3.00 j + 4.00k , B = 3.00i + 1.00 j − 3.00k
Use Eq.(1.8) to find the magnitudes of the vectors.
EXECUTE: 2
2
(a) A = Ax + Ay + Az2 = (−2.00) 2 + (3.00) 2 + (4.00) 2 = 5.38 2
2
B = Bx + By + Bz2 = (3.00) 2 + (1.00) 2 + (−3.00) 2 = 4.36
""
ˆ
ˆ
ˆ
ˆ
(b) A − B = ( −2.00i + 3.00 ˆ + 4.00k ) − (3.00i + 1.00 ˆ − 3.00k )
j
j
""
ˆ
ˆ
ˆ
ˆ
A − B = ( −2.00 − 3.00) i + (3.00 − 1.00) ˆ + (4.00 − ( −3.00))k = −5.00i + 2.00 ˆ + 7.00k .
j
j
"""
(c) Let C = A − B, so Cx = −5.00, C y = +2.00, Cz = +7.00 1.90. 1.91. 1.92. 2
C = Cx2 + C y + C z2 = (−5.00) 2 + (2.00) 2 + (7.00) 2 = 8.83
""
""
""
""
B − A = −( A − B ), so A − B and B − A have the same magnitude but opposite directions.
EVALUATE: A, B and C are each larger than any of their components.
IDENTIFY: Calculate the scalar product and use Eq.(1.18) to determine φ .
SET UP: The unit vectors are perpendicular to each other.
EXECUTE: The direction vectors each have magnitude 3 , and their scalar product is
(1)(1) + (1)( −1) + (1) ( −1) = 21, so from Eq. (1.18) the angle between the bonds is ⎛ −1 ⎞
⎛ 1⎞
arccos ⎜
⎟ = arccos ⎜ − 3 ⎟ = 109°.
3 3⎠
⎝⎠
⎝
EVALUATE: The angle between the two vectors in the bond directions is greater than 90° .
IDENTIFY: Use the relation derived in part (a) of Problem 1.92: C 2 = A2 + B 2 + 2 AB cosφ , where φ is the angle
"
"
between A and B .
SET UP: cos φ = 0 for φ = 90° . cos φ < 0 for 90° < φ < 180° and cos φ > 0 for 0° < φ < 90° .
"
"
EXECUTE: (a) If C 2 = A2 + B 2 , cos φ = 0, and the angle between A and B is 90° (the vectors are
perpendicular).
"
"
(b) If C 2 < A2 + B 2 , cosφ < 0, and the angle between A and B is greater than 90° .
"
"
(c) If C 2 > A2 + B 2 , cosφ > 0, and the angle between A and B is less than 90°.
EVALUATE: It is easy to verify the expression from Problem 1.92 for the special cases φ = 0 , where C = A + B ,
and for φ = 180° , where C = A − B .
""
"""
IDENTIFY: Let C = A + B and calculate the scalar product C ⋅ C .
""
" ""
SET UP: For any vector V , V ⋅V = V 2 . A ⋅ B = AB cos φ .
""
EXECUTE: (a) Use the linearity of the dot product to show that the square of the magnitude of the sum A + B is
""
""
"" "" "" "" "" ""
""
""
A+ B ⋅ A+ B = A ⋅ A + A ⋅ B + B ⋅ A + B ⋅ B = A ⋅ A + B ⋅ B + 2 A ⋅ B = A2 + B 2 + 2 A ⋅ B ( )( ) = A2 + B 2 + 2 AB cos φ Units, Physical Quantities and Vectors 1-27 (b) Using the result of part (a), with A = B, the condition is that A2 = A2 + A2 + 2 A2cos φ , which solves for
1 = 2 + 2cos φ , cos φ = − 1 , and φ = 120°.
2
1.93. EVALUATE: The expression C 2 = A2 + B 2 + 2 AB cos φ is called the law of cosines.
IDENTIFY: Find the angle between specified pairs of vectors.
""
A⋅ B
SET UP: Use cos φ =
AB
"
ˆ (along line ab)
EXECUTE: (a) A = k
"
ˆjˆ
B = i + ˆ + k (along line ad ) A = 1, B = 12 + 12 + 12 = 3
""
ˆˆjˆ
A⋅ B = k ⋅ i + ˆ + k =1
""
A⋅ B
= 1/ 3; φ = 54.7°
So cos φ =
AB
"
ˆjˆ
(b) A = i + ˆ + k (along line ad )
"
B = ˆ + k (along line ac)
jˆ ( A = 12 + 12 + 12 = 3; B = 12 + 12 = 2
""
ˆjˆˆj
A⋅ B = i + ˆ + k ⋅ i + ˆ =1+1 = 2
""
A⋅ B
2
2
So cos φ =
=
=
; φ = 35.3°
AB
32
6
EVALUATE: Each angle is computed to be less than 90°, in agreement with what is deduced from Fig. 1.43 in
the textbook.
"
""
"
IDENTIFY: The cross product A × B is perpendicular to both A and B .
""
SET UP: Use Eq.(1.27) to calculate the components of A × B .
EXECUTE: The cross product is
⎡
ˆ
ˆ
ˆ ⎛ 6.00 ⎞ ˆ − 11.00 k ⎤ . The magnitude of the vector in square
ˆ
(−13.00) i + (6.00) ˆ + (−11.00) k = 13 ⎢ −(1.00) i + ⎜
j
⎟j
⎥
⎝ 13.00 ⎠ 13.00 ⎦
⎣ ( 1.94. ) brackets is )( ) 1.93, and so a unit vector in this direction is
ˆ
ˆ
⎡ −(1.00) i + (6.00 /13.00) ˆ − (11.00 /13.00) k ⎤
j
⎢
⎥.
1.93
⎢
⎥
⎣
⎦ The negative of this vector, 1.95. 1.96. ˆ
ˆ
⎡ (1.00) i − (6.00 /13.00) ˆ + (11.00 /13.00) k ⎤
j
⎢
⎥,
1.93
⎢
⎥
⎣
⎦
"
"
is also a unit vector perpendicular to A and B .
EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular to both
vectors is perpendicular to this plane.
"
"
"
IDENTIFY and SET UP: The target variables are the components of C . We are given A and B. We also know
""
""
A ⋅ C and B ⋅ C , and this gives us two equations in the two unknowns C x and C y .
""
"
"
EXECUTE: A and C are perpendicular, so A ⋅ C = 0. AxC x + AyC y = 0, which gives 5.0C x − 6.5C y = 0.
""
B ⋅ C = 15.0, so −3.5Cx + 7.0C y = 15.0
We have two equations in two unknowns C x and C y . Solving gives Cx = 8.0 and C y = 6.1
"
""
EVALUATE: We can check that our result does give us a vector C that satisfies the two equations A ⋅ C = 0 and
""
B ⋅ C = 15.0.
IDENTIFY: Calculate the magnitude of the vector product and then use Eq.(1.22).
SET UP: The magnitude of a vector is related to its components by Eq.(1.12). 1-28 Chapter 1 EXECUTE: ""
A× B
""
A × B = AB sinθ . sinθ =
=
AB ( −5.00 ) + ( 2.00 )
( 3.00 )( 3.00 )
2 2 = 0.5984 and θ = sin −1 ( 0.5984 ) = 36.8°. 1.97. "
"
EVALUATE: We haven't found A and B , just the angle between them.
"""
"""
(a) IDENTIFY: Prove that A ⋅ B × C = A × B ⋅ C . ( )( ) SET UP: Express the scalar and vector products in terms of components.
EXECUTE:
"""
""
""
""
A ⋅ B × C = Ax B × C + Ay B × C + Az B × C ( ) ( ) ( x ) ( y ) z """
A ⋅ B × C = Ax ( By Cz − BzC y ) + Ay ( BzC x − BxCz ) + Az ( BxC y − By Cx ) ( ) " " " " " " " ( A× B) ⋅C = ( A× B) C + ( A× B)
x x y ""
C y + A × B Cz ( ) z """
A × B ⋅ C = ( Ay Bz − Az By ) Cx + ( Az Bx − Ax Bz ) C y + ( Ax By − Ay Bx ) Cz
"""
"""
Comparison of the expressions for A ⋅ B × C and A × B ⋅ C shows they contain the same terms, so
"""
"""
A ⋅ B × C = A × B ⋅ C.
""
"""
"
(b) IDENTIFY: Calculate A× B ⋅ C , given the magnitude and direction of A, B, and C .
""
""
SET UP: Use Eq.(1.22) to find the magnitude and direction of A × B. Then we know the components of A × B
"
and of C and can use an expression like Eq.(1.21) to find the scalar product in terms of components.
EXECUTE: A = 5.00; θ A = 26.0°; B = 4.00, θ B = 63.0°
""
A × B = AB sin φ .
"
"
The angle φ between A and B is equal to φ = θ B − θ A = 63.0° − 26.0° = 37.0°. So
""
""
A × B = (5.00)(4.00)sin 37.0° = 12.04, and by the right hand-rule A × B is in the + z -direction. Thus
"""
A × B ⋅ C = (12.04)(6.00) = 72.2
"
"""
""
EVALUATE: A × B is a vector, so taking its scalar product with C is a legitimate vector operation. A × B ⋅ C ( ( ( 1.98. )( ) ) ( ( ) ( ) ) ) ( ) is a scalar product between two vectors so the result is a scalar.
IDENTIFY: Use the maximum and minimum values of the dimensions to find the maximum and minimum areas
and volumes.
SET UP: For a rectangle of width W and length L the area is LW. For a rectangular solid with dimensions L, W
and H the volume is LWH.
EXECUTE: (a) The maximum and minimum areas are ( L + l ) (W + w ) = LW + lW + Lw, ( L − l )(W − w ) = LW − lW − Lw, where the common terms wl have been omitted. The area and its uncertainty are
then WL ± (lW + Lw), so the uncertainty in the area is a = lW + Lw.
(b) The fractional uncertainty in the area is 1.99. a lW + Wl l
w
=
=+
, the sum of the fractional uncertainties in the
A
WL
LW length and width.
(c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and
Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the
v lWH + LwH + LWh l
w
h
volume is =
=+
+ , the sum of the fractional uncertainties in the length, width and
V
LWH
LWH
height.
EVALUATE: The calculation assumes the uncertainties are small, so that terms involving products of two or more
uncertainties can be neglected.
IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to the
receiver.
SET UP: Add the x-components and the y-components. Units, Physical Quantities and Vectors 1-29 EXECUTE: The receiver's position is
ˆ
ˆ
[( +1.0 + 9.0 − 6.0 + 12.0 ) yd]i + [( −5.0 + 11.0 + 4.0 + 18.0 ) yd] ˆ = (16.0 yd ) i + ( 28.0 yd ) ˆ .
j
j The vector from the quarterback to the receiver is the receiver's position minus the quarterback's position, or
2
2
j
(16.0 yd ) iˆ + ( 35.0 yd ) ˆ , a vector with magnitude (16.0 yd ) + ( 35.0 yd ) = 38.5 yd . The angle is 1.100. ⎛ 16.0 ⎞
arctan ⎜
⎟ = 24.6° to the right of downfield.
⎝ 35.0 ⎠
EVALUATE: The vector from the quarterback to receiver has positive x-component and positive y-component.
IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the distance
between two objects is the magnitude of this vector. Use the scalar product to find the angle between two vectors.
"
SET UP: If object A has coordinates ( x A , y A ) and object B has coordinates ( xB , yB ) , the vector rAB from A to B
has x-component xB − x A and y-component yB − y A .
EXECUTE: (a) The diagram is sketched in Figure 1.100.
(b) (i) In AU, (0.3182) 2 + (0.9329) 2 = 0.9857. (ii) In AU, (1.3087) 2 + ( −0.4423) 2 + (−0.0414) 2 = 1.3820.
(iii) In AU (0.3182 − 1.3087) 2 + (0.9329 − (−0.4423))2 + (0.0414)2 = 1.695.
(c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product.
Combining Equations (1.18) and (1.21),
⎛ (−0.3182)(1.3087 − 0.3182) + ( −0.9329)( −0.4423 − 0.9329) + (0) ⎞
⎟ = 54.6°.
(0.9857)(1.695)
⎝
⎠ φ = arccos ⎜ (d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o.
EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that on this
date Mars was farther from the earth than it is from the Sun. Figure 1.100
1.101. IDENTIFY: Draw the vector addition diagram for the position vectors.
"
SET UP: Use coordinates in which the Sun to Merak line lies along the x-axis. Let A be the position vector of
"
"
Alkaid relative to the Sun, M is the position vector of Merak relative to the Sun, and R is the position vector for
Alkaid relative to Merak. A = 138 ly and M = 77 ly .
"""
EXECUTE: The relative positions are shown in Figure 1.101. M + R = A . Ax = M x + Rx so Rx = Ax − M x = (138 ly)cos 25.6° − 77 ly = 47.5 ly . Ry = Ay − M y = (138 ly)sin 25.6° − 0 = 59.6 ly . R = 76.2 ly is
the distance between Alkaid and Merak.
Rx 47.5 ly
=
and θ = 51.4° . Then φ = 180° − θ = 129° .
R 76.2 ly
The concepts of vector addition and components make these calculations very simple. (b) The angle is angle φ in Figure 1.101. cosθ =
EVALUATE: Figure 1.101 1-30 1.102. Chapter 1 "
""
ˆj
ˆ
IDENTIFY: Define S = Ai + Bˆ + Ck . Show that r ⋅ S = 0 if Ax + By + Cz = 0 .
SET UP: Use Eq.(1.21) to calculate the scalar product.
""
ˆjˆ
ˆj
ˆ
EXECUTE: r ⋅ S = ( xi + yˆ + zk ) ⋅ ( Ai + Bˆ + Ck ) = Ax + By + Cz
"
""
"
If the points satisfy Ax + By + Cz = 0, then r ⋅ S = 0 and all points r are perpendicular to S . The vector and plane
are sketched in Figure 1.102.
EVALUATE: If two vectors are perpendicular their scalar product is zero. Figure 1.102 2 MOTION ALONG A STRAIGHT LINE 2.1. IDENTIFY: The average velocity is vav-x = Δx
.
Δt Let + x be upward.
1000 m − 63 m
EXECUTE: (a) vav-x =
= 197 m/s
4.75 s
1000 m − 0
(b) vav-x =
= 169 m/s
5.90 s SET UP: 63 m − 0
= 54.8 m/s . When the velocity isn’t constant the
1.15 s
average velocity depends on the time interval chosen. In this motion the velocity is increasing.
Δx
IDENTIFY: vav-x =
Δt
SET UP: 13.5 days = 1.166 × 105 s . At the release point, x = +5.150 × 106 m .
EVALUATE: 2.2. x2 − x1 5.150 × 106 m
=
= −4.42 m/s
1.166 × 106 s
Δt
(b) For the round trip, x2 = x1 and Δx = 0 . The average velocity is zero.
EVALUATE: The average velocity for the trip from the nest to the release point is positive.
IDENTIFY: Target variable is the time Δt it takes to make the trip in heavy traffic. Use Eq.(2.2) that relates the
average velocity to the displacement and average time.
Δx
Δx
so Δx = vav-x Δt and Δt =
.
SET UP: vav-x =
Δt
vav-x
EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two
cities:
EXECUTE: 2.3. For the first 1.15 s of the flight, vav-x = (a) vav-x = Δx = vav-x Δt = (105 km/h)(1 h/60 min)(140 min) = 245 km.
Now use vav-x for heavy traffic to calculate Δt ; Δx is the same as before: Δt = Δx
245 km
=
= 3.50 h = 3 h and 30 min.
vav-x 70 km/h The trip takes an additional 1 hour and 10 minutes.
EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is
(105/ 70)(140 m) = 210 min.
2.4. Δx
. Use the average speed for each segment to find the time traveled
Δt
in that segment. The average speed is the distance traveled by the time.
SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m + 280 m = 480 m .
200 m
280 m
EXECUTE: (a) The eastward run takes time
= 40.0 s and the westward run takes
= 70.0 s . The
5.0 m/s
4.0 m/s
480 m
= 4.4 m/s .
average speed for the entire trip is
110.0 s
Δx −80 m
(b) vav-x =
=
= −0.73 m/s . The average velocity is directed westward.
Δt 110.0 s
IDENTIFY: The average velocity is vav-x = 2-1 2-2 2.5. 2.6. Chapter 2 EVALUATE: The displacement is much less than the distance traveled and the magnitude of the average velocity
is much less than the average speed. The average speed for the entire trip has a value that lies between the average
speed for the two segments.
IDENTIFY: When they first meet the sum of the distances they have run is 200 m.
SET UP: Each runs with constant speed and continues around the track in the same direction, so the distance each
runs is given by d = vt . Let the two runners be objects A and B.
200 m
EXECUTE: (a) d A + d B = 200 m , so (6.20 m/s)t + (5.50 m/s)t = 200 m and t =
= 17.1 s .
11.70 m/s
(b) d A = v At = (6.20 m/s)(17.1 s) = 106 m . d B = vBt = (5.50 m/s)(17.1 s) = 94 m . The faster runner will be 106 m
from the starting point and the slower runner will be 94 m from the starting point. These distances are measured
around the circular track and are not straight-line distances.
EVALUATE: The faster runner runs farther.
IDENTIFY: To overtake the slower runner the first time the fast runner must run 200 m farther. To overtake the
slower runner the second time the faster runner must run 400 m farther.
SET UP: t and x0 are the same for the two runners.
(a) Apply x − x0 = v0 xt to each runner: ( x − x0 )f = (6.20 m/s)t and ( x − x0 )s = (5.50 m/s)t . EXECUTE: ( x − x0 )f = ( x − x0 )s + 200 m gives (6.20 m/s)t = (5.50 m/s)t + 200 m and t = 200 m
= 286 s .
6.20 m/s − 5.50 m/s ( x − x0 )f = 1770 m and ( x − x0 )s = 1570 m . 2.7. (b) Repeat the calculation but now ( x − x0 )f = ( x − x0 )s + 400 m . t = 572 s . The fast runner has traveled 3540 m.
He has made 17 full laps for 3400 m and 140 m past the starting line in this 18th lap.
EVALUATE: In part (a) the fast runner will have run 8 laps for 1600 m and will be 170 m past the starting line in
his 9th lap.
IDENTIFY: In time tS the S-waves travel a distance d = vStS and in time tP the P-waves travel a distance
d = vPtP .
SET UP: tS = tP + 33 s dd
1
1
⎛
⎞
−
= + 33 s . d ⎜
⎟ = 33 s and d = 250 km .
vS vP
⎝ 3.5 km/s 6.5 km/s ⎠
EVALUATE: The times of travel for each wave are tS = 71 s and tP = 38 s .
EXECUTE: 2.8. IDENTIFY:
SET UP: Δx
. Use x(t ) to find x for each t.
Δt
x (0) = 0 , x(2.00 s) = 5.60 m , and x(4.00 s) = 20.8 m
The average velocity is vav-x = (a) vav-x = EXECUTE: 5.60 m − 0
= +2.80 m/s
2.00 s 20.8 m − 0
= +5.20 m/s
4.00 s
20.8 m − 5.60 m
(c) vav-x =
= +7.60 m/s
2.00 s
EVALUATE: The average velocity depends on the time interval being considered.
(a) IDENTIFY: Calculate the average velocity using Eq.(2.2).
Δx
SET UP: vav-x =
so use x(t ) to find the displacement Δx for this time interval.
Δt
EXECUTE: t = 0 : x = 0
t = 10.0 s: x = (2.40 m/s 2 )(10.0 s) 2 − (0.120 m/s3 )(10.0 s)3 = 240 m − 120 m = 120 m.
(b) vav-x = 2.9. Δx 120 m
=
= 12.0 m/s.
Δt 10.0 s
(b) IDENTIFY: Use Eq.(2.3) to calculate vx (t ) and evaluate this expression at each specified t.
Then vav-x = dx
= 2bt − 3ct 2 .
dt
EXECUTE: (i) t = 0 : vx = 0
SET UP: vx = (ii) t = 5.0 s: vx = 2(2.40 m/s 2 )(5.0 s) − 3(0.120 m/s3 )(5.0 s)2 = 24.0 m/s − 9.0 m/s = 15.0 m/s.
(iii) t = 10.0 s: vx = 2(2.40 m/s 2 )(10.0 s) − 3(0.120 m/s3 )(10.0 s)2 = 48.0 m/s − 36.0 m/s = 12.0 m/s. Motion Along a Straight Line (c) IDENTIFY:
SET UP: 2-3 Find the value of t when vx (t ) from part (b) is zero. vx = 2bt − 3ct 2 vx = 0 at t = 0.
vx = 0 next when 2bt − 3ct 2 = 0
EXECUTE: 2.10. 2.11. 2b = 3ct so t = 2b 2(2.40 m/s 2 )
=
= 13.3 s
3c 30(.120 m/s3 ) EVALUATE: vx (t ) for this motion says the car starts from rest, speeds up, and then slows down again.
IDENTIFY and SET UP: The instantaneous velocity is the slope of the tangent to the x versus t graph.
EXECUTE: (a) The velocity is zero where the graph is horizontal; point IV.
(b) The velocity is constant and positive where the graph is a straight line with positive slope; point I.
(c) The velocity is constant and negative where the graph is a straight line with negative slope; point V.
(d) The slope is positive and increasing at point II.
(e) The slope is positive and decreasing at point III.
EVALUATE: The sign of the velocity indicates its direction.
Δx
IDENTIFY: The average velocity is given by vav-x =
. We can find the displacement Δt for each constant
Δt
velocity time interval. The average speed is the distance traveled divided by the time.
SET UP: For t = 0 to t = 2.0 s , vx = 2.0 m/s . For t = 2.0 s to t = 3.0 s , vx = 3.0 m/s . In part (b), vx = −3.0 m/s for t = 2.0 s to t = 3.0 s . When the velocity is constant, Δx = vx Δt .
EXECUTE: (a) For t = 0 to t = 2.0 s , Δx = (2.0 m/s)(2.0 s) = 4.0 m . For t = 2.0 s to t = 3.0 s ,
Δx = (3.0 m/s)(1.0 s) = 3.0 m . For the first 3.0 s, Δx = 4.0 m + 3.0 m = 7.0 m . The distance traveled is also 7.0 m.
Δx 7.0 m
=
= 2.33 m/s . The average speed is also 2.33 m/s.
Δt 3.0 s
(b) For t = 2.0 s to 3.0 s, Δx = ( −3.0 m/s)(1.0 s) = −3.0 m . For the first 3.0 s, Δx = 4.0 m + (−3.0 m) = +1.0 m .
The dog runs 4.0 m in the + x -direction and then 3.0 m in the − x -direction, so the distance traveled is still 7.0 m.
Δx 1.0 m
7.00 m
=
= 0.33 m/s . The average speed is
= 2.33 m/s .
vav-x =
Δt 3.0 s
3.00 s
EVALUATE: When the motion is always in the same direction, the displacement and the distance traveled are
equal and the average velocity has the same magnitude as the average speed. When the motion changes direction
during the time interval, those quantities are different.
Δv
IDENTIFY and SET UP: aav,x = x . The instantaneous acceleration is the slope of the tangent to the vx versus
Δt
t graph.
EXECUTE: (a) 0 s to 2 s: aav,x = 0 ; 2 s to 4 s: aav,x = 1.0 m/s 2 ; 4 s to 6 s: aav,x = 1.5 m/s 2 ; 6 s to 8 s:
The average velocity is vav-x = 2.12. aav,x = 2.5 m/s 2 ; 8 s to 10 s: aav,x = 2.5 m/s 2 ; 10 s to 12 s: aav,x = 2.5 m/s 2 ; 12 s to 14 s: aav,x = 1.0 m/s 2 ; 14 s to
16 s: aav,x = 0 . The acceleration is not constant over the entire 16 s time interval. The acceleration is constant
between 6 s and 12 s.
(b) The graph of vx versus t is given in Fig. 2.12. t = 9 s : ax = 2.5 m/s 2 ; t = 13 s : ax = 1.0 m/s 2 ; t = 15 s : ax = 0 . 2-4 Chapter 2 EVALUATE: The acceleration is constant when the velocity changes at a constant rate. When the velocity is
constant, the acceleration is zero. Figure 2.12
2.13. Δvx
.
Δt
SET UP: Assume the car is moving in the + x direction. 1 mi/h = 0.447 m/s , so 60 mi/h = 26.82 m/s ,
200 mi/h = 89.40 m/s and 253 mi/h = 113.1 m/s .
EXECUTE: (a) The graph of vx versus t is sketched in Figure 2.13. The graph is not a straight line, so the
acceleration is not constant.
26.82 m/s − 0
89.40 m/s − 26.82 m/s
(b) (i) aav-x =
= 12.8 m/s 2 (ii) aav-x =
= 3.50 m/s 2 (iii)
2.1 s
20.0 s − 2.1 s
113.1 m/s − 89.40 m/s
aav-x =
= 0.718 m/s 2 . The slope of the graph of vx versus t decreases as t increases. This is
53 s − 20.0 s
consistent with an average acceleration that decreases in magnitude during each successive time interval.
EVALUATE: The average acceleration depends on the chosen time interval. For the interval between 0 and 53 s,
113.1 m/s − 0
aav-x =
= 2.13 m/s 2 .
53 s
IDENTIFY: The average acceleration for a time interval Δt is given by aav-x = Figure 2.13 Motion Along a Straight Line 2.14. 2-5 Δvx
. ax (t ) is the slope of the vx versus t graph.
Δt
SET UP: 60 km/h = 16.7 m/s
16.7 m/s − 0
0 − 16.7 m/s
EXECUTE: (a) (i) aav-x =
= 1.7 m/s 2 . (ii) aav-x =
= −1.7 m/s 2 .
10 s
10 s
(iii) Δvx = 0 and aav-x = 0 . (iv) Δvx = 0 and aav-x = 0 .
IDENTIFY: aav-x = (b) At t = 20 s , vx is constant and ax = 0 . At t = 35 s , the graph of vx versus t is a straight line and ax = aav-x = −1.7 m/s 2 . 2.15. EVALUATE: When aav-x and vx have the same sign the speed is increasing. When they have opposite sign the
speed is decreasing.
dx
dv
and ax = x to calculate vx (t ) and ax (t ).
IDENTIFY and SET UP: Use vx =
dt
dt
dx
2
= 2.00 cm/s − (0.125 cm/s )t
EXECUTE: vx =
dt
dv
ax = x = −0.125 cm/s 2
dt
(a) At t = 0, x = 50.0 cm, vx = 2.00 cm/s, ax = −0.125 cm/s 2 .
(b) Set vx = 0 and solve for t: t = 16.0 s.
(c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0 s. The turtle returns to the starting point after
32.0 s.
(d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm.
Set x = 60.0 cm and solve for t: t = 6.20 s and t = 25.8 s.
At t = 6.20 s, vx = +1.23 cm/s. At t = 25.8 s, vx = −1.23 cm/s.
Set x = 40.0 cm and solve for t: t = 36.4 s (other root to the quadratic equation is negative and hence
nonphysical).
At t = 36.4 s, vx = −2.55 cm/s.
(e) The graphs are sketched in Figure 2.15. Figure 2.15 2.16. EVALUATE: The acceleration is constant and negative. vx is linear in time. It is initially positive, decreases to
zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin
but then stops and moves in the − x -direction.
IDENTIFY: Use Eq.(2.4), with Δt = 10 s in all cases.
SET UP: vx is negative if the motion is to the right. ( ( 5.0 m/s ) − (15.0 m/s ) ) / (10 s ) = −1.0 m/s
(b) ( ( −15.0 m/s ) − ( −5.0 m/s ) ) / (10 s ) = −1.0 m/s
(c) ( ( −15.0 m/s ) − ( +15.0 m/s ) ) / (10 s ) = −3.0 m/s
EXECUTE: (a) 2 2 2 In all cases, the negative acceleration indicates an acceleration to the left.
Δv
IDENTIFY: The average acceleration is aav-x = x
Δt
SET UP: Assume the car goes from rest to 65 mi/h (29 m/s) in 10 s. In braking, assume the car goes from 65 mi/h
to zero in 4.0 s. Let + x be in the direction the car is traveling.
29 m/s − 0
EXECUTE: (a) aav-x =
= 2.9 m/s 2
10 s
0 − 29 m/s
(b) aav-x =
= −7.2 m/s 2
4.0 s
EVALUATE: 2.17. 2-6 2.18. Chapter 2 (c) In part (a) the speed increases so the acceleration is in the same direction as the velocity. If the velocity
direction is positive, then the acceleration is positive. In part (b) the speed decreases so the acceleration is in the
direction opposite to the direction of the velocity. If the velocity direction is positive then the acceleration is
negative, and if the velocity direction is negative then the acceleration direction is positive.
EVALUATE: The sign of the velocity and of the acceleration indicate their direction.
Δv
IDENTIFY: The average acceleration is aav-x = x . Use vx (t ) to find vx at each t. The instantaneous acceleration
Δt
dvx
.
is ax =
dt
SET UP: vx (0) = 3.00 m/s and vx (5.00 s) = 5.50 m/s .
EXECUTE: (a) aav-x = Δvx 5.50 m/s − 3.00 m/s
=
= 0.500 m/s 2
Δt
5.00 s dvx
= (0.100 m/s3 )(2t ) = (0.200 m/s3 )t . At t = 0 , ax = 0 . At t = 5.00 s , ax = 1.00 m/s 2 .
dt
(c) Graphs of vx (t ) and ax (t ) are given in Figure 2.18.
(b) ax = ax (t ) is the slope of vx (t ) and increases at t increases. The average acceleration for t = 0 to EVALUATE: t = 5.00 s equals the instantaneous acceleration at the midpoint of the time interval, t = 2.50 s , since ax (t ) is a
linear function of t. Figure 2.18
2.19. (a) IDENTIFY and SET UP:
EXECUTE: vx is the slope of the x versus t curve and ax is the slope of the vx versus t curve. t = 0 to t = 5 s : x versus t is a parabola so ax is a constant. The curvature is positive so ax is positive. vx versus t is a straight line with positive slope. v0 x = 0.
t = 5 s to t = 15 s : x versus t is a straight line so vx is constant and ax = 0. The slope of x versus t is positive so
vx is positive.
t = 15 s to t = 25 s: x versus t is a parabola with negative curvature, so ax is constant and negative. vx versus t is a
straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s.
t = 25 s to t = 35 s: x versus t is a straight line so vx is constant and ax = 0. The slope of x versus t is negative so
vx is negative.
t = 35 s to t = 40 s: x versus t is a parabola with positive curvature, so ax is constant and positive. vx versus t is a
straight line with positive slope. The velocity reaches zero at t = 40 s. Motion Along a Straight Line 2-7 The graphs of vx (t ) and ax (t ) are sketched in Figure 2.19a. Figure 2.19a
(b) The motions diagrams are sketched in Figure 2.19b. Figure 2.19b 2.20. EVALUATE: The spider speeds up for the first 5 s, since vx and ax are both positive. Starting at t = 15 s the
spider starts to slow down, stops momentarily at t = 20 s, and then moves in the opposite direction. At t = 35 s the
spider starts to slow down again and stops at t = 40 s.
dx
dv
and ax (t ) = x
IDENTIFY: vx (t ) =
dt
dt
dn
n −1
SET UP:
(t ) = nt for n ≥ 1 .
dt
EXECUTE: (a) vx (t ) = (9.60 m/s 2 )t − (0.600 m/s6 )t 5 and ax (t ) = 9.60 m/s 2 − (3.00 m/s 6 )t 4 . Setting vx = 0 gives t = 0 and t = 2.00 s . At t = 0 , x = 2.17 m and ax = 9.60 m/s 2 . At t = 2.00 s , x = 15.0 m and ax = −38.4 m/s 2 .
(b) The graphs are given in Figure 2.20. 2-8 Chapter 2 EVALUATE: For the entire time interval from t = 0 to t = 2.00 s , the velocity vx is positive and x increases. While ax is also positive the speed increases and while ax is negative the speed decreases. Figure 2.20
2.21. IDENTIFY: Use the constant acceleration equations to find v0 x and ax .
(a) SET UP: The situation is sketched in Figure 2.21. x − x0 = 70.0 m
t = 7.00 s
vx = 15.0 m/s
v0 x = ?
Figure 2.21 2( x − x0 )
2(70.0 m)
⎛v +v ⎞
− vx =
− 15.0 m/s = 5.0 m/s.
Use x − x0 = ⎜ 0 x x ⎟ t , so v0 x =
t
7.00 s
2⎠
⎝
v −v
15.0 m/s − 5.0 m/s
= 1.43 m/s 2 .
(b) Use vx = v 0 x + axt , so ax = x 0 x =
t
7.00 s
EVALUATE: The average velocity is (70.0 m)/(7.00 s) = 10.0 m/s. The final velocity is larger than this, so the
EXECUTE: 2.22. antelope must be speeding up during the time interval; v0 x < vx and ax > 0.
IDENTIFY: Apply the constant acceleration kinematic equations.
SET UP: Let + x be in the direction of the motion of the plane. 173 mi/h = 77.33 m/s . 307 ft = 93.57 m .
2
2
EXECUTE: (a) v0 x = 0 , vx = 77.33 m/s and x − x0 = 93.57 m . vx = v0 x + 2ax ( x − x0 ) gives
ax = 2
2
vx − v0 x
(77.33 m/s) 2 − 0
=
= 32.0 m/s 2 .
2( x − x0 )
2(93.57 m) 2( x − x0 )
2(93.57 m)
⎛v +v ⎞
(b) x − x0 = ⎜ 0 x x ⎟ t gives t =
=
= 2.42 s
v0 x + vx
0 + 77.33 m/s
2⎠
⎝ 2.23. EVALUATE: Either vx = v0 x + axt or x − x0 = v0 xt + 1 axt 2 could also be used to find t and would give the same
2
result as in part (b).
IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply.
SET UP: Assume the ball starts from rest and moves in the + x -direction.
2
2
EXECUTE: (a) x − x0 = 1.50 m , vx = 45.0 m/s and v0 x = 0 . vx = v0 x + 2ax ( x − x0 ) gives ax = 2
2
vx − v0 x
(45.0 m/s) 2
=
= 675 m/s 2 .
2( x − x0 ) 2(1.50 m) 2( x − x0 ) 2(1.50 m)
⎛v +v ⎞
(b) x − x0 = ⎜ 0 x x ⎟ t gives t =
=
= 0.0667 s
v0 x + vx
45.0 m/s
2⎠
⎝
v
45.0 m/s
EVALUATE: We could also use vx = v0 x + axt to find t = x =
= 0.0667 s which agrees with our
ax 675 m/s 2
previous result. The acceleration of the ball is very large. Motion Along a Straight Line 2.24. 2-9 IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply.
SET UP: Assume the ball moves in the + x direction.
EXECUTE: (a) vx = 73.14 m/s , v0 x = 0 and t = 30.0 ms . vx = v0 x + axt gives vx − v0 x 73.14 m/s − 0
=
= 2440 m/s 2 .
t
30.0 × 10−3 s
⎛ v + v ⎞ ⎛ 0 + 73.14 m/s ⎞
−3
(b) x − x0 = ⎜ 0 x x ⎟ t = ⎜
⎟ (30.0 × 10 s) = 1.10 m
2
⎝2⎠⎝
⎠
ax = EVALUATE: 2.25. x − x0 = 1 (2440 m/s 2 )(30.0 × 10−3 s) 2 = 1.10 m , which agrees with our previous result. The acceleration of the ball
2
is very large.
IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set
ax equal to its maximum allowed value.
SET UP: Let + x be the direction of the initial velocity of the car. ax = −250 m/s 2 . 105 km/h = 29.17 m/s . 2
2
vx − v0 x 0 − (29.17 m/s) 2
=
= 1.70 m .
2ax
2(−250 m/s 2 )
EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your
1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while
stopping.
IDENTIFY: Apply constant acceleration equations to the motion of the car.
SET UP: Let + x be the direction the car is moving.
2
vx
(20 m s) 2
EXECUTE: (a) From Eq. (2.13), with v0 x = 0, ax =
=
= 1.67 m s 2 .
2( x − x0 ) 2(120 m) EXECUTE: 2.26. We could also use x − x0 = v0 xt + 1 axt 2 to calculate x − x0 :
2 2
2
v0 x = +29.17 m/s . vx = 0 . vx = v0 x + 2ax ( x − x0 ) gives x − x0 = (b) Using Eq. (2.14), t = 2( x − x0 ) v x = 2(120 m) (20 m s) = 12 s. 2.27. (c) (12 s)(20 m s) = 240 m.
EVALUATE: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as
far.
Δv
IDENTIFY: The average acceleration is aav-x = x . For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14)
Δt
apply.
SET UP: Assume the shuttle travels in the + x direction. 161 km/h = 44.72 m/s and 1610 km/h = 447.2 m/s .
1.00 min = 60.0 s
Δv
44.72 m/s − 0
= 5.59 m/s 2
EXECUTE: (a) (i) aav-x = x =
Δt
8.00 s
447.2 m/s − 44.72 m/s
(ii) aav-x =
= 7.74 m/s 2
60.0 s − 8.00 s
⎛ v + v ⎞ ⎛ 0 + 44.72 m/s ⎞
(b) (i) t = 8.00 s , v0 x = 0 , and vx = 44.72 m/s . x − x0 = ⎜ 0 x x ⎟ t = ⎜
⎟ (8.00 s) = 179 m .
2
⎝2⎠⎝
⎠ (ii) Δt = 60.0 s − 8.00 s = 52.0 s , v0 x = 44.72 m/s , and vx = 447.2 m/s . 2.28. ⎛ v + v ⎞ ⎛ 44.72 m/s + 447.2 m/s ⎞
4
x − x0 = ⎜ 0 x x ⎟ t = ⎜
⎟ (52.0 s) = 1.28 × 10 m .
2⎠⎝
2
⎝
⎠
EVALUATE: When the acceleration is constant the instantaneous acceleration throughout the time interval equals
the average acceleration for that time interval. We could have calculated the distance in part (a) as
x − x0 = v0 xt + 1 axt 2 = 1 (5.59 m/s 2 )(8.00 s) 2 = 179 m , which agrees with our previous calculation.
2
2
IDENTIFY: Apply the constant acceleration kinematic equations to the motion of the car.
SET UP: 0.250 mi = 1320 ft . 60.0 mph = 88.0 ft/s . Let + x be the direction the car is traveling.
EXECUTE: ax = 2
2
(a) braking: v0 x = 88.0 ft/s , x − x0 = 146 ft , vx = 0 . vx = v0 x + 2ax ( x − x0 ) gives 2
2
vx − v0 x
0 − (88.0 ft/s) 2
=
= −26.5 ft/s 2
2( x − x0 )
2(146 ft) Speeding up: v0 x = 0 , x − x0 = 1320 ft , t = 19.9 s . x − x0 = v0 xt + 1 axt 2 gives
2
ax = 2( x − x0 ) 2(1320 ft)
=
= 6.67 ft/s 2
t2
(19.9 s) 2 2-10 Chapter 2 (b) vx = v0 x + axt = 0 + (6.67 ft/s 2 )(19.9 s) = 133 ft/s = 90.5 mph vx − v0 x 0 − 88.0 ft/s
=
= 3.32 s
ax
−26.5 ft/s 2
EVALUATE: The magnitude of the acceleration while braking is much larger than when speeding up. That is why
it takes much longer to go from 0 to 60 mph than to go from 60 mph to 0.
IDENTIFY: The acceleration ax is the slope of the graph of vx versus t.
(c) t = 2.29. SET UP: The signs of vx and of ax indicate their directions. EXECUTE:
to the left. (a) Reading from the graph, at t = 4.0 s , vx = 2.7 cm/s , to the right and at t = 7.0 s , vx = 1.3 cm/s , 8.0 cm/s
= −1.3 cm/s 2 . The acceleration is constant and equal to
6.0 s
1.3 cm/s 2 , to the left. It has this value at all times.
(c) Since the acceleration is constant, x − x0 = v0 xt + 1 axt 2 . For t = 0 to 4.5 s,
2
(b) vx versus t is a straight line with slope − x − x0 = (8.0 cm/s)(4.5 s) + 1 (−1.3 cm/s 2 )(4.5 s) 2 = 22.8 cm . For t = 0 to 7.5 s,
2
x − x0 = (8.0 cm/s)(7.5 s) + 1 (−1.3 cm/s 2 )(7.5 s) 2 = 23.4 cm
2
(d) The graphs of ax and x versus t are given in Fig. 2.29.
EVALUATE: ⎛v +v ⎞
In part (c) we could have instead used x − x0 = ⎜ 0 x x ⎟ t .
⎝2⎠ Figure 2.29
2.30. IDENTIFY: Use the constant acceleration equations to find x, v0 x , vx and ax for each constant-acceleration
segment of the motion.
SET UP: Let + x be the direction of motion of the car and let x = 0 at the first traffic light.
⎛ v + v ⎞ ⎛ 0 + 20 m/s ⎞
EXECUTE: (a) For t = 0 to t = 8 s : x = ⎜ 0 x x ⎟ t = ⎜
⎟ (8 s) = 80 m .
2
⎝2⎠⎝
⎠ vx − v0 x 20 m/s
=
= +2.50 m/s 2 . The car moves from x = 0 to x = 80 m . The velocity vx increases linearly
t
8s
from zero to 20 m/s. The acceleration is a constant 2.50 m/s 2 .
Constant speed for 60 m: The car moves from x = 80 m to x = 140 m . vx is a constant 20 m/s. ax = 0 . This
ax = 60 m
+ 8 s = 11 s .
20 m/s
Slowing from 20 m/s until stopped: The car moves from x = 140 m to x = 180 m . The velocity decreases linearly
2(40 m)
⎛v +v ⎞
2
2
= 4 s . vx = v0 x + 2ax ( x − x0 ) gives
from 20 m/s to zero. x − x0 = ⎜ 0 x x ⎟ t gives t =
20 m/s + 0
⎝2⎠
interval starts at t = 8 s and continues until t = ax = −(20.0 m/s) 2
= −5.00 m/s 2 This segment is from t = 11 s to t = 15 s . The acceleration is a
2(40 m) constant −5.00 m/s 2 .
The graphs are drawn in Figure 2.30a.
(b) The motion diagram is sketched in Figure 2.30b. Motion Along a Straight Line 2-11 !
!
!
!
EVALUATE: When a and v are in the same direction, the speed increases ( t = 0 to t = 8 s ). When a and v are in
opposite directions, the speed decreases ( t = 11 s to t = 15 s ). When a = 0 the speed is constant t = 8 s to t = 11 s . Figure 2.30a-b
2.31. (a) IDENTIFY and SET UP: The acceleration ax at time t is the slope of the tangent to the vx versus t curve at
time t.
EXECUTE: At t = 3 s, the vx versus t curve is a horizontal straight line, with zero slope. Thus ax = 0. At t = 7 s, the vx versus t curve is a straight-line segment with slope 45 m/s − 20 m/s
= 6.3 m/s 2 .
9 s −5 s Thus ax = 6.3 m/s 2 .
At t = 11 s the curve is again a straight-line segment, now with slope −0 − 45 m/s
= −11.2 m/s 2 .
13 s − 9 s Thus ax = −11.2 m/s 2 .
EVALUATE: ax = 0 when vx is constant, ax > 0 when vx is positive and the speed is increasing, and ax < 0 when vx is positive and the speed is decreasing.
(b) IDENTIFY: Calculate the displacement during the specified time interval.
SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is
constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration
equations for each segment. For the time interval t = 0 to t = 5 s the acceleration is constant and equal to zero.
For the time interval t = 5 s to t = 9 s the acceleration is constant and equal to 6.25 m/s 2 . For the interval t = 9 s
to t = 13 s the acceleration is constant and equal to −11.2 m/s 2 .
EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas
can be used.
v0 x = 20 m/s ax = 0 t = 5 s x − x0 = ?
x − x0 = v0 xt (ax = 0 so no 1
2 axt 2 term) x − x0 = (20 m/s)(5 s) = 100 m; this is the distance the officer travels in the first 5 seconds.
During the interval t = 5 s to 9 s the acceleration is again constant. The constant acceleration formulas can be
applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time t = 0 and ends at
time t = 5 s. (Note that the acceleration is not constant over the entire t = 0 to t = 9 s interval.)
v0 x = 20 m/s ax = 6.25 m/s 2 t = 4 s x0 = 100 m x − x0 = ?
x − x0 = v0 xt + 1 axt 2
2
x − x0 = (20 m/s)(4 s) + 1 (6.25 m/s 2 )(4 s) 2 = 80 m + 50 m = 130 m.
2
Thus x − x0 + 130 m = 100 m + 130 m = 230 m. 2-12 Chapter 2 At t = 9 s the officer is at x = 230 m, so she has traveled 230 m in the first 9 seconds.
During the interval t = 9 s to t = 13 s the acceleration is again constant. The constant acceleration formulas can be
applied for this 4 second interval but not for the whole t = 0 to t = 13 s interval. To use the equations restart our
clock so this interval begins at time t = 0 and ends at time t = 4 s.
v0 x = 45 m/s (at the start of this time interval)
ax = −11.2 m/s 2 t = 4 s x0 = 230 m x − x0 = ?
x − x0 = v0 xt + 1 axt 2
2
x − x0 = (45 m/s)(4 s) + 1 ( −11.2 m/s 2 )(4 s) 2 = 180 m − 89.6 m = 90.4 m.
2
Thus x = x0 + 90.4 m = 230 m + 90.4 m = 320 m.
At t = 13 s the officer is at x = 320 m, so she has traveled 320 m in the first 13 seconds.
EVALUATE: The velocity vx is always positive so the displacement is always positive and displacement and
distance traveled are the same. The average velocity for time interval Δt is vav-x = Δx / Δt. For t = 0 to 5 s, 2.32. vav-x = 20 m/s. For t = 0 to 9 s, vav-x = 26 m/s. For t = 0 to 13 s, vav-x = 25 m/s. These results are consistent with
Fig. 2.33.
IDENTIFY: In each constant acceleration interval, the constant acceleration equations apply.
SET UP: When ax is constant, the graph of vx versus t is a straight line and the graph of x versus t is a parabola.
When ax = 0 , vx is constant and x versus t is a straight line.
EXECUTE: The graphs are given in Figure 2.32.
EVALUATE: The slope of the x versus t graph is vx (t ) and the slope of the vx versus t graph is ax (t ) . Figure 2.32
2.33. (a) IDENTIFY: The maximum speed occurs at the end of the initial acceleration period.
SET UP: ax = 20.0 m/s 2 t = 15.0 min = 900 s v0 x = 0 vx = ? vx = v0 x + axt
EXECUTE: vx = 0 + (20.0 m/s 2 )(900 s) = 1.80 × 104 m/s
(b) IDENTIFY: Use constant acceleration formulas to find the displacement Δx. The motion consists of three
constant acceleration intervals. In the middle segment of the trip ax = 0 and vx = 1.80 × 104 m/s, but we can’t
directly find the distance traveled during this part of the trip because we don’t know the time. Instead, find the
distance traveled in the first part of the trip (where ax = +20.0 m/s 2 ) and in the last part of the trip (where ax = −20.0 m/s 2 ). Subtract these two distances from the total distance of 3.84 × 108 m to find the distance traveled
in the middle part of the trip (where ax = 0).
first segment
SET UP: x − x0 = ? t = 15.0 min = 900 s ax = +20.0 m/s 2 v0 x = 0
x − x0 = v0 xt + 1 axt 2
2
EXECUTE: x − x 0 = 0 + 1 (20.0 m/s 2 )(900 s)2 = 8.10 × 106 m = 8.10 × 103 km
2 second segment
SET UP: x − x0 = ? t = 15.0 min = 900 s ax = −20.0 m/s 2
v0 x = 1.80 × 104 m/s
x − x0 = v0 xt + 1 axt 2
2
EXECUTE: x − x0 = (1.80 × 104 s)(900 s) + 1 (−20.0 m/s 2 )(900 s) 2 = 8.10 × 106 m = 8.10 × 103 km (The same
2
distance as traveled as in the first segment.) Motion Along a Straight Line 2-13 Therefore, the distance traveled at constant speed is
3.84 × 108 m − 8.10 × 106 m − 8.10 × 106 m = 3.678 × 108 m = 3.678 × 105 km.
3.678 × 108 m
= 0.958.
The fraction this is of the total distance is
3.84 × 108 m
(c) IDENTIFY: We know the time for each acceleration period, so find the time for the constant speed segment.
SET UP: x − x0 = 3.678 × 108 m vx = 1.80 × 104 m/s ax = 0 t = ?
x − x0 = v0 xt + 1 axt 2
2
x − x0 3.678 × 108 m
=
= 2.043 × 104 s = 340.5 min.
v0 x
1.80 × 104 m/s
The total time for the whole trip is thus 15.0 min + 340.5 min + 15.0 min = 370min.
EVALUATE: If the speed was a constant 1.80 × 104 m/s for the entire trip, the trip would take
(3.84 × 108 m)/(1.80 × 104 m/s) = 356 min. The trip actually takes a bit longer than this since the average velocity is
EXECUTE: 2.34. t= less than 1.80 × 108 m/s during the relatively brief acceleration phases.
IDENTIFY: Use constant acceleration equations to find x − x0 for each segment of the motion.
SET UP: Let + x be the direction the train is traveling.
EXECUTE: t = 0 to 14.0 s: x − x0 = v0 xt + 1 axt 2 = 1 (1.60 m/s 2 )(14.0 s) 2 = 157 m .
2
2
At t = 14.0 s , the speed is vx = v0 x + axt = (1.60 m/s 2 )(14.0 s) = 22.4 m/s . In the next 70.0 s, ax = 0 and
x − x0 = v0 xt = (22.4 m/s)(70.0 s) = 1568 m .
For the interval during which the train is slowing down, v0 x = 22.4 m/s , ax = −3.50 m/s 2 and vx = 0 .
2
2
vx − v0 x 0 − (22.4 m/s) 2
=
= 72 m .
2a x
2(−3.50 m/s 2 )
The total distance traveled is 157 m + 1568 m + 72 m = 1800 m .
EVALUATE: The acceleration is not constant for the entire motion but it does consist of constant acceleration
segments and we can use constant acceleration equations for each segment.
IDENTIFY: vx (t ) is the slope of the x versus t graph. Car B moves with constant speed and zero acceleration.
Car A moves with positive acceleration; assume the acceleration is constant.
SET UP: For car B, vx is positive and ax = 0 . For car A, ax is positive and vx increases with t.
EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.35a.
(b) The two cars have the same position at times when their x-t graphs cross. The figure in the problem shows this
occurs at approximately t = 1 s and t = 3 s .
(c) The graphs of vx versus t for each car are sketched in Figure 2.35b.
(d) The cars have the same velocity when their x-t graphs have the same slope. This occurs at approximately
t=2s.
(e) Car A passes car B when x A moves above xB in the x-t graph. This happens at t = 3 s .
2
2
vx = v0 x + 2ax ( x − x0 ) gives x − x0 = 2.35 (f) Car B passes car A when xB moves above x A in the x-t graph. This happens at t = 1 s .
EVALUATE: When ax = 0 , the graph of vx versus t is a horizontal line. When ax is positive, the graph of vx versus t is a straight line with positive slope. Figure 2.35a-b
2.36. IDENTIFY: Apply the constant acceleration equations to the motion of each vehicle. The truck passes the car
when they are at the same x at the same t > 0 . 2-14 Chapter 2 The truck has ax = 0 . The car has v0 x = 0 . Let + x be in the direction of motion of the vehicles. Both SET UP: vehicles start at x0 = 0 . The car has aC = 3.20 m/s 2 . The truck has vx = 20.0 m/s .
EXECUTE: (a) x − x0 = v0 xt + 1 axt 2 gives xT = v0Tt and xC = 1 aCt 2 . Setting xT = xC gives t = 0 and v0T = 1 aCt , so
2
2
2 2v0T 2(20.0 m/s)
=
= 12.5 s . At this t, xT = (20.0 m/s)(12.5 s) = 250 m and x = 1 (3.20 m/s 2 )(12.5 s) 2 = 250 m .
2
aC
3.20 m/s 2
The car and truck have each traveled 250 m.
(b) At t = 12.5 s , the car has vx = v0 x + axt = (3.20 m/s 2 )(12.5 s) = 40 m/s .
t= (c) xT = v0Tt and xC = 1 aCt 2 . The x-t graph of the motion for each vehicle is sketched in Figure 2.36a.
2
(d) vT = v0T . vC = aCt . The vx -t graph for each vehicle is sketched in Figure 2.36b.
EVALUATE: When the car overtakes the truck its speed is twice that of the truck. Figure 2.36a-b
2.37. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply.
SET UP: Take + y to be downward, so the motion is in the + y direction. 19,300 km/h = 5361 m/s ,
1600 km/h = 444.4 m/s , and 321 km/h = 89.2 m/s . 4.0 min = 240 s .
EXECUTE: (a) Stage A: t = 240 s , v0 y = 5361 m/s , v y = 444.4 m/s . v y = v0 y + a y t gives v y − v0 y 444.4 m/s − 5361 m/s
= −20.5 m/s 2 .
240 s
Stage B: t = 94 s , v0 y = 444.4 m/s , v y = 89.2 m/s . v y = v0 y + a y t gives
ay = t v y − v0 y = 89.2 m/s − 444.4 m/s
= −3.8 m/s 2 .
94 s
2
2
Stage C: y − y0 = 75 m , v0 y = 89.2 m/s , v y = 0 . v y = v0 y + 2a y ( y − y0 ) gives
ay = ay = t = 2
2
v y − v0 y 2( y − y0 )
upward. = 0 − (89.2 m/s) 2
= −53.0 m/s 2 . In each case the negative sign means that the acceleration is
2(75 m) ⎛ v + v y ⎞ ⎛ 5361 m/s + 444.4 m/s ⎞
(b) Stage A: y − y0 = ⎜ 0 y
⎟t = ⎜
⎟ (240 s) = 697 km .
2
⎠
⎝2⎠⎝ 2.38. ⎛ 444.4 m/s + 89.2 m/s ⎞
Stage B: y − y0 = ⎜
⎟ (94 s) = 25 km .
2
⎝
⎠
Stage C: The problem states that y − y0 = 75 m = 0.075 km .
The total distance traveled during all three stages is 697 km + 25 km + 0.075 km = 722 km .
EVALUATE: The upward acceleration produced by friction in stage A is calculated to be greater than the upward
acceleration due to the parachute in stage B. The effects of air resistance increase with increasing speed and in
reality the acceleration was probably not constant during stages A and B.
IDENTIFY: Assume an initial height of 200 m and a constant acceleration of 9.80 m/s 2 .
SET UP: Let + y be downward. 1 km/h = 0.2778 m/s and 1 mi/h = 0.4470 m/s . Motion Along a Straight Line EXECUTE: 2-15 2
2
(a) y − y0 = 200 m , a y = 9.80 m/s 2 , v0 y = 0 . v y = v0 y + 2a y ( y − y0 ) gives v y = 2(9.80 m/s 2 )(200 m) = 60 m/s = 200 km/h = 140 mi/h . 2.39. (b) Raindrops actually have a speed of about 1 m/s as they strike the ground.
(c) The actual speed at the ground is much less than the speed calculated assuming free-fall, so neglect of air
resistance is a very poor approximation for falling raindrops.
EVALUATE: In the absence of air resistance raindrops would land with speeds that would make them very
dangerous.
IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground,
a y = g , downward. Take the origin at the ground and the positive direction to be upward.
(a) SET UP: At the maximum height v y = 0. v y = 0 y − y0 = 0.440 m a y = −9.80 m/s 2 v0 y = ?
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE:
(b) SET UP: v0 y = −2a y ( y − y0 ) = −2(−9.80 m/s 2 )(0.440 m) = 2.94 m/s
When the flea has returned to the ground y − y0 = 0. y − y0 = 0 v0 y = +2.94 m/s a y = −9.80 m/s 2 t = ?
y − y0 = v0 yt + 1 a yt 2
2
EXECUTE:
EVALUATE:
2.40. With y − y0 = 0 this gives t = − 2v0 y
ay =− 2(2.94 m/s)
= 0.600 s.
−9.80 m/s 2 We can use v y = v0 y + a y t to show that with v0 y = 2.94 m/s, v y = 0 after 0.300 s. IDENTIFY: Apply constant acceleration equations to the motion of the lander.
SET UP: Let + y be positive. Since the lander is in free-fall, a y = +1.6 m/s 2 .
EXECUTE: 2
2
v0 y = 0.8 m/s , y − y0 = 5.0 m , a y = +1.6 m/s 2 in v y = v0 y + 2a y ( y − y0 ) gives 2
v y = v0 y + 2a y ( y − y0 ) = (0.8 m/s) 2 + 2(1.6 m/s 2 )(5.0 m) = 4.1 m/s . 2.41. EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to
gravity on earth is much larger than on the moon.
IDENTIFY: Apply constant acceleration equations to the motion of the meterstick. The time the meterstick falls is
your reaction time.
SET UP: Let + y be downward. The meter stick has v0 y = 0 and a y = 9.80 m/s 2 . Let d be the distance the meterstick falls.
EXECUTE: (a) y − y0 = v0 yt + 1 a yt 2 gives d = (4.90 m/s 2 )t 2 and t =
2 d
.
4.90 m/s 2 0.176 m
= 0.190 s
4.90 m/s 2
EVALUATE: The reaction time is proportional to the square of the distance the stick falls.
IDENTIFY: Apply constant acceleration equations to the vertical motion of the brick.
SET UP: Let + y be downward. a y = 9.80 m/s 2
(b) t = 2.42. EXECUTE: (a) v0 y = 0 , t = 2.50 s , a y = 9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 = 1 (9.80 m/s 2 )(2.50 s) 2 = 30.6 m . The
2
2 building is 30.6 m tall.
(b) v y = v0 y + a yt = 0 + (9.80 m/s 2 )(2.50 s) = 24.5 m/s
(c) The graphs of a y , v y and y versus t are given in Fig. 2.42. Take y = 0 at the ground. 2-16 Chapter 2 EVALUATE: ⎛ v + vy ⎞
2
2
We could use either y − y0 = ⎜ 0 y
⎟ t or v y = v0 y + 2a y ( y − y0 ) to check our results.
2⎠
⎝ Figure 2.42
2.43. IDENTIFY: When the only force is gravity the acceleration is 9.80 m/s 2 , downward. There are two intervals of
constant acceleration and the constant acceleration equations apply during each of these intervals.
SET UP: Let + y be upward. Let y = 0 at the launch pad. The final velocity for the first phase of the motion is the
initial velocity for the free-fall phase.
EXECUTE: (a) Find the velocity when the engines cut off. y − y0 = 525 m , a y = +2.25 m/s 2 , v0 y = 0 .
2
2
v y = v0 y + 2a y ( y − y0 ) gives v y = 2(2.25 m/s 2 )(525 m) = 48.6 m/s . Now consider the motion from engine cut off to maximum height: y0 = 525 m , v0 y = +48.6 m/s , v y = 0 (at the
2
2
maximum height), a y = −9.80 m/s 2 . v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = 0 − (48.6 m/s) 2
= 121 m and
2(−9.80 m/s 2 ) y = 121 m + 525 m = 646 m .
(b) Consider the motion from engine failure until just before the rocket strikes the ground: y − y0 = −525 m ,
2
2
a y = −9.80 m/s 2 , v0 y = +48.6 m/s . v y = v0 y + 2a y ( y − y0 ) gives v y = − (48.6 m/s) 2 + 2( −9.80 m/s 2 )(−525 m) = −112 m/s . Then v y = v0 y + a y t gives
t= v y − v0 y
ay = −112 m/s − 48.6 m/s
= 16.4 s .
−9.80 m/s 2 (c) Find the time from blast-off until engine failure: y − y0 = 525 m , v0 y = 0 , a y = +2.25 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2(525 m)
=
= 21.6 s . The rocket strikes the launch pad
ay
2.25 m/s 2 21.6 s + 16.4 s = 38.0 s after blast off. The acceleration a y is +2.25 m/s 2 from t = 0 to t = 21.6 s . It is
−9.80 m/s 2 from t = 21.6 s to 38.0 s . v y = v0 y + a y t applies during each constant acceleration segment, so the
graph of v y versus t is a straight line with positive slope of 2.25 m/s 2 during the blast-off phase and with negative
slope of −9.80 m/s 2 after engine failure. During each phase y − y0 = v0 yt + 1 a yt 2 . The sign of a y determines the
2
curvature of y (t ) . At t = 38.0 s the rocket has returned to y = 0 . The graphs are sketched in Figure 2.43.
EVALUATE: In part (b) we could have found the time from y − y0 = v0 yt + 1 a yt 2 , finding v y first allows us to
2 avoid solving for t from a quadratic equation. Figure 2.43 Motion Along a Straight Line 2.44. 2-17 IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag.
SET UP: Take + y upward. a y = −9.80 m/s 2 . The initial velocity of the sandbag equals the velocity of the balloon, so v0 y = +5.00 m/s . When the balloon reaches the ground, y − y0 = −40.0 m . At its maximum height the
sandbag has v y = 0 .
(a) t = 0.250 s : y − y0 = v0 yt + 1 a yt 2 = (5.00 m/s)(0.250 s) + 1 (−9.80 m/s 2 )(0.250 s) 2 = 0.94 m . The
2
2 EXECUTE: sandbag is 40.9 m above the ground. v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(0.250 s) = 2.55 m/s .
t = 1.00 s : y − y0 = (5.00 m/s)(1.00 s) + 1 (−9.80 m/s 2 )(1.00 s) 2 = 0.10 m . The sandbag is 40.1 m above the
2
ground. v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(1.00 s) = −4.80 m/s .
(b) y − y0 = −40.0 m , v0 y = 5.00 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives
2 −40.0 m = (5.00 m/s)t − (4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (5.00 m/s)t − 40.0 m = 0 and ) ( 1
5.00 ± (−5.00) 2 − 4(4.90)(−40.0) s = (0.51 ± 2.90) s . t must be positive, so t = 3.41 s .
9.80
(c) v y = v0 y + a yt = +5.00 m/s + ( −9.80 m/s 2 )(3.41 s) = −28.4 m/s
t= 2
2
(d) v0 y = 5.00 m/s , a y = −9.80 m/s 2 , v y = 0 . v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = 0 − (5.00 m/s) 2
= 1.28 m . The maximum height is 41.3 m above the ground.
2(−9.80 m/s 2 ) (e) The graphs of a y , v y , and y versus t are given in Fig. 2.44. Take y = 0 at the ground .
EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with
increasing speed. Figure 2.44
2.45. IDENTIFY:
(a) SET UP: The balloon has constant acceleration a y = g , downward.
Take the + y direction to be upward. t = 2.00 s, v0 y = −6.00 m/s, a y = −9.80 m/s 2 , v y = ?
EXECUTE:
(b) SET UP:
EXECUTE:
(c) SET UP: v y = v0 y + a yt = −6.00 m/s + ( −9.80 m/s 2 )(2.00 s) = −25.5 m/s
y − y0 = ?
y − y0 = v0 yt + 1 a yt 2 = (−6.00 m/s)(2.00 s) + 1 (−9.80 m/s 2 )(2.00 s) 2 = −31.6 m
2
2
y − y0 = −10.0 m, v0 y = −6.00 m/s, a y = −9.80 m/s 2 , v y = ? 2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: 2
v y = − v0 y + 2a y ( y − y0 ) = − (−6.00 m/s) 2 + 2( −9.80 m/s 2 )(−10.0 m) = −15.2 m/s (d) The graphs are sketched in Figure 2.45. Figure 2.45
EVALUATE: The speed of the balloon increases steadily since the acceleration and velocity are in the same
direction. v y = 25.5 m/s when y − y0 = 31.6 m, so v y is less than this (15.2 m/s) when y − y0 is less (10.0 m). 2-18 2.46. Chapter 2 IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of
magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the egg.
SET UP: Take + y to be upward. At the maximum height, v y = 0 .
(a) y − y0 = −50.0 m , t = 5.00 s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives
2 EXECUTE: y − y0 1
−50.0 m 1
− 2 ayt =
− 2 ( −9.80 m/s 2 )(5.00 s) = +14.5 m/s .
t
5.00 s
2
2
(b) v0 y = +14.5 m/s , v y = 0 (at the maximum height), a y = −9.80 m/s 2 . v y = v0 y + 2a y ( y − y0 ) gives
v0 y = y − y0 = 2
2
v y − v0 y 2a y = 0 − (14.5 m/s) 2
= 10.7 m .
2(−9.80 m/s 2 ) (c) At the maximum height v y = 0 .
(d) The acceleration is constant and equal to 9.80 m/s 2 , downward, at all points in the motion, including at the
maximum height.
(e) The graphs are sketched in Figure 2.46.
v y − v0 y −14.5 m/s
=
= 1.48 s . The egg has
EVALUATE: The time for the egg to reach its maximum height is t =
ay
−9.8 m/s 2 returned to the level of the cornice after 2.96 s and after 5.00 s it has traveled downward from the cornice for
2.04 s. Figure 2.46
2.47. Use the constant acceleration equations to calculate ax and x − x 0 . IDENTIFY: vx = 224 m/s, v0 x = 0, t = 0.900 s, ax = ? (a) SET UP: vx = v0 x + axt
vx − v0 x 224 m/s − 0
=
= 249 m/s 2
t
0.900 s
(b) ax / g = (249 m/s 2 ) /(9.80 m/s 2 ) = 25.4
ax = EXECUTE: (c) x − x0 = v0 xt + 1 axt 2 = 0 + 1 (249 m/s 2 )(0.900 s) 2 = 101 m
2
2
(d) SET UP: Calculate the acceleration, assuming it is constant:
t = 1.40 s, v0 x = 283 m/s, vx = 0 (stops), ax = ? vx = v0 x + axt
vx − v0 x 0 − 283 m/s
=
= −202 m/s 2
t
1.40 s
ax / g = (−202 m/s 2 ) /(9.80 m/s 2 ) = −20.6; ax = −20.6 g
If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is only 20.6g. But if
the acceleration is not constant it is certainly possible that at some point the instantaneous acceleration could be as
large as 40g.
EVALUATE: It is reasonable that for this motion the acceleration is much larger than g.
IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of
magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the boulder.
SET UP: Take + y to be upward.
EXECUTE: 2.48. ax = EXECUTE: (a) v0 y = +40.0 m/s , v y = +20.0 m/s , a y = −9.80 m/s 2 . v y = v0 y + a y t gives t= v y − v0 y
ay = 20.0 m/s − 40.0 m/s
= +2.04 s .
−9.80 m/s 2 Motion Along a Straight Line (b) v y = −20.0 m/s . t = v y − v0 y
ay = 2-19 −20.0 m/s − 40.0 m/s
= +6.12 s .
−9.80 m/s 2 (c) y − y0 = 0 , v0 y = +40.0 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives t = 0 and
2 t=− 2v0 y
ay 2(40.0 m/s)
= +8.16 s .
−9.80 m/s 2 =− (d) v y = 0 , v0 y = +40.0 m/s , a y = −9.80 m/s 2 . v y = v0 y + a y t gives t = v y − v0 y
ay = 0 − 40.0 m/s
= 4.08 s .
−9.80 m/s 2 2 (e) The acceleration is 9.80 m/s , downward, at all points in the motion.
(f) The graphs are sketched in Figure 2.48.
EVALUATE: v y = 0 at the maximum height. The time to reach the maximum height is half the total time in the air, so the answer in part (d) is half the answer in part (c). Also note that 2.04 s < 4.08 s < 6.12 s . The boulder is going
upward until it reaches its maximum height and after the maximum height it is traveling downward. Figure 2.48
2.49. IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall
and acceleration due to gravity and setting up a ratio involving time of fall and acceleration due to gravity.
SET UP: Let g En be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the common height from which the object is dropped. Let + y be downward, so y − y0 = h . v0 y = 0
2
2
y − y0 = v0 yt + 1 a yt 2 gives h = 1 gtE and h = 1 g En tEn . Combining these two equations gives
2
2
2 EXECUTE: 2 2.50. 2 ⎛t ⎞
⎛ 1.75 s ⎞
2
gt = g t and g En = g ⎜ E ⎟ = (9.80 m/s 2 ) ⎜
⎟ = 0.0868 m/s .
tEn ⎠
18.6 s ⎠
⎝
⎝
EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall.
IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use
Eqs.(2.17) and (2.18). Use the values of vx and of x at t = 1.0 s to evaluate v0 x and x0 .
2
E 2
En En SET UP:
EXECUTE: 1 ∫ t dt = n + 1t
n n +1 , for n ≥ 0 .
t (a) vx = v0 x + ∫ α tdt = v0 x + 1 α t 2 = v0 x + (0.60 m/s3 )t 2 . vx = 5.0 m/s when t = 1.0 s gives
2
0 v0 x = 4.4 m/s . Then, at t = 2.0 s , vx = 4.4 m/s + (0.60 m/s3 )(2.0 s) 2 = 6.8 m/s .
t
1
(b) x = x0 + ∫ (v0 x + 1 α t 2 ) dt = x0 + v0 xt + α t 3 . x = 6.0 m at t = 1.0 s gives x0 = 1.4 m . Then, at t = 2.0 s ,
2
0
6
1
x = 1.4 m + (4.4 m/s)(2.0 s) + (1.24 m/s3 )(2.0 s)3 = 11.8 m .
6
(c) x (t ) = 1.4 m + (4.4 m/s)t + (0.20 m/s3 )t 3 . vx (t ) = 4.4 m/s + (0.60 m/s3 )t 2 . ax (t ) = (1.20m/s3 )t . The graphs are
sketched in Figure 2.50. 2-20 Chapter 2 We can verify that ax = EVALUATE: dvx
dx
and vx =
.
dt
dt Figure 2.50
2.51. ax = At − Bt 2 with A = 1.50 m/s3 and B = 0.120 m/s 4
(a) IDENTIFY:
SET UP: Integrate ax (t ) to find vx (t ) and then integrate vx (t ) to find x(t ).
t vx = v0 x + ∫ ax dt
0 t vx = v0 x + ∫ ( At − Bt 2 ) dt = v0 x + 1 At 2 − 1 Bt 3
2
3 EXECUTE: 0 At rest at t = 0 says that v0 x = 0, so
vx = 1 At 2 − 1 Bt 3 = 1 (1.50 m/s3 )t 2 − 1 (0.120 m/s 4 )t 3
2
3
2
3
vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3
SET UP: t x − x0 + ∫ vx dt EXECUTE: 0 x = x0 + ∫ t
0 ( 1
2 1
At 2 − 1 Bt 3 ) dt = x0 + 1 At 3 − 12 Bt 4
3
6 At the origin at t = 0 says that x0 = 0, so
1
1
x = 1 At 3 − 12 Bt 4 = 1 (1.50 m/s3 )t 3 − 12 (0.120 m/s 4 )t 4
6
6 x = (0.25 m/s3 )t 3 − (0.010 m/s 4 )t 4
dx
dv
and ax (t ) = x .
dt
dt
dvx
dvx
(b) IDENTIFY and SET UP: At time t, when vx is a maximum,
= 0. (Since ax =
, the maximum velocity
dt
dt
is when ax = 0. For earlier times ax is positive so vx is still increasing. For later times ax is negative and vx is
decreasing.)
dv
EXECUTE: ax = x = 0 so At − Bt 2 = 0
dt
One root is t = 0, but at this time vx = 0 and not a maximum.
EVALUATE: We can check our results by using them to verify that vx (t ) = A 1.50 m/s3
=
= 12.5 s
B 0.120 m/s 4
At this time vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3 gives
The other root is t = vx = (0.75 m/s3 )(12.5 s) 2 − (0.040 m/s 4 )(12.5 s)3 = 117.2 m/s − 78.1 m/s = 39.1 m/s.
2.52. EVALUATE: For t < 12.5 s, ax > 0 and vx is increasing. For t > 12.5 s, ax < 0 and vx is decreasing.
IDENTIFY: a (t ) is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph.
SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2.52.
EXECUTE: (a) Slope = a = 0 for t ≥ 1.3 ms .
(b)
1
hmax = Area under v-t graph ≈ ATriangle + ARectangle ≈ (1.3 ms) (133 cm/s ) + (2.5 ms − 1.3 ms)(133 cm s) ≈ 0.25 cm
2
133cm s
2
(c) a = slope of v-t graph. a (0.5 ms) ≈ a (1.0 ms) ≈
= 1.0 × 105 cm s .
1.3ms
a (1.5 ms) = 0 because the slope is zero. Motion Along a Straight Line 2-21 1
(d) h = area under v-t graph. h(0.5 ms) ≈ ATriangle = (0.5 ms) ( 33 cm/s ) = 8.3 × 10−3 cm .
2
1
h(1.0 ms) ≈ ATriangle = (1.0 ms)(100 cm s) = 5.0 × 10−2 cm .
2
1
h(1.5 ms) ≈ ATriangle + ARectangle = (1.3 ms) (133 cm/s ) (0.2 ms)(1.33) = 0.11 cm
2
EVALUATE: The acceleration is constant until t = 1.3 ms , and then it is zero. g = 980 cm/s 2 . The acceleration
during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are
considering. Figure 2.52
2.53. (a) IDENTIFY and SET UP: The change in speed is the area under the ax versus t curve between vertical lines at
t = 2.5 s and t = 7.5 s.
EXECUTE: This area is 1 (4.00 cm/s 2 + 8.00 cm/s 2 )(7.5 s − 2.5 s) = 30.0 cm/s
2
This acceleration is positive so the change in velocity is positive.
(b) Slope of vx versus t is positive and increasing with t. The graph is sketched in Figure 2.53. Figure 2.53
EVALUATE:
2.54. 2.55. The calculation in part (a) is equivalent to Δvx = ( aav-x )Δt. Since ax is linear in t, aav-x = (a0 x + ax ) / 2. Thus aav-x = 1 (4.00 cm/s 2 + 8.00 cm/s 2 ) for the time interval t = 2.5 s to t = 7.5 s.
2
IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the
distance traveled divided by the average speed.
SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that first
10 mi
10 miles is
= 1.25 h .
8 mi/h
20 mi
= 5.0 h . The second 10 mi must
EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of
4 mi/h
10 mi
be covered in 5.0 h − 1.25 h = 3.75 h . This corresponds to an average speed of
= 2.7 mi/h .
3.75 h
20 mi
= 1.67 h . The second 10 mi must be
(b) An average speed of 12 mi/h for 20 mi gives a total time of
12 mi/h
10 mi
covered in 1.67 h − 1.25 h = 0.42 h . This corresponds to an average speed of
= 24 mi/h .
0.42 h
20 mi
= 1.25 h . But 1.25 h was already spent
(c) An average speed of 16 mi/h for 20 mi gives a total time of
16 mi/h
during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an
average speed of 16 mi/h for the 20-mile ride is not possible.
EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile
segment. The rider spends a different amount of time traveling at each of the two average speeds.
dx
dv
and ax = x .
IDENTIFY: vx (t ) =
dt
dt
dn
SET UP:
(t ) = nt n −1 , for n ≥ 1 .
dt
EXECUTE: (a) vx (t ) = (9.00 m/s3 )t 2 − (20.0 m/s 2 )t + 9.00 m/s . ax (t ) = (18.0 m/s 3 )t − 20.0 m/s 2 . The graphs are
sketched in Figure 2.55. 2-22 Chapter 2 (b) The particle is instantaneously at rest when vx (t ) = 0 . v0 x = 0 and the quadratic formula gives 1
(20.0 ± (20.0) 2 − 4(9.00)(9.00)) s = 1.11 s ± 0.48 s . t = 0.63 s and t = 1.59 s . These results agree with the
18.0
vx -t graphs in part (a). t= (c) For t = 0.63 s , ax = (18.0 m/s3 )(0.63 s) − 20.0 m/s 2 = −8.7 m/s 2 . For t = 1.59 s , ax = +8.6 m/s 2 . At t = 0.63 s the slope of the vx -t graph is negative and at t = 1.59 s it is positive, so the same answer is deduced from the
vx (t ) graph as from the expression for ax (t ) .
20.0 m/s 2
= 1.11 s .
18.0m/s3
(e) When the particle is at its greatest distance from the origin, vx = 0 and ax < 0 (so the particle is starting to
move back toward the origin). This is the case for t = 0.63 s , which agrees with the x-t graph in part (a) . At
t = 0.63 s , x = 2.45 m .
(f) The particle’s speed is changing at its greatest rate when ax has its maximum magnitude. The ax -t graph in part
(d) vx (t ) is instantaneously not changing when ax = 0 . This occurs at t = (a) shows this occurs at t = 0 and at t = 2.00 s . Since vx is always positive in this time interval, the particle is
speeding up at its greatest rate when ax is positive, and this is for t = 2.00 s .
The particle is slowing down at its greatest rate when ax is negative and this is for t = 0 .
EVALUATE: Since ax (t ) is linear in t, vx (t ) is a parabola and is symmetric around the point where vx (t ) has its minimum value ( t = 1.11 s ). For this reason, the answer to part (d) is midway between the two times in part (c). Figure 2.55
2.56. IDENTIFY: The average velocity is vav-x = Δx
. The average speed is the distance traveled divided by the
Δt elapsed time.
SET UP: Let + x be in the direction of the first leg of the race. For the round trip, Δx ≥ 0 and the total distance
traveled is 50.0 m. For each leg of the race both the magnitude of the displacement and the distance traveled
are 25.0 m.
Δx 25.0 m
EXECUTE: (a) vav-x =
=
= 1.25 m/s . This is the same as the average speed for this leg of the race.
20.0 s
Δt
(b) vav-x = Δx 25.0 m
=
= 1.67 m/s . This is the same as the average speed for this leg of the race.
Δt
15.0 s (c) Δx = 0 so vav-x = 0 . 50.0 m
= 1.43 m/s .
35.0 s
EVALUATE: Note that the average speed for the round trip is not equal to the arithmetic average of the average
speeds for each leg.
IDENTIFY: Use information about displacement and time to calculate average speed and average velocity. Take
the origin to be at Seward and the positive direction to be west.
distance traveled
(a) SET UP: average speed =
time
EXECUTE: The distance traveled (different from the net displacement ( x − x0 ) ) is 76 km + 34 km = 110 km.
(d) The average speed is 2.57. Find the total elapsed time by using vav-x =
Seward to Auora: t = Δx x − x0
=
to find t for each leg of the journey.
Δt
t x − x0
76 km
=
= 0.8636 h
vav-x
88 km/h Motion Along a Straight Line 2-23 x − x0
−34 km
=
= 0.4722 h
vav-x
−72 km/h
Total t = 0.8636 h + 0.4722 h = 1.336 h.
110 km
= 82 km/h.
Then average speed =
1.336 h
Δx
(b) SET UP: vav-x =
, where Δx is the displacement, not the total distance traveled.
Δt
Auora to York: t = 42 km
= 31 km/h.
l.336 h
EVALUATE: The motion is not uniformly in the same direction so the displacement is less than the distance
traveled and the magnitude of the average velocity is less than the average speed.
IDENTIFY: The vehicles are assumed to move at constant speed. The speed (mi/h) divided by the frequency with
which vehicles pass a given point (vehicles/h) is the total space per vehicle (the length of the vehicle plus space to
the next vehicle).
SET UP: 96 km/h = 96 × 103 m/h
96 × 103 m/h
EXECUTE: (a) The total space per vehicle is
= 40 m/vehicle . Since the average length of a
2400 vehicles/h
vehicle is 4.6 m, the average space between vehicles is 40 m − 4.6 m = 35 m .
96 × 103 m/h
(b) The frequency of vehicles (vehicles/h) is
= 7000 vehicles/h .
(4.6 + 9.2) m/vehicle
EVALUATE: The traffic flow rate per lane would nearly triple. Note that the traffic flow rate is directly
proportional to the traffic speed.
Δv
v −v
(a) IDENTIFY: Calculate the average acceleration using aav-x = x = x 0 x Use the information about the time
Δt
t
and total distance to find his maximum speed.
SET UP: v0 x = 0 since the runner starts from rest.
For the whole trip he ends up 76 km − 34 km = 42 km west of his starting point. vav-x = 2.58. 2.59. t = 4.0 s, but we need to calculate vx , the speed of the runner at the end of the acceleration period.
EXECUTE: For the last 9.1 s − 4.0 s = 5.1 s the acceleration is zero and the runner travels a distance of
d1 = (5.1 s)vx (obtained using x − x0 = v0 xt + 1 axt 2 )
2
During the acceleration phase of 4.0 s, where the velocity goes from 0 to vx , the runner travels a distance
⎛v +v ⎞ v
d 2 = ⎜ 0 x x ⎟ t = x (4.0 s) = (2.0 s)vx
2
⎝2⎠
The total distance traveled is 100 m, so d1 + d 2 = 100 m. This gives (5.1 s)vx + (2.0 s)vx = 100 m.
vx = 100 m
= 14.08 m/s.
7.1 s vx − v0 x 14.08 s − 0
=
= 3.5 m/s 2 .
t
4.0 s
(b) For this time interval the velocity is constant, so aav − x = 0. Now we can calculate aav-x : aav-x =
EVALUATE: Now that we have vx we can calculate d1 = (5.1 s)(14.08 m/s) = 71.9 m and d 2 = (2.0 s)(14.08 m/s) = 28.2 m. So, d1 + d 2 = 100 m, which checks.
vx − v0 x
, where now the time interval is the full 9.1 s of the race.
t
We have calculated the final speed to be 14.08 m/s, so
(c) IDENTIFY and SET UP: aav-x = 14.08 m/s
= 1.5 m/s 2 .
9.1 s
EVALUATE: The acceleration is zero for the last 5.1 s, so it makes sense for the answer in part (c) to be less than
half the answer in part (a).
(d) The runner spends different times moving with the average accelerations of parts (a) and (b).
IDENTIFY: Apply the constant acceleration equations to the motion of the sled. The average velocity for a time
Δx
interval Δt is vav-x =
.
Δt
aav-x = 2.60. 2-24 Chapter 2 SET UP: Let + x be parallel to the incline and directed down the incline. The problem doesn’t state how much
time it takes the sled to go from the top to 14.4 m from the top.
25.6 m − 14.4 m
EXECUTE: (a) 14.4 m to 25.6 m: vav-x =
= 5.60 m/s . 25.6 to 40.0 m:
2.00 s
40.0 m − 25.6 m
57.6 m − 40.0 m
vav-x =
= 7.20 m/s . 40.0 m to 57.6 m: vav-x =
= 8.80 m/s .
2.00 s
2.00 s
(b) For each segment we know x − x0 and t but we don’t know v0 x or vx . Let x1 = 14.4 m and x2 = 25.6 m . For x −x
⎛v +v ⎞ x −x
this interval ⎜ 1 2 ⎟ = 2 1 and at = v2 − v1 . Solving for v2 gives v2 = 1 at + 2 1 . Let x2 = 25.6 m and
2
t
t
⎝2⎠
⎛v +v ⎞ x −x
x3 = 40.0 m . For this second interval, ⎜ 2 3 ⎟ = 3 2 and at = v3 − v2 . Solving for v2 gives
t
⎝2⎠
x3 − x2
v2 = − 1 at +
. Setting these two expressions for v2 equal to each other and solving for a gives
2
t
1
1
a = 2 [( x3 − x2 ) − ( x2 − x1 )] =
[(40.0 m − 25.6 m) − (25.6 m − 14.4 m)] = 0.80 m/s 2 .
t
(2.00 s) 2
Note that this expression for a says a = vav-23 − vav-12
, where vav-12 and vav-23 are the average speeds for successive
t 2.00 s intervals.
(c) For the motion from x = 14.4 m to x = 25.6 m , x − x0 = 11.2 m , ax = 0.80 m/s 2 and t = 2.00 s .
x − x0 1
11.2 m 1
− 2 axt =
− (0.80 m/s 2 )(2.00 s) = 4.80 m/s .
t
2.00 s 2
(d) For the motion from x = 0 to x = 14.4 m , x − x0 = 14.4 m , v0 x = 0 , and vx = 4.8 m/s .
x − x0 = v0 xt + 1 axt 2 gives v0 x =
2 2( x − x0 ) 2(14.4 m)
⎛v +v ⎞
x − x0 = ⎜ 0 x x ⎟ t gives t =
=
= 6.0 s .
v0 x + vx
4.8 m/s
⎝2⎠
(e) For this 1.00 s time interval, t = 1.00 s , v0 x = 4.8 m/s , ax = 0.80 m/s 2 . x − x0 = v0 xt + 1 axt 2 = (4.8 m/s)(1.00 s) + 1 (0.80 m/s 2 )(1.00 s) 2 = 5.2 m .
2
2 2.61. EVALUATE: With x = 0 at the top of the hill, x (t ) = v0 xt + 1 axt 2 = (0.40 m/s 2 )t 2 . We can verify that
2
t = 6.0 s gives x = 14.4 m , t = 8.0 s gives 25.6 m, t = 10.0 s gives 40.0 m, and t = 12.0 s gives 57.6 m.
IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion consists of
two constant acceleration segments and the constant acceleration equations can be used for each segment. Since
vx is always positive the motion is always in the + x direction and the total distance moved equals the magnitude of the displacement. The acceleration ax is the slope of the vx versus t graph.
SET UP: For the t = 0 to t = 10.0 s segment, v0 x = 4.00 m/s and vx = 12.0 m/s . For the t = 10.0 s to 12.0 s segment, v0 x = 12.0 m/s and vx = 0 .
EXECUTE: ⎛ v + v ⎞ ⎛ 4.00 m/s + 12.0 m/s ⎞
(a) For t = 0 to t = 10.0 s , x − x0 = ⎜ 0 x x ⎟ t = ⎜
⎟ (10.0 s) = 80.0 m . For
2
⎝2⎠⎝
⎠ ⎛ 12.0 m/s + 0 ⎞
t = 10.0 s to t = 12.0 s , x − x0 = ⎜
⎟ (2.00 s) = 12.0 m . The total distance traveled is 92.0 m.
2
⎝
⎠
(b) x − x0 = 80.0 m + 12.0 m = 92.0 m
(c) For t = 0 to 10.0 s, ax = ax = 12.0 m/s − 4.00 m/s
= 0.800 m/s 2 . For t = 10.0 s to 10.2 s,
10.0 s 0 − 12.0 m/s
= −6.00 m/s 2 . The graph of ax versus t is given in Figure 2.61.
2.00 s Motion Along a Straight Line 2-25 EVALUATE: When vx and ax are both positive, the speed increases. When vx is positive and ax is negative, the
speed decreases. Figure 2.61
2.62. IDENTIFY: Since light travels at constant speed, d = ct
SET UP: The distance from the earth to the sun is 1.50 × 1011 m . The distance from the earth to the moon is
3.84 × 108 m . c = 186,000 mi/s .
EXECUTE: ⎛ 365 1 d ⎞ ⎛ 24 h ⎞⎛ 3600 s ⎞
15
4
(a) d = ct = (3.0 × 108 m/s)(1 y) ⎜
⎟⎜
⎟⎜
⎟ = 9.5 × 10 m
1 y ⎠ ⎝ 1 d ⎠⎝ 1 h ⎠
⎝ (b) d = ct = (3.0 × 108 m/s)(10−9 s) = 0.30 m
(c) t = d
1.5 × 1011 m
=
= 500 s = 8.33 min
c 3.0 × 108 m s (d) t = d 2(3.84 × 108 m)
=
= 2.6 s
3.0 × 108 m s
c d
3 × 109 mi
=
= 16,100 s = 4.5 h
c 186,000 mi s
EVALUATE: The speed of light is very large but it still takes light a measurable length of time to travel a large
distance.
IDENTIFY: Speed is distance d divided by time t. The distance around a circular path is d = 2π R , where R is the
radius of the circular path.
SET UP: The radius of the earth is RE = 6.38 × 106 m . The earth rotates once in 1 day = 86,400 s . The radius of
(e) t = 2.63. the earth’s orbit around the sun is 1.50 × 1011 m and the earth completes this orbit in 1 year = 3.156 × 107 s . The
speed of light in vacuum is c = 3.00 × 108 m/s .
d 2π RE 2π (6.38 × 106 m)
EXECUTE: (a) v = =
=
= 464 m/s .
t
t
86,400 s
2π R 2π (1.50 × 1011 m)
=
= 2.99 × 104 m/s .
t
3.156 × 107 s
d 2π RE 2π (6.38 × 106 m)
(c) The time for light to go around once is t = =
=
= 0.1336 s . In 1.00 s light would go
c
c
3.00 × 108 m/s
1.00 s
= 7.49 times .
around the earth
0.1336 s
EVALUATE: All these speeds are large compared to speeds of objects in our everyday experience.
IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion consists of
two constant acceleration segments and the constant acceleration equations can be used for each segment. For
t = 0 to 5.0 s, vx is positive and the ball moves in the + x direction. For t = 5.0 s to 20.0 s, vx is negative and the
(b) v = 2.64. ball moves in the − x direction. The acceleration ax is the slope of the vx versus t graph.
SET UP: For the t = 0 to t = 5.0 s segment, v0 x = 0 and vx = 30.0 m/s . For the t = 5.0 s to t = 20.0 s segment, v0 x = −20.0 m/s and vx = 0 . 2-26 Chapter 2 EXECUTE: ⎛ v + v ⎞ ⎛ 0 + 30.0 m/s ⎞
(a) For t = 0 to 5.0 s, x − x0 = ⎜ 0 x x ⎟ t = ⎜
⎟ (5.0 m/s) = 75.0 m . The ball travels a
2
⎝2⎠⎝
⎠ ⎛ −20.0 m/s + 0 ⎞
distance of 75.0 m. For t = 5.0 s to 20.0 s, x − x0 = ⎜
⎟ (15.0 m/s) = −150.0 m . The total distance
2
⎝
⎠
traveled is 75.0 m + 150.0 m = 225.0 m .
(b) The total displacement is x − x0 = 75.0 m +( − 150.0 m) = −75.0 m . The ball ends up 75.0 m in the negative xdirection from where it started.
30.0 m/s − 0
0 − ( −20.0 m/s)
(c) For t = 0 to 5.0 s, ax =
= 6.00 m/s 2 . For t = 5.0 s to 20.0 s, ax =
= +1.33 m/s 2 .
5.0 s
15.0 s
The graph of ax versus t is given in Figure 2.64.
(d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity
doesn't change instantaneously. So, no, the actual graph of vx (t ) is not really vertical at 5.00 s.
EVALUATE: For t = 0 to 5.0 s, both vx and ax are positive and the speed increases. For t = 5.0 s to 20.0 s, vx is negative and ax is positive and the speed decreases. Since the direction of motion is not the same throughout, the
displacement is not equal to the distance traveled. Figure 2.64
2.65. IDENTIFY and SET UP: Apply constant acceleration equations.
Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find x − x0 for
the first 5.0 s.
EXECUTE: For the first 5.0 s of the motion, v0 x = 0, t = 5.0 s. vx = v0 x + axt gives vx = ax (5.0 s).
This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s:
v0 x = ax (5.0 s), t = 5.0 s, x − x0 = 150 m.
x − x0 = v0 xt + 1 axt 2 gives 150 m = (25 s 2 )ax + (12.5 s 2 )ax and ax = 4.0 m/s 2
2
Use this ax and consider the first 5.0 s of the motion:
x − x0 = v0 xt + 1 axt 2 = 0 + 1 (4.0 m/s 2 )(5.0 s) 2 = 50.0 m.
2
2
EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact,
x − x0 is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go
four times as far. In 10.0 s we calculated it went 50 m + 150 m = 200 m, which is four times 50 m.
2.66. IDENTIFY: Apply x − x0 = v0 xt + 1 axt 2 to the motion of each train. A collision means the front of the passenger
2
train is at the same location as the caboose of the freight train at some common time.
SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train x0 = 0 and for the caboose of the freight train x0 = 200 m . For the freight train vF = 15.0 m/s and aF = 0 . For the passenger train
vP = 25.0 m/s and aP = −0.100 m/s 2 .
EXECUTE: (a) x − x0 = v0 xt + 1 axt 2 for each object gives xP = vPt + 1 aPt 2 and xF = 200 m + vFt . Setting
2
2 xP = xF gives vPt + 1 aPt 2 = 200 m + vFt . (0.0500 m/s 2 )t 2 − (10.0 m/s)t + 200 m = 0 . The
2 ( ) 1
+10.0 ± (10.0) 2 − 4(0.0500)(200) s = (100 ± 77.5) s . The collision occurs at
0.100
t = 100 s − 77.5 s = 22.5 s . The equations that specify a collision have a physical solution (real, positive t), so a
collision does occur. quadratic formula gives t = Motion Along a Straight Line 2-27 (b) xP = (25.0 m/s)(22.5 s) + 1 (−0.100 m/s 2 )(22.5 s) 2 = 537 m . The passenger train moves 537 m before the
2
collision. The freight train moves (15.0 m/s)(22.5 s) = 337 m .
(c) The graphs of xF and xP versus t are sketched in Figure 2.66.
EVALUATE: The second root for the equation for t, t = 177.5 s is the time the trains would meet again if they
were on parallel tracks and continued their motion after the first meeting. Figure 2.66
2.67. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, you and the cockroach. You
catch up with the roach when both objects are at the same place at the same time. Let T be the time when you catch
up with the cockroach.
SET UP: Take x = 0 to be at the t = 0 location of the roach and positive x to be in the direction of motion of the
two objects.
roach:
v0 x = 1.50 m/s, ax = 0, x0 = 0, x = 1.20 m, t = T
you:
v0 x = 0.80 m/s, x0 = −0.90 m, x = 1.20 m, t = T , ax = ? Apply x − x0 = v0 xt + 1 axt 2 to both objects:
2
EXECUTE: roach: 1.20 m = (1.50 m/s)T , so T = 0.800 s.
you: 1.20 m − (−0.90 m) = (0.80 m/s)T + 1 axT 2
2
2.10 m = (0.80 m/s)(0.800 s) + 1 ax (0.800 s) 2
2
2.10 m = 0.64 m + (0.320 s 2 )ax
ax = 4.6 m/s 2 .
⎛v +v ⎞
Your final velocity is vx = v0 x + axt = 4.48 m/s. Then x − x0 = ⎜ 0 x x ⎟ t = 2.10 m, which checks.
⎝2⎠
You have to accelerate to a speed greater than that of the roach so you will travel the extra 0.90 m you are initially
behind.
IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together.
SET UP: Each car has moved 100 m when they hit.
100 m
= 10 s . During this time the grasshopper travels a distance of
EXECUTE: The time until the cars hit is
10 m/s
(15 m/s)(10 s) = 150 m .
EVALUATE: The grasshopper ends up 100 m from where it started, so the magnitude of his final displacement is
100 m. This is less than the total distance he travels since he spends part of the time moving in the opposite
direction.
IDENTIFY: Apply constant acceleration equations to each object.
Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.69a
Let d be the distance that the auto initially is behind the truck, so x0 (auto) = −d and x0 (truck) = 0. Let T be the
EVALUATE: 2.68. 2.69. time it takes the auto to catch the truck. Thus at time T the truck has undergone a displacement x − x0 = 40.0 m, so
is at x = x0 + 40.0 m = 40.0 m. The auto has caught the truck so at time T is also at x = 40.0 m. Figure 2.69a 2-28 Chapter 2 (a) SET UP: Use the motion of the truck to calculate T:
x − x0 = 40.0 m, v0 x = 0 (starts from rest), ax = 2.10 m/s 2 , t = T x − x0 = v0 xt + 1 axt 2
2
Since v0 x = 0, this gives t = 2( x − x0 )
ax 2(40.0 m)
= 6.17 s
2.10 m/s 2
(b) SET UP: Use the motion of the auto to calculate d:
x − x0 = 40.0 m + d , v0 x = 0, ax = 3.40 m/s 2 , t = 6.17 s
EXECUTE: T= x − x0 = v0 xt + 1 axt 2
2
EXECUTE: d + 40.0 m = 1 (3.40 m/s 2 )(6.17 s) 2
2
d = 64.8 m − 40.0 m = 24.8 m
(c) auto: vx = v0 x + axt = 0 + (3.40 m/s 2 )(6.17 s) = 21.0 m/s truck: vx = v0 x + axt = 0 + (2.10 m/s 2 )(6.17 s) = 13.0 m/s
(d) The graph is sketched in Figure 2.69b. Figure 2.69b 2.70. EVALUATE: In part (c) we found that the auto was traveling faster than the truck when they come abreast. The
graph in part (d) agrees with this: at the intersection of the two curves the slope of the x-t curve for the auto is
greater than that of the truck. The auto must have an average velocity greater than that of the truck since it must
travel farther in the same time interval.
IDENTIFY: Apply the constant acceleration equations to the motion of each car. The collision occurs when the
cars are at the same place at the same time.
SET UP: Let + x be to the right. Let x = 0 at the initial location of car 1, so x01 = 0 and x02 = D . The cars collide when x1 = x2 . v0 x1 = 0 , ax1 = ax , v0 x 2 = −v0 and ax 2 = 0 .
EXECUTE:
1
2 (a) x − x0 = v0 xt + 1 axt 2 gives x1 = 1 axt 2 and x2 = D − v0t . x1 = x2 gives
2
2 axt 2 + v0t − D = 0 . The quadratic formula gives t = so t = ( ) ( ) 1
2 axt 2 = D − v0t . 1
2
−v0 ± v0 + 2ax D . Only the positive root is physical,
ax 1
2
−v0 + v0 + 2ax D .
ax 2
(b) v1 = axt = v0 + 2ax D − v0 (c) The x-t and vx -t graphs for the two cars are sketched in Figure 2.70. Motion Along a Straight Line EVALUATE: 2-29 In the limit that ax = 0 , D − v0t = 0 and t = D / v0 , the time it takes car 2 to travel distance D. In the limit that v0 = 0 , t = 2D
, the time it takes car 1 to travel distance D.
ax Figure 2.70 The average speed is the distance traveled divided by the time. The average velocity is vav-x = Δx
.
Δt 2.71. IDENTIFY: 2.72. SET UP: The distance the ball travels is half the circumference of a circle of diameter 50.0 cm so is
1
π d = 1 π (50.0 cm) = 78.5 cm . Let + x be horizontally from the starting point toward the ending point, so
2
2
Δx equals the diameter of the bowl.
1
π d 78.5 cm
EXECUTE: (a) The average speed is 2
=
= 7.85 cm/s .
t
10.0 s
Δx 50.0 cm
=
= 5.00 cm/s .
(b) The average velocity is vav-x =
Δt
10.0 s
EVALUATE: The average speed is greater than the magnitude of the average velocity, since the distance traveled
is greater than the magnitude of the displacement.
IDENTIFY: ax is the slope of the vx versus t graph. x is the area under the vx versus t graph.
SET UP: The slope of vx is positive and decreasing in magnitude. As vx increases, the displacement in a given
amount of time increases.
EXECUTE: The ax -t and x-t graphs are sketched in Figure 2.72.
EVALUATE: vx is the slope of the x versus t graph. The x (t ) graph we sketch has zero slope at t = 0 , the slope is
always positive, and the slope initially increases and then approaches a constant. This behavior agrees with the
vx (t ) that is given in the graph in the problem. Figure 2.72
2.73. IDENTIFY: Apply constant acceleration equations to each vehicle.
SET UP: (a) It is very convenient to work in coordinates attached to the truck.
Note that these coordinates move at constant velocity relative to the earth. In these coordinates the truck is at rest,
and the initial velocity of the car is v0 x = 0. Also, the car’s acceleration in these coordinates is the same as in
coordinates fixed to the earth.
EXECUTE: First, let’s calculate how far the car must travel relative to the truck: The situation is sketched in
Figure 2.73. Figure 2.73 2-30 Chapter 2 The car goes from x0 = −24.0 m to x = 51.5 m. So x − x0 = 75.5 m for the car.
Calculate the time it takes the car to travel this distance:
ax = 0.600 m/s 2 , v0 x = 0, x − x0 = 75.5 m, t = ?
x − x0 = v0 xt + 1 axt 2
2
t= 2( x − x0 )
2(75.5 m)
=
= 15.86 s
ax
0.600 m/s 2 It takes the car 15.9 s to pass the truck.
(b) Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. In these
coordinates v0 x = 20.0 m/s for the car. Take the origin to be at the initial position of the car.
v0 x = 20.0 m/s, ax = 0.600 m/s 2 , t = 15.86 s, x − x0 = ?
x − x0 = v0 xt + 1 axt 2 = (20.0 m/s)(15.86 s) + 1 (0.600 m/s 2 )(15.86 s) 2
2
2
x − x0 = 317.2 m + 75.5 m = 393 m.
(c) In coordinates fixed to the earth:
vx = v0 x + axt = 20.0 m/s + (0.600 m/s 2 )(15.86 s) = 29.5 m/s 2.74. EVALUATE: In 15.9 s the truck travels x − x0 = (20.0 m/s)(15.86 s) = 317.2 m. The car travels
392.7 m − 317.2 m = 75 m farther than the truck, which checks with part (a). In coordinates attached to the truck,
⎛v +v ⎞
for the car v0 x = 0, vx = 9.5 m/s and in 15.86 s the car travels x − x0 = ⎜ 0 x x ⎟ t = 75 m, which checks with
⎝2⎠
part (a).
IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use
t
dv
ax (t ) = x and x = x0 + ∫ vx (t )dt .
0
dt
1 n +1
SET UP: ∫ t n dt =
t
for n ≥ 0 .
n +1
EXECUTE: t (a) x (t ) = x0 + ∫ [α − β t 2 ]dt = x0 + α t − 1 β t 3 . x = 0 at t = 0 gives x0 = 0 and
3
0 dvx
= −2 β t = −(4.00 m/s3 )t .
dt
(b) The maximum positive x is when vx = 0 and ax < 0 . vx = 0 gives α − β t 2 = 0 and
x (t ) = α t − 1 β t 3 = (4.00 m/s)t − (0.667 m/s3 )t 3 . ax (t ) =
3 t= 2.75. α
4.00 m/s
=
= 1.41 s . At this t, ax is negative. For t = 1.41 s ,
β
2.00 m/s3 x = (4.00 m/s)(1.41 s) − (0.667 m/s 3 )(1.41 s)3 = 3.77 m .
EVALUATE: After t = 1.41 s the object starts to move in the − x direction and goes to x = −∞ as t → ∞ .
a (t ) = α + β t , with α = −2.00 m/s 2 and β = 3.00 m/s3
(a) IDENTIFY and SET UP:
EXECUTE: Integrage ax (t ) to find vx (t ) and then integrate vx (t ) to find x(t ). t t 0 0 vx = v0 x + ∫ ax dt = v0 x + ∫ (α + β ) dt = v0 x + α t + 1 β t 2
2 t t 0 0 x = x0 + ∫ vx dt = x0 + ∫ (v0 x + α t + 1 β t 2 ) dt = x0 + v0 xt + 1 α t 2 + 1 β t 3
2
2
6 At t = 0, x = x0 .
To have x = x0 at t1 = 4.00 s requires that v0 xt1 + 1 α t12 + 1 β t13 = 0.
2
6
Thus v0 x = − 1 β t12 − 1 α t1 = − 1 (3.00 m/s3 )(4.00 s) 2 − 1 (−2.00 m/s 2 )(4.00 s) = −4.00 m/s.
6
2
6
2
(b) With v0 x as calculated in part (a) and t = 4.00 s, v0 = v0 x + α t + 1 β t 2 = −4.00 s + (−2.00 m/s 2 )(4.00 s) + 1 (3.00 m/s3 )(4.00 s) 2 = +12.0 m/s.
2
2
EVALUATE: ax = 0 at t = 0.67 s. For t > 0.67 s, ax > 0. At t = 0, the particle is moving in the − x -direction
and is speeding up. After t = 0.67 s, when the acceleration is positive, the object slows down and then starts to
move in the + x -direction with increasing speed. Motion Along a Straight Line 2.76. 2-31 IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head.
SET UP: Let + y be downward. The egg has v0 y = 0 and a y = 9.80 m/s 2 . At the height of the professor’s head, the egg has y − y0 = 44.2 m .
EXECUTE: 2.77. y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2(44.2 m)
=
= 3.00 s . The professor walks a distance
ay
9.80 m/s 2 x − x0 = v0 xt = (1.20 m/s)(3.00 s) = 3.60 m . Release the egg when your professor is 3.60 m from the point directly
below you.
EVALUATE: Just before the egg lands its speed is (9.80 m/s 2 )(3.00s) = 29.4 m/s . It is traveling much faster than
the professor.
IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and
acceleration due to gravity and between time in the air and acceleration due to gravity.
SET UP: Let + y be upward. At the maximum height, v y = 0 . When the rock returns to the surface, y − y0 = 0 .
EXECUTE: 2
2
2
(a) v y = v0 y + 2a y ( y − y0 ) gives a y H = − 1 v0 y , which is constant, so aE H E = aM H M .
2 ⎛a ⎞
⎛ 9.80 m/s 2 ⎞
HM = HE ⎜ E ⎟ = H ⎜
= 2.64 H .
2⎟
⎝ 3.71 m/s ⎠
⎝ aM ⎠
(b) y − y0 = v0 yt + 1 a yt 2 with y − y0 = 0 gives a y t = −2v0 y , which is constant, so aETE = aMTM .
2 2.78. ⎡a ⎤
TM = TE ⎢ E ⎥ = 2.64T .
⎣ aM ⎦
EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in
the air for a longer time.
IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown
ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected.
SET UP: For the thrown ball, let + y be upward. a y = −9.80 m/s 2 . y − y0 = 0 when the ball returns to its original position.
5.50 m
= 2.20 s to reach the table and an equal time to return. For the ball,
2.50 m/s
y − y0 = 0 , t = 4.40 s and a y = −9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives
2 EXECUTE: (a) It takes her v0 y = − 1 a y t = − 1 (−9.80 m/s 2 )(4.40 s) = 21.6 m/s .
2
2
(b) Find y − y0 when t = 2.20 s . y − y0 = v0 yt + 1 a yt 2 = (21.6 m/s)(2.20 s) + 1 (−9.80 m/s 2 )(2.20 s) 2 = 23.8 m
2
2 2.79. EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its
maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum
height requires that the act either be done outdoors, or in a building with a very high ceiling.
(a) IDENTIFY: Use constant acceleration equations, with a y = g , downward, to calculate the speed of the diver when she reaches the water.
SET UP: Take the origin of coordinates to be at the platform, and take the + y -direction to be downward.
y − y0 = +21.3 m, a y = +9.80 m/s 2 , v0 y = 0 (since diver just steps off), v y = ?
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: v y = + 2a y ( y − y0 ) = + 2(9.80 m/s 2 )(31.3 m) = +20.4 m/s. We know that v y is positive because the diver is traveling downward when she reaches the water.
The announcer has exaggerated the speed of the diver.
EVALUATE: We could also use y − y0 = v0 yt + 1 a yt 2 to find t = 2.085 s. The diver gains 9.80 m/s of speed each
2
second, so has v y = (9.80 m/s 2 )(2.085 s) = 20.4 m/s when she reaches the water, which checks.
(b) IDENTIFY: Calculate the initial upward velocity needed to give the diver a speed of 25.0 m/s when she
reaches the water. Use the same coordinates as in part (a).
SET UP: v0 y = ?, v y = +25.0 m/s, a y = +9.80 m/s 2 , y − y0 = +21.3 m
2
2
v y = v0 y + 2a y ( y − y0 ) 2-32 Chapter 2
2
v0 y = − v y − 2a y ( y − y0 ) = − (25.0 m/s) 2 − 2(9.80 m/s 2 )(21.3 m) = −14.4 m/s EXECUTE: (v0 y is negative since the direction of the initial velocity is upward.)
EVALUATE: One way to decide if this speed is reasonable is to calculate the maximum height above the platform
it would produce:
v0 y = −14.4 m/s, v y = 0 (at maximum height), a y = +9.80 m/s 2 , y − y0 = ?
2
2
v y = v0 y + 2a y ( y − y0 ) y − y0 =
2.80. 2
2
v y − v0 y 2a y = 0 − (−14.4 s) 2
= −10.6 m
2(+9.80 m/s) This is not physically attainable; a vertical leap of 10.6 m upward is not possible.
IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the
window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the
windowsill to the top of the window.
SET UP: Let + y be downward. Throughout the motion a y = +9.80 m/s 2 .
EXECUTE: Motion past the window: y − y0 = 1.90 m , t = 0.420 s , a y = +9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives
2 y − y0 1
1.90 m 1
− 2 a yt =
− (9.80 m/s 2 )(0.420 s) = 2.466 m/s . This is the velocity of the flowerpot when it is
0.420 s 2
t
at the top of the window.
Motion from the windowsill to the top of the window: v0 y = 0 , v y = 2.466 m/s , a y = +9.80 m/s 2 .
v0 y = 2
2
v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y = 2a y (2.466 m/s) 2 − 0
= 0.310 m . The top of the window is 0.310 m
2(9.80 m/s 2 ) below the windowsill.
EVALUATE: It takes the flowerpot t = v y − v0 y
ay = 2.466 m/s
= 0.252 s to fall from the sill to the top of the
9.80 m/s 2 window. Our result says that from the windowsill the pot falls 0.310 m + 1.90 m = 2.21 m in
0.252 s + 0.420 s = 0.672 s . y − y0 = v0 yt + 1 a yt 2 = 1 (9.80 m/s 2 )(0.672 s) 2 = 2.21 m , which checks.
2
2
2.81. IDENTIFY: For parts (a) and (b) apply the constant acceleration equations to the motion of the bullet. In part (c)
neglect air resistance, so the bullet is free-fall. Use the constant acceleration equations to establish a relation
between initial speed v0 and maximum height H.
SET UP: For parts (a) and (b) let + x be in the direction of motion of the bullet. For part (c) let + y be upward, so a y = − g . At the maximum height, v y = 0 .
EXECUTE: ax = 2
2
(a) x − x0 = 0.700 m , v0 x = 0 , vx = 965 m/s . vx = v0 x + 2ax ( x − x0 ) gives 2
2
vx − v0 x
(965 m/s) 2 − 0
a
=
= 6.65 × 105 m/s 2 . x = 6.79 × 104 , so ax = (6.79 × 104 ) g .
g
2( x − x0 )
2(0.700 m) 2( x − x0 ) 2(0.700 m)
⎛v +v ⎞
=
= 1.45 ms .
(b) x − x0 = ⎜ 0 x x ⎟ t gives t =
v0 x + vx
0 + 965 m/s
⎝2⎠
2
2
(c) v y = v0 y + 2a y ( y − y0 ) and v y = 0 gives 2
v0 y y − y0 = −2a y , which is constant. 2
2
v01 v02
=
.
H1 H 2 2 ⎛ v2 ⎞
⎛ 1v ⎞
H 2 = H1 ⎜ 02 ⎟ = H ⎜ 2 01 ⎟ = H / 4 .
2
⎝ v01 ⎠
⎝ v01 ⎠
v2 − v2
−(965 m/s) 2
EVALUATE: H = y 0 y =
= 47.5 km . Rifle bullets fired vertically don't actually reach such a
2a y
2( −9.80 m/s 2 )
2.82. large height; it is not an accurate approximation to ignore air resistance.
IDENTIFY: Assume the firing of the second stage lasts a very short time, so the rocket is in free-fall after 25.0 s.
The motion consists of two constant acceleration segments.
SET UP: Let + y be upward. After t = 25.0 s , a y = −9.80 m/s 2 .
EXECUTE: (a) Find the height of the rocket at t = 25.0 s : v0 y = 0 , a y = +3.50 m/s 2 , t = 25.0 s . y − y0 = v0 yt + 1 a yt 2 = 1 (3.50 m/s)(25.0 s) 2 = 1.0938 × 103 m . Find the displacement of the rocket from firing of the
2
2 Motion Along a Straight Line 2-33 second stage until the maximum height is reached: v0 y = 132.5 m/s , v y = 0 (at maximum height), a y = −9.80 m/s 2 .
2
2
v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = 0 − (132.5 m/s) 2
= 896 m . The total height is
2(−9.80 m/s 2 ) 1094 m + 896 m = 1990 m .
(b) v0 y = +132.5 m/s , a y = −9.80 m/s 2 , y − y0 = −1094 m . y − y0 = v0 yt + 1 a yt 2 gives
2
−1093.8 m = (132.5 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 33.7 s as the positive root. The rocket
returns to the launch pad 33.7 s after the second stage fires.
(c) v y = v0 y + a yt = +132.5 m/s + ( −9.80 m/s 2 )(33.7 s) = −198 m/s . The rocket has speed 198 m/s as it reaches the 2.83. launch pad.
EVALUATE: The speed when the rocket returns to the launch pad is greater than 132.5 m/s. When the rocket
returns to the height where the second stage fired, its velocity is 132.5 m/s downward and it continues to speed up
during the rest of the descent.
Take positive y to be upward.
(a) IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand.
SET UP: v0 y = 0, v y = ?, a y = 45.0 m/s 2 , y − y0 = 0.640 m
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: v y = 2a y ( y − y0 ) = 2(45.0 m/s 2 )(0.640 m) = 7.59 m/s (b) IDENTIFY: Consider the motion of the shot from the point where he releases it to its maximum height, where
v = 0. Take y = 0 at the ground. y0 = 2.20 m, y = ?, a y = −9.80 m/s 2 (free fall), v0 y = 7.59 m/s SET UP: (from part (a), v y = 0 at maximum height)
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: y − y0 = 2
2
v y − v0 y 2a y = 0 − (7.59 m/s) 2
= 2.94 m
2(−9.80 m/s 2 ) y = 2.20 m + 2.94 m = 5.14 m.
(c) IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height
of his head. Take y = 0 at the ground.
y0 = 2.20 m, y = 1.83 m, a y = −9.80 m/s 2 , v0 y = +7.59 m/s, t = ? y − y0 = v0 yt + 1 a yt 2
2 SET UP:
EXECUTE: 1.83 m − 2.20 m = (7.59 m/s)t + 1 ( −9.80 m/s 2 )t 2
2 = (7.59 m/s)t − (4.90 m/s 2 )t 2 4.90t 2 − 7.59t − 0.37 = 0, with t in seconds.
Use the quadratic formula to solve for t:
1
7.59 ± (7.59) 2 − 4(4.90)(−0.37) = 0.774 ± 0.822
t=
9.80
t must be positive, so t = 0.774 s + 0.822 s = 1.60 s
EVALUATE: Calculate the time to the maximum height: v y = v0 y + a yt , so ( ) t = (v y − v0 y ) / a y = −(7.59 m/s)/( − 9.80 m/s 2 ) = 0.77 s. It also takes 0.77 s to return to 2.2 m above the ground, for a 2.84. total time of 1.54 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his
head a little later than 1.54 s after being thrown; the answer of 1.60 s in part (c) makes sense.
IDENTIFY: The teacher is in free-fall and falls with constant acceleration 9.80 m/s 2 , downward. The sound from
her shout travels at constant speed. The sound travels from the top of the cliff, reflects from the ground and then
travels upward to her present location. If the height of the cliff is h and she falls a distance y in 3.0 s, the sound
must travel a distance h + (h − y ) in 3.0 s.
SET UP:
EXECUTE: Let + y be downward, so for the teacher a y = 9.80 m/s 2 and v0 y = 0 . Let y = 0 at the top of the cliff.
(a) For the teacher, y = 1 (9.80 m/s 2 )(3.0 s) 2 = 44.1 m . For the sound, h + ( h − y ) = vst .
2 h = 1 (vst + y ) = 1 ([340 m/s][3.0 s] + 44.1 m) = 532 m , which rounds to 530 m.
2
2
2
2
(b) v y = v0 y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(532 m) = 102 m/s . 2-34 Chapter 2 EVALUATE:
2.85. She is in the air for t = IDENTIFY and SET UP: v y − v0 y
ay = 102 m/s
= 10.4 s and strikes the ground at high speed.
9.80 m/s 2 Let + y be upward. Each ball moves with constant acceleration a y = −9.80 m/s 2 . In parts (c) and (d) require that the two balls be at the same height at the same time.
EXECUTE: (a) At ceiling, v y = 0, y − y0 = 3.0 m, a y = −9.80 m/s 2 . Solve for v0 y .
2
2
v y = v0 y + 2a y ( y − y0 ) gives v0 y = 7.7 m/s. (b) v y = v0 y + a yt with the information from part (a) gives t = 0.78 s.
(c) Let the first ball travel downward a distance d in time t. It starts from its maximum height, so v0 y = 0. y − y0 = v0 y t = 1 a yt 2 gives d = (4.9 m/s 2 )t 2
2
2
The second ball has v0 y = 3 (7.7 m/s) = 5.1 m/s. In time t it must travel upward 3.0 m d to be at the same place as the first ball.
y − y0 = v0 yt + 1 a yt 2 gives 3.0 m − d = (5.1 m/s)t − (4.9 m/s 2 )t 2 .
2 2.86. We have two equations in two unknowns, d and t. Solving gives t = 0.59 s and d = 1.7 m.
(d) 3.0 m − d = 1.3 m
EVALUATE: In 0.59 s the first ball falls d = (4.9 m/s 2 )(0.59 s) 2 = 1.7 m, so is at the same height as the
second ball.
IDENTIFY: The helicopter has two segments of motion with constant acceleration: upward acceleration for 10.0 s
and then free-fall until it returns to the ground. Powers has three segments of motion with constant acceleration:
upward acceleration for 10.0 s, free-fall for 7.0 s and then downward acceleration of 2.0 m/s 2 .
SET UP: Let + y be upward. Let y = 0 at the ground.
EXECUTE: (a) When the engine shuts off both objects have upward velocity
v y = v0 y + a yt = (5.0 m/s 2 )(10.0 s) = 50.0 m/s and are at y = v0 yt + 1 a yt 2 = 1 (5.0 m/s 2 )(10.0 s) 2 = 250 m . For the
2
2
helicopter, v y = 0 (at the maximum height), v0 y = +50.0 m/s , y0 = 250 m , and a y = −9.80 m/s 2 .
2
2
v y = v0 y + 2a y ( y − y0 ) gives y = 2
2
v y − v0 y 2a y + y0 = 0 − (50.0 m/s) 2
+ 250 m = 378 m , which rounds to 380 m.
2( −9.80 m/s 2 ) (b) The time for the helicopter to crash from the height of 250 m where the engines shut off can be found using
v0 y = +50.0 m/s , a y = −9.80 m/s 2 , and y − y0 = −250 m . y − y0 = v0 yt + 1 a yt 2 gives
2 −250 m = (50.0 m/s)t − (4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (50.0 m/s)t − 250 m = 0 . The quadratic formula gives ( ) 1
50.0 ± (50.0) 2 + 4(4.90)(250) s . Only the positive solution is physical, so t = 13.9 s . Powers therefore
9.80
has free-fall for 7.0 s and then downward acceleration of 2.0 m/s 2 for 13.9 s − 7.0 s = 6.9 s . After 7.0 s of free-fall
he is at y − y0 = v0 yt + 1 a yt 2 = 250 m + (50.0 m/s)(7.0 s) + 1 ( −9.80 m/s 2 )(7.0 s) 2 = 360 m and has velocity
2
2
t= vx = v0 x + axt = 50.0 m/s + ( −9.80 m/s 2 )(7.0 s) = −18.6 m/s . After the next 6.9 s he is at
y − y0 = v0 yt + 1 a yt 2 = 360 m + (−18.6 m/s)(6.9 s) + 1 ( −2.00 m/s 2 )(6.9 s) 2 = 184 m . Powers is 184 m above the
2
2 2.87. ground when the helicopter crashes.
EVALUATE: When Powers steps out of the helicopter he retains the initial velocity he had in the helicopter but
his acceleration changes abruptly from 5.0 m/s 2 upward to 9.80 m/s 2 downward. Without the jet pack he would
have crashed into the ground at the same time as the helicopter. The jet pack slows his descent so he is above the
ground when the helicopter crashes.
IDENTIFY: Apply the constant acceleration equations to his motion. Consider two segments of the motion: the
last 1.0 s and the motion prior to that. The final velocity for the first segment is the initial velocity for the second
segment.
SET UP: Let + y be downward, so a y = +9.80 m/s 2 .
EXECUTE: Motion from the roof to a height of h / 4 above ground: y − y0 = 3h / 4 , a y = +9.80 m/s 2 , v0 y = 0 . 2
2
v y = v0 y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 3.834 h y − y0 = h / 4 , a y = +9.80 m/s 2 , v0 y = 3.834 h m / s . Motion from height of h / 4 to the ground: m / s , t = 1.00 s . y − y0 = v0 yt + 1 a yt 2 gives
2 Motion Along a Straight Line h
= 3.834 h
4 m + 4.90 m . Let h = u 2 and solve for u. 1 u 2 − 3.834u
4 ( u = 2 3.834 ± ( −3.834) 2 + 4.90 ) 2-35 m − 4.90 m = 0 . m . Only the positive root is physical, so u = 16.52 m and h = u 2 = 273 m , which rounds to 270 m. The building is 270 m tall.
EVALUATE: 2.88. With h = 273 m the total time of fall is t = 2h
= 7.46 s . In 7.47 s − 1.00 s = 6.46 s Spider-Man
ay falls a distance y − y0 = 1 (9.80 m/s 2 )(6.46 s) 2 = 204 m . This leaves 69 m for the last 1.0 s of fall, which is h / 4 .
2
IDENTIFY: Apply constant acceleration equations to the motion of the rock. Sound travels at constant speed.
SET UP: Let tfall be the time for the rock to fall to the ground and let ts be the time it takes the sound to travel
from the impact point back to you. tfall + ts = 10.0 s . Both the rock and sound travel a distance d that is equal to the
height of the cliff. Take + y downward for the motion of the rock. The rock has v0 y = 0 and a y = 9.80 m/s 2 .
EXECUTE: (a) For the rock, y − y0 = v0 yt + 1 a yt 2 gives tfall =
2 2d
.
9.80 m/s 2 d
= 10.0 s . Let α 2 = d . 0.00303α 2 + 0.4518α − 10.0 = 0 . α = 19.6 and d = 384 m .
330 m/s
(b) You would have calculated d = 1 (9.80 m/s 2 )(10.0 s) 2 = 490 m . You would have overestimated the height of
2
the cliff. It actually takes the rock less time than 10.0 s to fall to the ground.
EVALUATE: Once we know d we can calculate that tfall = 8.8 s and ts = 1.2 s . The time for the sound of impact
to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it
strikes the ground.
(a) IDENTIFY: Let + y be upward. The can has constant acceleration a y = − g . The initial upward velocity of the For the sound, ts = 2.89. can equals the upward velocity of the scaffolding; first find this speed.
SET UP: y − y0 = −15.0 m, t = 3.25 s, a y = −9.80 m/s 2 , v0 y = ?
EXECUTE: y − y0 = v0 yt + 1 a yt 2 gives v0 y = 11.31 m/s
2 Use this v0 y in v y = v0 y + a y t to solve for v y : v y = −20.5 m/s
(b) IDENTIFY: Find the maximum height of the can, above the point where it falls from the scaffolding:
SET UP: v y = 0, v0 y = +11.31 m/s, a y = −9.80 m/s 2 , y − y0 = ?
EXECUTE: 2.90. 2
2
v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 6.53 m The can will pass the location of the other painter. Yes, he gets a chance.
EVALUATE: Relative to the ground the can is initially traveling upward, so it moves upward before stopping
momentarily and starting to fall back down.
IDENTIFY: Both objects are in free-fall. Apply the constant acceleration equations to the motion of each person.
SET UP: Let + y be downward, so a y = +9.80 m/s 2 for each object.
EXECUTE: (a) Find the time it takes the student to reach the ground: y − y0 = 180 m , v0 y = 0 , a y = 9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2(180 m)
=
= 6.06 s . Superman must reach the ground in
ay
9.80 m/s 2 6.06 s − 5.00 s = 1.06 s : t = 1.06 s , y − y0 = 180 m , a y = +9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives
2
y − y0 1
180 m 1
− 2 a yt =
− (9.80 m/s 2 )(1.06 s) = 165 m/s . Superman must have initial speed v0 = 165 m/s .
1.06 s 2
t
(b) The graphs of y-t for Superman and for the student are sketched in Figure 2.90.
(c) The minimum height of the building is the height for which the student reaches the ground in 5.00 s, before
Superman jumps. y − y0 = v0 yt + 1 a yt 2 = 1 (9.80 m/s 2 )(5.00 s) 2 = 122 m . The skyscraper must be at least
2
2
v0 y = 122 m high. 2-36 Chapter 2 165 m/s = 369 mi/h , so only Superman could jump downward with this initial speed. EVALUATE: Figure 2.90
2.91. IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister
after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of
the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground.
SET UP: Let + y be upward. At the instant that the canister is released, it has the same velocity as the rocket. After it is released, the canister has a y = −9.80 m/s 2 . At its maximum height the canister has v y = 0 .
EXECUTE: (a) Find the speed of the rocket when the canister is released: v0 y = 0 , a y = 3.30 m/s 2 , 2
2
y − y0 = 235 m . v y = v0 y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 2(3.30 m/s 2 )(235 m) = 39.4 m/s . For the motion of the canister after it is released, v0 y = +39.4 m/s , a y = −9.80 m/s 2 , y − y0 = −235 m .
y − y0 = v0 yt + 1 a yt 2 gives −235 m = (39.4 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 12.0 s as the
2 positive solution. Then for the motion of the rocket during this 12.0 s,
y − y0 = v0 yt + 1 a yt 2 = 235 m + (39.4 m/s)(12.0 s) + 1 (3.30 m/s 2 )(12.0 s) 2 = 945 m .
2
2
(b) Find the maximum height of the canister above its release point: v0 y = +39.4 m/s , v y = 0 , a y = −9.80 m/s 2 .
2
2
v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = 0 − (39.4 m/s) 2
= 79.2 m . After its release the canister travels
2(−9.80 m/s 2 ) upward 79.2 m to its maximum height and then back down 79.2 m + 235 m to the ground. The total distance it
travels is 393 m.
EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is
v y = v0 y + a yt = 39.4 m/s + (3.30 m/s 2 )(12.0 s) = 79.0 m/s . We can calculate its height at this instant by
2
2
v y = v0 y + 2a y ( y − y0 ) with v0 y = 0 and v y = 79.0 m/s . y − y0 = 2.92. 2
2
v y − v0 y 2a y = (79.0 m/s) 2
= 946 m , which agrees
2(3.30 m/s 2 ) with our previous calculation.
IDENTIFY: Both objects are in free-fall and move with constant acceleration 9.80 m/s 2 , downward. The two
balls collide when they are at the same height at the same time.
SET UP: Let + y be upward, so a y = −9.80 m/s 2 for each ball. Let y = 0 at the ground. Let ball A be the one
thrown straight up and ball B be the one dropped from rest at height H. y0 A = 0 , y0 B = H .
EXECUTE: (a) y − y0 = v0 yt + 1 a yt 2 applied to each ball gives y A = v0t − 1 gt 2 and yB = H − 1 gt 2 . y A = yB gives
2
2
2 v0t − 1 gt 2 = H − 1 gt 2 and t =
2
2 H
.
v0 (b) For ball A at its highest point, v yA = 0 and v y = v0 y + a y t gives t = part (a) gives H v0
v2
= and H = 0 .
g
v0 g EVALUATE: In part (a), using t = v0
. Setting this equal to the time in
g less than ⎛ gH ⎞
H
in the expressions for y A and yB gives y A = yB = H ⎜1 − 2 ⎟ . H must be
v0
⎝ 2v0 ⎠ 2
2v0
in order for the balls to collide before ball A returns to the ground. This is because it takes ball A
g Motion Along a Straight Line 2-37 2v0
2v 2
2v 2
to return to the ground and ball B falls a distance 1 gt 2 = 0 during this time. When H = 0 the
2
g
g
g
two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before
they collide.
IDENTIFY and SET UP: Use vx = dx / dt and ax = dvx / dt to calculate vx (t ) and ax (t ) for each car. Use these
equations to answer the questions about the motion.
dx
dv
EXECUTE: x A = α t + β t 2 , v Ax = A = α + 2 β t , a Ax = Ax = 2 β
dt
dt
dxB
dvBx
xB = γ t 2 − δ t 3 , vBx =
= 2γ t − 3δ t 2 , aBx =
− 2γ − 6δ t
dt
dt
(a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger v0 x .
time t = 2.93. EXECUTE: At t = 0, v Ax = α and vBx = 0. So initially car A moves ahead. (b) IDENTIFY and SET UP: Cars at the same point implies x A = xB . αt + βt = γ t − δ t
EXECUTE: One solution is t = 0, which says that they start from the same point. To find the other solutions,
2 2 3 divide by t: α + β t = γ t − δ t 2 δ t 2 + ( β − γ )t + α = 0 ) ( ) ( 1
1
−( β − γ ) ± ( β − γ ) 2 − 4δα =
+1.60 ± (1.60) 2 − 4(0.20)(2.60) = 4.00 s ± 1.73 s
2δ
0.40
So x A = xB for t = 0, t = 2.27 s and t = 5.73 s.
t= EVALUATE: Car A has constant, positive ax . Its vx is positive and increasing. Car B has v0 x = 0 and ax that is
initially positive but then becomes negative. Car B initially moves in the + x -direction but then slows down and
finally reverses direction. At t = 2.27 s car B has overtaken car A and then passes it. At t = 5.73 s, car B is
moving in the − x-direction as it passes car A again.
d ( xB − x A )
. If this
(c) IDENTIFY: The distance from A to B is xB − x A . The rate of change of this distance is
dt
d ( xB − x A )
distance is not changing,
= 0. But this says vBx − v Ax = 0. (The distance between A and B is neither
dt
decreasing nor increasing at the instant when they have the same velocity.)
SET UP: v Ax = vBx requires α + 2 β t = 2γ t − 3δ t 2
EXECUTE: 3δ t 2 + 2( β − γ )t + α = 0 ) ( ( 1
1
3.20 ± 4(−1.60)2 − 12(0.20)(2.60)
−2( β − γ ) ± 4( β − γ ) 2 − 12δα =
6δ
1.20
t = 2.667 s ± 1.667 s, so v Ax = vBx for t = 1.00 s and t = 4.33 s.
t= ) EVALUATE: At t = 1.00 s, v Ax = vBx = 5.00 m/s. At t = 4.33 s, v Ax = vBx = 13.0 m/s. Now car B is slowing down
while A continues to speed up, so their velocities aren’t ever equal again.
(d) IDENTIFY and SET UP: a Ax = aBx requires 2β = 2γ − 6δ t
EXECUTE:
EVALUATE:
2.94. t= γ − β 2.80 m/s 2 − 1.20 m/s 2
=
= 2.67 s.
3δ
3(0.20 m/s3 ) At t = 0, aBx > a Ax , but aBx is decreasing while a Ax is constant. They are equal at t = 2.67 s but for all times after that aBx < a Ax .
IDENTIFY: The apple has two segments of motion with constant acceleration. For the motion from the tree to the
top of the grass the acceleration is g, downward and the apple falls a distance H − h . For the motion from the top
of the grass to the ground the acceleration is a, upward, the apple travels downward a distance h, and the final
speed is zero.
SET UP: Let + y be upward and let y = 0 at the ground. The apple is initially a height H + h above the ground.
EXECUTE: 2
2
(a) Motion from y0 = H + h to y = H : y − y0 = − H , v0 y = 0 , a y = − g . v y = v0 y + 2a y ( y − y0 ) gives v y = − 2 gH . The speed of the apple is 2 gH as it enters the grass. 2-38 Chapter 2
2
2
(b) Motion from y0 = h to y = 0 : y − y0 = − h , v0 y = − 2 gH . v y = v0 y + 2a y ( y − y0 ) gives ay = 2
2
v y − v0 y 2( y − y0 ) = −2 gH gH
=
. The acceleration of the apple while it is in the grass is gH / h , upward.
h
2( −h) (c) Graphs of y-t, v y -t and a y -t are sketched in Figure 2.94.
EVALUATE:
increases. The acceleration a produced by the grass increases when H increases and decreases when h Figure 2.94
2.95. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, the student and the bus.
SET UP: For convenience, let the student's (constant) speed be v0 and the bus's initial position be x0 . Note that
these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be
zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as functions of time are then x1 = v0t and x2 = x0 + (1 2)at 2 .
EXECUTE: )
(
(5.0 m s) − 2(0.170 m s )(40.0 m) ) = 9.55 s and 49.3 s . (a) Setting x1 = x2 and solving for the times t gives t = ( 1
2
v0 ± v0 − 2ax0 .
a 1
2
2
(5.0 m s) ±
(0.170 m s 2 )
The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later
time). During this time, the student has run a distance v0t = (5 m s)(9.55 s) = 47.8 m.
t= (b) The speed of the bus is (0.170 m/s 2 )(9.55 s) = 1.62 m/s .
(c) The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown
in Figure 2.95a.
(d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then
catches up to her. At this later time the bus's velocity is ( 0.170 m s 2 ) ( 49.3 s ) = 8.38 m s.
2
(e) No; v0 < 2ax0 , and the roots of the quadratic are imaginary. When the student runs at 3.5 m s, Figure 2.95b
shows that the two lines do not intersect:
2
(f) For the student to catch the bus, v0 > 2ax0 . and so the minimum speed is 2 ( 0.170 m s 2 ) ( 40 m s ) = 3.688 m s. She would be running for a time
distance ( 3.688 m s ) ( 21.7 s ) = 80.0 m. 3.69 m s
= 21.7 s, and covers a
0.170 m/s 2 However, when the student runs at 3.688 m s, the lines intersect at one point, at x = 80 m , as shown in
Figure 2.95c.
EVALUATE: The graph in part (c) shows that the student is traveling faster than the bus the first time they meet
but at the second time they meet the bus is traveling faster.
t2 = ttot − t1 Figure 2.95 Motion Along a Straight Line 2.96. 2-39 Apply y − y0 = v0 yt + 1 a yt 2 to the motion from the maximum height, where v0 y = 0 . The time spent
2 IDENTIFY: above ymax / 2 on the way down equals the time spent above ymax / 2 on the way up.
SET UP: Let + y be downward. a y = g . y − y0 = ymax / 2 when he is a distance ymax / 2 above the floor.
The time from the maximum height to ymax / 2 above the floor is given by ymax / 2 = 1 gt12 . The time
2 EXECUTE: 2
from the maximum height to the floor is given by ymax = 1 gt tot and the time from a height of ymax / 2 to the floor is .
2 t1
=
t2 ymax / 2
ymax − ymax / 2 EVALUATE:
2.97. = 1
= 2.4 .
2 −1 The person spends over twice as long above ymax / 2 as below ymax / 2 . His average speed is less above ymax / 2 than it is when he is below this height.
IDENTIFY: Apply constant acceleration equations to both objects.
SET UP: Let + y be upward, so each ball has a y = − g . For the purpose of doing all four parts with the least
1
1
2
repetition of algebra, quantities will be denoted symbolically. That is, let y1 = h + v0t − gt 2 , y2 = h − g ( t − t0 ) .
2
2
In this case, t0 = 1.00 s .
EXECUTE:
1
2 (a) Setting y1 = y2 = 0, expanding the binomial ( t − t0 ) and eliminating the common term
2 2
gt 2 yields v0t = gt0t − 1 gt0 . Solving for t: t =
2 2
⎞
gt0
1
t⎛
= 0⎜
⎟.
gt0 − v0 2 ⎝ 1 − v0 /( gt0 ) ⎠
1
2 Substitution of this into the expression for y1 and setting y1 = 0 and solving for h as a function of v0 yields, after ( 1 gt0 − v0 )
2
2
( gt0 − v0 ) 2 2
some algebra, h = 1 gt0
2 . Using the given value t0 = 1.00 s and g = 9.80 m s 2 , 2
⎛ 4.9 m s − v0 ⎞
h = 20.0 m = ( 4.9 m ) ⎜
⎟.
⎝ 9.8 m s − v0 ⎠
This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see
part (c)). The physical solution involves taking the negative square root before solving for v0 , and yields 8.2 m s.
The graph of y versus t for each ball is given in Figure 2.97.
(b) The above expression gives for (i), 0.411 m and for (ii) 1.15 km.
(c) As v0 approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the
second ball is thrown that approaches zero. If v0 > 9.8 m s, the first ball can never catch the second ball.
(d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If v0 < 4.9 m s, the first ball will already
have passed the roof on the way down before the second ball is released, and the second ball can never catch up.
EVALUATE: Note that the values of v0 in parts (a) and (b) are all greater than vmin and less than vmax . Figure 2.97
2.98. IDENTIFY: Apply constant acceleration equations to the motion of the boulder.
SET UP: Let + y be downward, so a y = + g . 2-40 Chapter 2 EXECUTE: (a) Let the height be h and denote the 1.30-s interval as Δt ; the simultaneous equations h = 1 gt 2 , 2 h = 1 g (t − Δt ) 2 can be solved for t. Eliminating h and taking the square root,
2
3
2 t
3
=
, and
t − Δt
2 Δt
, and substitution into h = 1 gt 2 gives h = 246 m.
2
1 − 2/3
(b) The above method assumed that t > 0 when the square root was taken. The negative root (with Δt = 0) gives
an answer of 2.51 m, clearly not a “cliff”. This would correspond to an object that was initially near the bottom of
this “cliff” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically
possible, the conditions of the problem preclude this answer.
EVALUATE: For the first two-thirds of the distance, y − y0 = 164 m , v0 y = 0 , and a y = 9.80 m/s 2 .
t= v y = 2a y ( y − y0 ) = 56.7 m/s . Then for the last third of the distance, y − y0 = 82.0 m , v0 y = 56.7 m/s and
a y = 9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives (4.90 m/s 2 )t 2 + (56.7 m/s)t − 82.0 m = 0 .
2
t= ( ) 1
−56.7 + (56.7) 2 + 4(4.9)(82.0) s = 1.30 s , as required.
9.8 3 MOTION IN TWO OR THREE DIMENSIONS 3.1. IDENTIFY and SET UP: Use Eq.(3.2), in component form.
Δx x2 − x1 5.3 m − 1.1 m
=
=
= 1.4 m/s
EXECUTE: ( vav ) x =
Δt t2 − t1
3.0 s − 0 ( vav ) y = Δy y2 − y1 −0.5 m − 3.4 m
=
=
= −1.3 m/s
Δt
t2 − t1
3.0 s − 0
tan α = ( vav ) y
( vav ) x = −1.3 m/s
= −0.9286
1.4 m/s α = 360° − 42.9° = 317°
vav = ( vav ) x + ( vav ) y
2 2 vav = (1.4 m/s) 2 + (−1.3 m/s) 2 = 1.9 m/s
Figure 3.1 3.2. !
EVALUATE: Our calculation gives that vav is in the 4th quadrant. This corresponds to increasing x and
decreasing y.
!
IDENTIFY: Use Eq.(3.2), written in component form. The distance from the origin is the magnitude of r .
SET UP: At time t1 , x1 = y1 = 0 .
EXECUTE: (a) x = (vav-x )Δt = ( −3.8m s)(12.0 s) = −45.6 m and y = (vav-y )Δt = (4.9m s)(12.0 s) = 58.8 m . (b) r = x 2 + y 2 = (−45.6 m) 2 + (58.8 m) 2 = 74.4 m.
!
!
EVALUATE: Δr is in the direction of vav . Therefore, Δx is negative since vav-x is negative and Δy is positive since vav-y is positive. 3.3. !
(a) IDENTIFY and SET UP: From r we can calculate x and y for any t.
Then use Eq.(3.2), in component form.
!
ˆ
EXECUTE: r = ⎡ 4.0 cm + ( 2.5 cm/s 2 ) t 2 ⎤ i + ( 5.0 cm/s ) tˆ
j
⎣
⎦
!
ˆ
At t = 0, r = ( 4.0 cm ) i .
!
ˆ
j
At t = 2.0 s, r = (14.0 cm ) i + (10.0 cm ) ˆ. Δx 10.0 cm
=
= 5.0 cm/s.
Δt
2.0 s
Δy 10.0 cm
=
=
= 5.0 cm/s.
Δt
2.0 s ( vav ) x =
( vav ) y 3-1 3-2 Chapter 3 vav = ( vav ) x + ( vav ) y tan α = 2 ( vav ) y
( vav ) x 2 = 7.1 cm/s = 1.00 θ = 45°.
Figure 3.3a !
EVALUATE: Both x and y increase, so vav is in the 1st quadrant.
!
!
(b) IDENTIFY and SET UP: Calculate r by taking the time derivative of r (t ).
!
! dr
ˆ
= ⎡5.0 cm/s 2 ⎤ t i + ( 5.0 cm/s ) ˆ
EXECUTE: v =
j
⎣
⎦
dt
t = 0 : vx = 0, v y = 5.0 cm/s; v = 5.0 cm/s and θ = 90° ( ) t = 1.0 s: vx = 5.0 cm/s, v y = 5.0 cm/s; v = 7.1 cm/s and θ = 45°
t = 2.0 s: vx = 10.0 cm/s, v y = 5.0 cm/s; v = 11 cm/s and θ = 27°
(c) The trajectory is a graph of y versus x.
x = 4.0 cm + (2.5 cm/s 2 )t 2 , y = (5.0 cm/s)t
For values of t between 0 and 2.0 s, calculate x and y and plot y versus x. Figure 3.3b
3.4. 3.5. EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory.
!
!
IDENTIFY: v = dr/dt . This vector will make a 45° -angle with both axes when its x- and y-components are equal.
d (t n )
SET UP:
= nt n −1 .
dt
!
ˆ
EXECUTE: v = 2bti + 3ct 2 ˆ . vx = v y gives t = 2b 3c .
j
!
EVALUATE: Both components of v change with t.
IDENTIFY and SET UP: Use Eq.(3.8) in component form to calculate ( aav ) x and ( aav ) y . Motion in Two or Three Dimensions 3-3 EXECUTE: (a) The velocity vectors at t1 = 0 and t2 = 30.0 s are shown in Figure 3.5a. Figure 3.5a
(b) ( aav ) x = ( aav ) y = Δv y
Δt Δvx v2 x − v1x −170 m/s − 90 m/s
=
=
= −8.67 m/s 2
Δt
t2 − t1
30.0 s
= v2 y − v1 y
t2 − t1 = 40 m/s − 110 m/s
= −2.33 m/s 2
30.0 s (c) a= ( aav ) x + ( aav ) y tan α = 2 ( aav ) y
( aav ) x 2 = = 8.98 m/s 2 −2.33 m/s 2
= 0.269
−8.67 m/s 2 α = 15° + 180° = 195°
Figure 3.5b
EVALUATE:
3.6. !
The changes in vx and v y are both in the negative x or y direction, so both components of aav are in the 3rd quadrant.
IDENTIFY: Use Eq.(3.8), written in component form.
2
2
2
2
SET UP: ax = (0.45m s )cos31.0° = 0.39m s , a y = (0.45m s )sin 31.0° = 0.23m s
EXECUTE: (a) aav-x = Δv
Δvx
2
and vx = 2.6 m s + (0.39 m s )(10.0 s) = 6.5 m s . aav-y = y and
Δt
Δt
2 v y = −1.8 m s + (0.23 m s )(10.0 s) = 0.52 m s .
⎛ 0.52 ⎞
(b) v = (6.5m s) 2 + (0.52 m s) 2 = 6.48m s , at an angle of arctan ⎜
⎟ = 4.6° above the horizontal.
⎝ 6.5 ⎠
!
!
(c) The velocity vectors v1 and v2 are sketched in Figure 3.6. The two velocity vectors differ in magnitude and
direction.
!
EVALUATE: v1 is at an angle of 35° below the + x -axis and has magnitude v1 = 3.2 m/s , so v2 > v1 and the
!
!
direction of v2 is rotated counterclockwise from the direction of v1 . Figure 3.6
3.7. IDENTIFY and SET UP: Use Eqs.(3.4) and (3.12) to find vx , v y , ax , and a y as functions of time. The magnitude
!
!
and direction of r and a can be found once we know their components. 3-4 Chapter 3 EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are given in
Figure 3.7a. Figure 3.7a dx
dy
= α vy =
= −2 β t
dt
dt
dv
dv
a y = x = 0 a y = y = −2 β
dt
dt
!
!
ˆ
Thus v = ai − 2 β tˆ a = −2 β ˆ
j
j
(b) vx = (c) velocity: At t = 2.0 s, vx = 2.4 m/s, v y = −2(1.2 m/s 2 )(2.0 s) = −4.8 m/s 2
2
v = vx + v y = 5.4 m/s −4.8 m/s
= −2.00
2.4 m/s
vx
α = −63.4° + 360° = 297° tan α = vy = Figure 3.7b acceleration: At t = 2.0 s, ax = 0, a y = −2(1.2 m/s 2 ) = −2.4 m/s 2
2
2
a = ax + a y = 2.4 m/s 2 tan β = ay
ax = −2.4 m/s 2
= −∞
0 β = 270°
Figure 3.7c !
EVALUATE: (d) a has a component a" in the same direction as
!
v , so we know that v is increasing (the bird is speeding up.)
!
!
a also has a component a⊥ perpendicular to v , so that the
!
direction of v is changing; the bird is turning toward the
− y -direction (toward the right)
Figure 3.7d Motion in Two or Three Dimensions 3-5 3.8. !
!
v is always tangent to the path; v at t = 2.0 s shown in part (c) is tangent to the path at this t, conforming to this
!
!
general rule. a is constant and in the − y -direction; the direction of v is turning toward the − y -direction.
!
!
!
IDENTIFY: The component a⊥ of a perpendicular to the path is related to the change in direction of v and the
!
!
!
component a" of a parallel to the path is related to the change in the magnitude of v .
!
!
!
SET UP: When the speed is increasing, a" is in the direction of v and when the speed is decreasing, a" is opposite
!
to the direction of v . When v is constant, a" is zero and when the path is a straight line, a⊥ is zero.
EXECUTE: The acceleration vectors in each case are sketched in Figure 3.8a-c.
!
EVALUATE: a⊥ is toward the center of curvature of the path. Figure 3.8a-c
3.9. IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal.
SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position of the
book, at the point where it leaves the table top. x-component:
ax = 0, v0 x = 1.10 m/s,
t = 0.350 s
y-component:
a y = −9.80 m/s 2 ,
v0 y = 0,
t = 0.350 s Figure 3.9a Use constant acceleration equations for the x and y components of the motion, with ax = 0 and a y = − g .
EXECUTE: (a) y − y0 = ? y − y0 = v0 yt + 1 a yt 2 = 0 + 1 ( −9.80 m/s 2 )(0.350 s) 2 = −0.600 m. The table top is 0.600 m above the floor.
2
2
(b) x − x0 = ? x − x0 = v0 xt + 1 axt 2 = (1.10 m/s)(0.350 s) + 0 = 0.358 m.
2
(c) vx = v0 x + axt = 1.10 m/s (The x-component of the velocity is constant, since ax = 0.) v y = v0 y + a yt = 0 + ( −9.80 m/s 2 )(0.350 s) = −3.43 m/s 2
2
v = vx + v y = 3.60 m/s −3.43 m/s
=
= −3.118
vx 1.10 m/s
α = −72.2°
!
Direction of v is 72.2° below the horizontal tan α = Figure 3.9b vy 3-6 Chapter 3 (d) The graphs are given in Figure 3.9c Figure 3.9c
EVALUATE: In the x-direction, ax = 0 and vx is constant. In the y-direction, a y = −9.80 m/s 2 and v y is downward and increasing in magnitude since a y and v y are in the same directions. The x and y motions occur 3.10. independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it travels
horizontally.
IDENTIFY: The bomb moves in projectile motion. Treat the horizontal and vertical components of the motion
separately. The vertical motion determines the time in the air.
SET UP: The initial velocity of the bomb is the same as that of the helicopter. Take + y downward, so ax = 0 ,
a y = +9.80 m/s 2 , v0 x = 60.0 m/s and v0 y = 0 .
EXECUTE: (a) y − y0 = v0 yt + 1 a yt 2 with y − y0 = 300 m gives t =
2 2( y − y0 )
2(300 m)
=
= 7.82 s .
ay
9.80 m/s 2 (b) The bomb travels a horizontal distance x − x0 = v0 xt + 1 axt 2 = (60.0 m/s)(7.82 s) = 470 m .
2
(c) vx = v0 x = 60.0 m/s . v y = v0 y + a yt = (9.80 m/s 2 )(7.82 s) = 76.6 m/s .
(d) The graphs are given in Figure 3.10.
(e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will
be 300 m directly above the bomb at impact.
EVALUATE: The initial horizontal velocity of the bomb doesn’t affect its vertical motion. Figure 3.10
3.11. IDENTIFY: Each object moves in projectile motion.
SET UP: Take + y to be downward. For each cricket, ax = 0 and a y = +9.80 m/s 2 . For Chirpy, v0 x = v0 y = 0 . For Milada, v0 x = 0.950 m/s , v0 y = 0 3.12. EXECUTE: Milada's horizontal component of velocity has no effect on her vertical motion. She also reaches the
ground in 3.50 s. x − x0 = v0 xt + 1 axt 2 = (0.950 m/s)(3.50 s) = 3.32 m
2
EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that the
time is the same for both.
IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time she falls
9.00 m vertically.
SET UP: Take + y downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0 .
EXECUTE: Time to fall 9.00 m: y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2(9.00 m)
=
= 1.36 s .
ay
9.80 m/s 2 Speed needed to travel 1.75 m horizontally during this time: x − x0 = v0 xt + 1 axt 2 gives
2
x − x0 1.75 m
=
= 1.29 m/s .
t
1.36 s
EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has traveled
horizontally farther than 1.75 m.
IDENTIFY: The car moves in projectile motion. The car travels 21.3 m − 1.80 m = 19.5 m downward during the
time it travels 61.0 m horizontally.
SET UP: Take + y to be downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0 .
v0 = v0 x = 3.13. Motion in Two or Three Dimensions 3-7 Use the vertical motion to find the time in the air:
2( y − y0 )
2(19.5 m)
=
= 1.995 s
y − y0 = v0 yt + 1 a yt 2 gives t =
2
ay
9.80 m/s 2 EXECUTE: Then x − x0 = v0 xt + 1 axt 2 gives v0 = v0 x =
2 x − x0 61.0 m
=
= 30.6 m/s .
t
1.995 s 2
2
(b) vx = 30.6m s since ax = 0 . v y = v0 y + a yt = −19.6m s . v = vx + v y = 36.3m s . 3.14. EVALUATE: We calculate the final velocity by calculating its x and y components.
IDENTIFY: The marble moves with projectile motion, with initial velocity that is horizontal and has
magnitude v0 . Treat the horizontal and vertical motions separately. If v0 is too small the marble will land to the left of the hole and if v0 is too large the marble will land to the right of the hole.
SET UP: Let + x be horizontal to the right and let + y be upward. v0 x = v0 , v0 y = 0 , ax = 0 , a y = −9.80 m/s 2 Use the vertical motion to find the time it takes the marble to reach the height of the level ground;
2( y − y0 )
2( −2.75 m)
y − y0 = −2.75 m . y − y0 = v0 yt + 1 a yt 2 gives t =
=
= 0.749 s . The time does not depend
2
ay
−9.80 m/s 2 EXECUTE: on v0 .
Minimum v0 : x − x0 = 2.00 m , t = 0.749 s . x − x0 = v0 xt + 1 axt 2 gives v0 =
2 x − x0 2.00 m
=
= 2.67 m/s .
t
0.749 s 3.50 m
= 4.67 m/s .
0.749 s
EVALUATE: The horizontal and vertical motions are independent and are treated separately. Their only
connection is that the time is the same for both.
IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has magnitude v0 .
Maximum v0 : x − x0 = 3.50 m and v0 = 3.15. The height h of the table and v0 are the same; the acceleration due to gravity changes from g E = 9.80 m/s 2 on earth
to g X on planet X.
SET UP: Let + x be horizontal and in the direction of the initial velocity of the marble and let + y be upward. v0 x = v0 , v0 y = 0 , ax = 0 , a y = − g , where g is either g E or g X .
EXECUTE: Use the vertical motion to find the time in the air: y − y0 = − h . y − y0 = v0 yt + 1 a yt 2 gives t =
2 Then x − x0 = v0 xt + 1 axt 2 gives x − x0 = v0 xt = v0
2 2h
. x − x0 = D on earth and 2.76D on Planet X.
g gE
= 0.131g E = 1.28 m/s 2 .
(2.76) 2
EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor, and it
travels farther horizontally.
IDENTIFY: The football moves in projectile motion.
SET UP: Let + y be upward. ax = 0 , a y = − g . At the highest point in the trajectory, v y = 0 .
( x − x0 ) g = v0 2h , which is constant, so D g E = 2.76 D g X . g X = 3.16. 2h
.
g v0 y 16.0m s
= 1.63 s .
2
9.80m s
(b) Different constant acceleration equations give different expressions but the same numerical result:
v2
1
gt 2 = 1 v y 0t = 0 y = 13.1 m .
2
2
2g
(c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s
(d) ax = 0 , so x − x0 = v0 xt = (20.0 m s)(3.27 s) = 65.3 m .
(e) The graphs are sketched in Figure 3.16.
EXECUTE: (a) v y = v0 y + a y t . The time t is g = 3-8 Chapter 3 EVALUATE: When the football returns to its original level, vx = 20.0 m/s and v y = −16.0 m/s . Figure 3.16
3.17. IDENTIFY: The shell moves in projectile motion.
SET UP: Let + x be horizontal, along the direction of the shell's motion, and let + y be upward. ax = 0 , a y = −9.80 m/s 2 .
EXECUTE: (a) v0 x = v0 cos α 0 = (80.0 m/s)cos60.0° = 40.0 m/s , v0 y = v0 sin α 0 = (80.0 m/s)sin60.0° = 69.3 m/s . (b) At the maximum height v y = 0 . v y = v0 y + a y t gives t =
2
2
(c) v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = v y − v0 y
ay = 0 − 69.3 m/s
= 7.07 s .
−9.80 m/s 2 0 − (69.3 m/s) 2
= 245 m .
2(−9.80 m/s 2 ) (d) The total time in the air is twice the time to the maximum height, so
x − x0 = v0 xt + 1 axt 2 = (40.0 m/s)(14.14 s) = 566 m .
2
(e) At the maximum height, vx = v0 x = 40.0 m/s and v y = 0 . At all points in the motion, ax = 0 and a y = −9.80 m/s 2 .
EVALUATE: The equation for the horizontal range R derived in Example 3.8 is R = 2
v0 sin 2α 0
. This gives
g (80.0 m/s) 2 sin(120.0°)
= 566 m , which agrees with our result in part (d).
9.80 m/s 2
IDENTIFY: The flare moves with projectile motion. The equations derived in Example 3.8 can be used to find the
maximum height h and range R.
v 2 sin 2 α 0
v 2 sin 2α 0
and R = 0
.
SET UP: From Example 3.8, h = 0
2g
g
R= 3.18. EXECUTE: (a) h = (125 m/s) 2 (sin 55.0°) 2
(125 m/s) 2 (sin110.0°)
= 535 m . R =
= 1500 m .
2(9.80 m/s 2 )
9.80 m/s 2 ⎛ 9.80 m/s 2 ⎞
(b) h and R are proportional to 1/ g , so on the Moon, h = ⎜
(535 m) = 3140 m and
2⎟
⎝ 1.67 m/s ⎠ 3.19. ⎛ 9.80 m/s 2 ⎞
R=⎜
(1500 m) = 8800 m .
2⎟
⎝ 1.67 m/s ⎠
EVALUATE: The projectile travels on a parabolic trajectory. It is incorrect to say that h = ( R / 2) tan α 0 .
IDENTIFY: The baseball moves in projectile motion. In part (c) first calculate the components of the velocity at
this point and then get the resultant velocity from its components.
SET UP: First find the x- and y-components of the initial velocity. Use coordinates where the + y -direction is
upward, the + x -direction is to the right and the origin is at the point where the baseball leaves the bat. v0 x = v0 cos α 0 = (30.0 m/s)cos36.9° = 24.0 m/s
v0 y = v0 sin α 0 = (30.0 m/s)sin 36.9° = 18.0 m/s Figure 3.19a Use constant acceleration equations for the x and y motions, with ax = 0 and a y = − g . Motion in Two or Three Dimensions 3-9 EXECUTE: (a) y-component (vertical motion):
y − y0 = +10.0 m/s, v0 y = 18.0 m/s, a y = −9.80 m/s 2 , t = ? y − y0 = v0 y + 1 a yt 2
2
10.0 m = (18.0 m/s)t − (4.90 m/s 2 )t 2
(4.90 m/s 2 )t 2 − (18.0 m/s)t + 10.0 m = 0
1
Apply the quadratic formula: t = 9.80 ⎡18.0 ± ( −18.0 ) − 4 ( 4.90 )(10.0 ) ⎤ s = (1.837 ± 1.154 ) s
⎢
⎥
⎣
⎦
The ball is at a height of 10.0 above the point where it left the bat at t1 = 0.683 s and at t2 = 2.99 s. At the earlier
time the ball passes through a height of 10.0 m as its way up and at the later time it passes through 10.0 m on its
way down.
(b) vx = v0 x = +24.0 m/s, at all times since ax = 0.
2 v y = v0 y + a y t
t1 = 0.683 s: v y = +18.0 m/s + (−9.80 m/s 2 )(0.683 s) = +11.3 m/s. ( v y is positive means that the ball is traveling
upward at this point.
t2 = 2.99 s: v y = +18.0 m/s + (−9.80 m/s 2 )(2.99 s) = −11.3 m/s. ( v y is negative means that the ball is traveling
downward at this point.)
(c) vx = v0 x = 24.0 m/s
Solve for v y :
v y = ?, y − y0 = 0 (when ball returns to height where motion started),
a y = −9.80 m/s 2 , v0 y = +18.0 m/s
2
2
v y = v0 y + 2a y ( y − y0 ) v y = −v0 y = −18.0 m/s (negative, since the baseball must be traveling downward at this point)
!
Now that have the components can solve for the magnitude and direction of v .
2
2
v = vx + v y v= ( 24.0 m/s ) 2 + ( −18.0 m/s ) = 30.0 m/s
2 −18.0 m/s
=
vx
24.0 m/s
α = −36.9°, 36.9° below the horizontal
tan α = vy Figure 3.19b 3.20. The velocity of the ball when it returns to the level where it left the bat has magnitude 30.0 m/s and is directed at
an angle of 36.9° below the horizontal.
EVALUATE: The discussion in parts (a) and (b) explains the significance of two values of t for which
y − y0 = +10.0 m. When the ball returns to its initial height, our results give that its speed is the same as its initial
speed and the angle of its velocity below the horizontal is equal to the angle of its initial velocity above the
horizontal; both of these are general results.
IDENTIFY: The shot moves in projectile motion.
SET UP: Let + y be upward.
EXECUTE: (a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and
2
− g = −9.80 m s vertically downward.
(b) The x-component of velocity is constant at vx = (12.0 m s)cos51.0° = 7.55 m s . The y-component is
2 v0 y = (12.0 m s)sin 51.0° = 9.32 m s at release and v y = v0 y − gt = (10.57 m s) − (9.80 m s )(2.08 s) = −11.06 m s
when the shot hits.
(c) x − x0 = v0 xt = (7.55 m s)(2.08 s) = 15.7 m .
(d) The initial and final heights are not the same.
(e) With y = 0 and v0 y as found above, Eq.(3.18) gives y0 = 1.81m .
(f) The graphs are sketched in Figure 3.20. 3-10 Chapter 3 EVALUATE: When the shot returns to its initial height, v y = −9.32 m/s . The shot continues to accelerate downward as it travels downward 1.81 m to the ground and the magnitude of v y at the ground is larger than
9.32 m/s. Figure 3.20
3.21. IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take positive y to
be upward. The quarter moves in projectile motion, with ax = 0, and a y = − g . It travels vertically for the time it takes it to travel horizontally 2.1 m.
v0 x = v0 cos α 0 = (6.4 m/s)cos60°
v0 x = 3.20 m/s
v0 y = v0 sin α 0 = (6.4 m/s)sin 60°
v0 y = 5.54 m/s
Figure 3.21
(a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels through the
air:
t = ?, x − x0 = 2.1 m, v0 x = 3.2 m/s, ax = 0 x − x0 = v0 xt + 1 axt 2 = v0 xt , since ax = 0
2
x − x0
2.1 m
=
= 0.656 s
v0 x
3.2 m/s
SET UP: Now find the vertical displacement of the quarter after this time:
y − y0 = ?, a y = −9.80 m/s 2 , v0 y = +5.54 m/s, t = 0.656 s
EXECUTE: t= y − y0 + v0 yt + 1 a yt 2
2
EXECUTE:
(b) SET UP: y − y0 = (5.54 m/s)(0.656 s) + 1 (−9.80 m/s 2 )(0.656 s) 2 = 3.63 m − 2.11 m = 1.5 m.
2
v y = ?, t = 0.656 s, a y = −9.80 m/s 2 , v0 y = +5.54 m/s v y = v0 y + a y t
EXECUTE:
EVALUATE: 3.22. v y = 5.54 m/s + (−9.80 m/s 2 )(0.656 s) = −0.89 m/s. !
The minus sign for v y indicates that the y-component of v is downward. At this point the quarter has passed through the highest point in its path and is on its way down. The horizontal range if it returned to its
original height (it doesn’t!) would be 3.6 m. It reaches its maximum height after traveling horizontally 1.8 m, so at
x − x0 = 2.1 m it is on its way down.
IDENTIFY: Use the analysis of Example 3.10.
d
SET UP: From Example 3.10, t =
and ydart = (v0 sin α 0 )t − 1 gt 2 .
2
v0 cos α 0
EXECUTE: Substituting for t in terms of d in the expression for ydart gives
⎛
⎞
gd
ydart = d ⎜ tan α 0 − 2
⎟.
2v0 cos 2 α 0 ⎠
⎝ Using the given values for d and α 0 to express this as a function of v0 ,
⎛
26.62 m 2 s 2 ⎞
y = (3.00 m) ⎜ 0.90 −
⎟.
2
v0
⎝
⎠
(a) v0 = 12.0 m/s gives y = 2.14 m .
(b) v0 = 8.0 m/s gives y = 1.45 m . Motion in Two or Three Dimensions 3-11 3.23. (c) v0 = 4.0 m/s gives y = −2.29 m . In this case, the dart was fired with so slow a speed that it hit the ground before
traveling the 3-meter horizontal distance.
EVALUATE: For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c)
the dart moves in a parabola and returns to the ground before it reaches the x-coordinate of the monkey.
IDENTIFY: Take the origin of coordinates at the roof and let the + y -direction be upward. The rock moves in projectile motion, with ax = 0 and a y = − g . Apply constant acceleration equations for the x and y components of
the motion.
SET UP: v0 x = v0 cos α 0 = 25.2 m/s
v0 y = v0 sin α 0 = 16.3 m/s Figure 3.23a
(a) At the maximum height v y = 0. a y = −9.80 m/s 2 , v y = 0, v0 y = +16.3 m/s, y − y0 = ?
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: y − y0 = 2
2
v y − v0 y 2a y = 0 − (16.3 m/s) 2
= +13.6 m
2(−9.80 m/s 2 ) (b) SET UP: Find the velocity by solving for its x and y components.
vx = v0 x = 25.2 m/s (since ax = 0 ) v y = ?, a y = −9.80 m/s 2 , y − y0 = −15.0 m (negative because at the ground the rock is below its initial position),
v0 y = 16.3 m/s
2
2
v y = v0 y + 2a y ( y − y0 )
2
v y = − v0 y + 2a y ( y − y0 ) ( v y is negative because at the ground the rock is traveling downward.) EXECUTE: v y = − (16.3 m/s) 2 + 2(−9.80 m/s 2 )( −15.0 m) = −23.7 m/s 2
2
Then v = vx + v y = (25.2 m/s) 2 + (−23.7 m/s) 2 = 34.6 m/s. (c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air:
t = ?, v y = −23.7 m/s (from part (b)), a y = −9.80 m/s 2 , v0 y = +16.3 m/s
EXECUTE: t= v y − v0 y
ay = −23.7 m/s − 16.3 m/s
= +4.08 s
−9.80 m/s 2 SET UP: Can use this t to calculate the horizontal range:
t = 4.08 s, v0 x = 25.2 m/s, ax = 0, x − x0 = ?
EXECUTE: x − x0 = v0 xt + 1 axt 2 = (25.2 m/s)(4.08 s) + 0 = 103 m
2 (d) Graphs of x versus t, y versus t, vx versus t, and v y versus t: Figure 3.23b 3-12 Chapter 3 EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally.
With v0 y = +16.3 m/s the time it takes the rock to return to the level of the roof ( y = 0) is t = 2v0 y / g = 3.33 s. The
3.24. time in the air is greater than this because the rock travels an additional 15.0 m to the ground.
IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m
horizontally in 3.00 s.
SET UP: Let + y be upward. ax = 0 , a y = −9.80 m/s 2 . v0 x = v0 cosθ 0 , v0 y = v0 sin θ 0 .
EXECUTE: (a) x − x0 = v0 xt + 1 axt 2 gives x − x0 = v0 (cosθ 0 )t and cosθ 0 =
2 45.0 m
= 0.600 ; θ 0 = 53.1°
(25.0 m/s)(3.00 s) 2
2
(b) At the highest point vx = v0 x = (25.0 m/s)cos53.1° = 15.0 m/s , v y = 0 and v = vx + v y = 15.0 m/s . At all points in the motion, a = 9.80 m/s 2 downward.
(c) Find y − y0 when t = 3.00s :
y − y0 = v0 yt + 1 a yt 2 = (25.0 m/s)(sin53.1°)(3.00 s) + 1 (−9.80 m/s 2 )(3.00 s)2 = 15.9 m
2
2
vx = v0 x = 15.0 m/s , v y = v0 y + a yt = (25.0 m/s)(sin53.1°) − (9.80m/s 2 )(3.00 s) = −9.41 m/s , and
v = vx 2 + v y 2 = (15.0 m/s) 2 + (−9.41 m/s) 2 = 17.7 m/s
EVALUATE: The acceleration is the same at all points of the motion. It takes the water
v
20.0 m/s
t = − 0y = −
= 2.04 s to reach its maximum height. When the water reaches the building it has passed
ay
−9.80 m/s 2
3.25. its maximum height and its vertical component of velocity is downward.
IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the balloon.
Use constant acceleration equations for the x and y components of its motion. Take + y to be upward.
EXECUTE: (a) Use the vertical motion of the rock to find the initial height.
t = 6.00 s, v0 y = +20.0 m/s, a y = +9.80 m/s 2 , y − y0 = ?
y − y0 = v0 yt + 1 a yt 2 gives y − y0 = 296 m
2
(b) In 6.00 s the balloon travels downward a distance y − y0 = (20.0 s)(6.00 s) = 120 m. So, its height above
ground when the rock hits is 296 m − 120 m = 176 m.
(c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is (176 m) 2 + (90 m) 2 = 198 m from the basket when it hits the ground.
(d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the basket.
Just before the rock hits the ground, its vertical component of velocity is v y = v0 y + a yt = 3.26. 20.0 m/s + (9.80 m/s 2 )(6.00 s) = 78.8 m/s, downward, relative to the ground. The basket is moving downward at
20.0 m/s, so relative to the basket the rock has downward component of velocity 58.8 m/s.
(e) horizontal: 15.0 m/s; vertical: 78.8 m/s
EVALUATE: The rock has a constant horizontal velocity and accelerates downward
IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have
y − y0 = 25.0 m when it has x − x0 = 60.0 m .
SET UP:
EXECUTE: Let + y be upward. ax = 0 , a y = − g . v0 x = v0 cos 43° , v0 y = v0 sin 43° .
(a) horizontal motion: x − x0 = v0 xt so t = 60.0 m
.
(v0 cos 43°)t vertical motion: y − y0 = v0 yt + 1 a yt 2 gives 25.0m = (v0 sin 43.0°)t + 1 ( −9.80m/s 2 )t 2 .
2
2
Solving these two simultaneous equations for v0 and t gives v0 = 3.26 m/s and t = 2.51 s .
(b) v y when shell reaches cliff: v y = v0 y + a yt = (32.6 m/s) sin 43.0° − (9.80 m/s 2 )(2.51 s) = −2.4 m/s
The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff.
v
EVALUATE: The shell reaches its maximum height at t = − 0 y = 2.27 s , which confirms that at t = 2.51 s it has
ay
3.27. passed its maximum height and is on its way down when it strikes the edge of the cliff.
IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity of the
airplane. Motion in Two or Three Dimensions 3-13 Take + y to be upward. ax = 0 , a y = − g . SET UP: EXECUTE: Use the vertical motion to find the time it takes the suitcase to reach the ground:
v0 y = v0 sin23°, a y = −9.80 m/s 2 , y − y0 = −114 m, t = ? y − y0 = v0 yt + 1 a yt 2 gives t = 9.60 s .
2 The distance the suitcase travels horizontally is x − x0 = v0 x = (v0 cos23.0°)t = 795 m .
EVALUATE:
3.28. An object released from rest at a height of 114 m strikes the ground at t = 2( y − y0 )
= 4.82 s . The
−g suitcase is in the air much longer than this since it initially has an upward component of velocity.
IDENTIFY: Determine how arad depends on the rotational period T.
4π 2 R
.
T2
EXECUTE: For any item in the washer, the centripetal acceleration will be inversely proportional to the square of
the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that
arad = SET UP: 3.29. the new period T ′ is given in terms of the previous period T by T ′ = T / 3 .
EVALUATE: The rotational period must be decreased in order to increase the rate of rotation and therefore
increase the centripetal acceleration.
IDENTIFY: Apply Eq. (3.30).
SET UP: T = 24 h .
4π 2 (6.38 × 106 m)
= 0.034 m/s 2 = 3.4 × 10−3 g .
EXECUTE: (a) arad =
((24 h)(3600 s/h)) 2
4π 2 (6.38 × 106 m)
= 5070 s =1.4 h.
9.80 m/s 2
1
EVALUATE: arad is proportional to 1/ T 2 , so to increase arad by a factor of
= 294 requires that T be
3.4 × 10−3
1
24 h
multiplied by a factor of
.
= 1.4 h .
294
294
IDENTIFY: Each blade tip moves in a circle of radius R = 3.40 m and therefore has radial acceleration
arad = v 2 / R .
(b) Solving Eq. (3.30) for the period T with arad = g , T = 3.30. 550 rev/min = 9.17 rev/s , corresponding to a period of T = SET UP: (a) v = EXECUTE: 1
= 0.109 s .
9.17 rev/s 2π R
= 196 m/s .
T v2
= 1.13 × 104 m/s 2 = 1.15 × 103 g .
R
4π 2 R
EVALUATE: arad =
gives the same results for arad as in part (b).
T2
IDENTIFY: Apply Eq.(3.30).
SET UP: R = 7.0 m . g = 9.80 m/s 2 .
(b) arad = 3.31. (a) Solving Eq. (3.30) for T in terms of R and arad , EXECUTE: T = 4π R / arad = 4π 2 (7.0 m)/(3.0)(9.80 m/s 2 ) = 3.07 s .
2 (b) arad = 10 g gives T = 1.68 s .
EVALUATE:
3.32. IDENTIFY:
SET UP: When arad increases, T decreases.
Each planet moves in a circular orbit and therefore has acceleration arad = v 2 / R . The radius of the earth’s orbit is r = 1.50 × 1011 m and its orbital period is T = 365 days = 3.16 × 107 s . For Mercury, r = 5.79 × 1010 m and T = 88.0 days = 7.60 × 106 s .
EXECUTE: (a) v = 2 r
= 2.98 × 104 m/s
T v2
= 5.91 × 10−3 m/s 2 .
r
(c) v = 4.79 × 104 m/s , and arad = 3.96 × 10−2 m/s 2 .
(b) arad = 3-14 3.33. 3.34. Chapter 3 EVALUATE: Mercury has a larger orbital velocity and a larger radial acceleration than earth.
IDENTIFY: Uniform circular motion.
!
dv
!
SET UP: Since the magnitude of v is constant. vtan =
= 0 and the resultant acceleration is equal to the radial
dt
component. At each point in the motion the radial component of the acceleration is directed in toward the center of
the circular path and its magnitude is given by v 2 / R.
v 2 (7.00 m/s) 2
EXECUTE: (a) arad = =
= 3.50 m/s 2 , upward.
R
14.0 m
(b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the
circle is downward. The acceleration at this point in the motion is 3.50 m/s 2 , downward.
(c) SET UP: The time to make one rotation is the period T, and the speed v is the distance for one revolution
divided by T.
2π R
2π R 2π (14.0 m)
EXECUTE: v =
so T =
=
= 12.6 s
T
v
7.00 m/s
EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in the
!
motion directed toward the center of the circular path. The acceleration is perpendicular to v and is nonzero
!
because the direction of v changes.
IDENTIFY: The acceleration is the vector sum of the two perpendicular components, arad and atan .
!
SET UP: atan is parallel to v and hence is associated with the change in speed; atan = 0.500 m/s 2 .
EXECUTE: (a) arad = v 2 / R = (3 m/s) 2 /(14 m) = 0.643 m/s 2 . a = ((0.643 m/s 2 ) 2 + (0.5 m/s 2 ) 2 )1/ 2 = 0.814 m/s 2 , 37.9° to the right of vertical.
(b) The sketch is given in Figure 3.34. Figure 3.34 v2
. The speed in rev/s is 1/ T ,
R
where T is the period in seconds (time for 1 revolution). The speed v increases with R along the length of his body
but all of him rotates with the same period T.
SET UP: For his head R = 8.84 m and for his feet R = 6.84 m .
IDENTIFY: Each part of his body moves in uniform circular motion, with arad = EXECUTE: 3.35. (a) v = Rarad = (8.84 m)(12.5)(9.80 m/s 2 ) = 32.9 m/s (b) Use arad = T = 2π 4π 2 R
. Since his head has arad = 12.5 g and R = 8.84 m ,
T2 R
8.84 m
R 4π 2 (6.84 m)
= 2π
= 1.688 s . Then his feet have arad = 2 =
= 94.8 m/s 2 = 9.67 g .
12.5(9.80 m/s 2 )
(1.688 s) 2
arad
T The difference between the acceleration of his head and his feet is 12.5 g − 9.67 g = 2.83 g = 27.7 m/s 2 .
(c) 1
1
=
= 0.592 rev/s = 35.5 rpm
T 1.69 s His feet have speed v = Rarad = (6.84 m)(94.8 m/s 2 ) = 25.5 m/s
!
!
IDENTIFY: The relative velocities are vS/F , the velocity of the scooter relative to the flatcar, vS/G , the scooter
!
!
!
!
relative to the ground and vF/G , the flatcar relative to the ground. vS/G = vS/F + vF/G . Carry out the vector addition by
drawing a vector addition diagram.
!
!
!
!
!
SET UP: vS/F = vS/G − vF/G . vF/G is to the right, so −vF/G is to the left.
EXECUTE: In each case the vector addition diagram gives
EVALUATE: 3.36. Motion in Two or Three Dimensions 3-15 3.37. (a) 5.0 m/s to the right
(b) 16.0 m/s to the left
(c) 13.0 m/s to the left.
EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative to the
!
!
flatcar, since in that case the two velocities vS/F and vF/G are in the same direction and their magnitudes add.
IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length divided
by the velocity of the woman relative to the ground.
SET UP: Let W stand for the woman, G for the ground, and S for the sidewalk. Take the positive direction to be
the direction in which the sidewalk is moving.
The velocities are vW/G (woman relative to the ground), vW/S (woman relative to the sidewalk), and vS/G (sidewalk
relative to the ground).
Eq.(3.33) becomes vW/G = vW/S + vS/G . The time to reach the other end is given by t =
EXECUTE: distance traveled relative to ground
vW/G (a) vS/G = 1.0 m/s vW/S = +1.5 m/s
vW/G = vW/S + vS/G = 1.5 m/s + 1.0 m/s = 2.5 m/s.
t= 35.0 m 35.0 m
=
= 14 s.
vW/G
2.5 m/s (b) vS/G = 1.0 m/s vW/S = −1.5 m/s 3.38. 3.39. 3.40. vW/G = vW/S + vS/G = −1.5 m/s + 1.0 m/s = −0.5 m/s. (Since vW/G now is negative, she must get on the moving
sidewalk at the opposite end from in part (a).)
−35.0 m −35.0 m
=
= 70 s.
t=
vW/G
−0.5 m/s
EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion of the
sidewalk.
IDENTIFY: Calculate the rower’s speed relative to the shore for each segment of the round trip.
SET UP: The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream.
EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths
of an hour (45.0 min).
1.5 km
1.5 km
+
= 1.47 h = 88.2 min.
The total time the rower takes is
6.8 km/h 1.2 km/h
EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h. The
rower spends more time at the slower speed.
IDENTIFY: Apply the relative velocity relation.
!
!
SET UP: The relative velocities are vC/E , the canoe relative to the earth, vR/E , the velocity of the river relative to
!
the earth and vC/R , the velocity of the canoe relative to the river.
!
!
!
!
!
!
!
EXECUTE: vC/E = vC/R + vR/E and therefore vC/R = vC/E − vR/E . The velocity components of vC/R are −0.50 m/s + (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south, for a velocity relative to the river of 0.36 m/s, at
52.5° south of west.
EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of the canoe
relative to the earth.
IDENTIFY: Use the relation that relates the relative velocities.
!
SET UP: The relative velocities are the velocity of the plane relative to the ground, vP/G , the velocity of the plane
!
!
!
!
relative to the air, vP/A , and the velocity of the air relative to the ground, vA/G . vP/G must due west and vA/G must
!
!
!
be south. vA/G = 80 km/h and vP/A = 320 km/h . vP/G = vP/A + vA/G . The relative velocity addition diagram is given
in Figure 3.40.
v
80 km/h
EXECUTE: (a) sin θ = A/G =
and θ = 14° , north of west.
vP/A 320 km/h
2
2
(b) vP/G = vP/A − vA/G = (320 km/h) 2 − (80.0 km/h) 2 = 310 km/h . 3-16 Chapter 3 EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward component and
also a component that is northward, opposite to the wind direction. Figure 3.40
3.41. IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time displacement)
depends on his velocity relative to the earth so we must solve for this velocity.
(a) SET UP: View the motion from above. The velocity vectors in the problem are:
!
vM/E , the velocity of the man relative to the earth
!
vW/E , the velocity of the water relative to the earth
!
vM/W , the velocity of the man relative to the water
The rule for adding these velocities is
!
!
!
vM/E = vM/W + vW/E
Figure 3.41a !
!
The problem tells us that vW/E has magnitude 2.0 m/s and direction due south. It also tells us that vM/W has
magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.41b
This diagram shows the vector addition
!
!
!
vM/E = vM/W + vW/E
!
!
and also has vM/W and vW/E in their
specified directions. Note that the
vector diagram forms a right triangle.
Figure 3.41b
2
2
2
The Pythagorean theorem applied to the vector addition diagram gives vM/E = vM/W + vW/E . vM/W 4.2 m/s
=
= 2.10; θ = 65°; or
vW/E 2.0 m/s
φ = 90° − θ = 25°. The velocity of the man relative to the earth has magnitude 4.7 m/s and direction 25° S of E.
(b) This requires careful thought. To cross the river the man must travel 800 m due east relative to the earth. The
!
man’s velocity relative to the earth is vM/E . But, from the vector addition diagram the eastward component of vM/E
EXECUTE: 2
2
vM/E = vM/W + vW/E = (4.2 m/s) 2 + (2.0 m/s) 2 = 4.7 m/s tan θ = equals vM/W = 4.2 m/s.
Thus t = x − x0 800 m
=
= 190 s.
vx
4.2 m/s !
(c) The southward component of vM/E equals vW/E = 2.0 m/s. Therefore, in the 190 s it takes him to cross the river
the distance south the man travels relative to the earth is
y − y0 = v yt = (2.0 m/s)(190 s) = 380 m. 3.42. EVALUATE: If there were no current he would cross in the same time, (800 m) /(4.2 m/s) = 190 s. The current
carries him downstream but doesn’t affect his motion in the perpendicular direction, from bank to bank.
IDENTIFY: Use the relation that relates the relative velocities.
!
!
SET UP: The relative velocities are the water relative to the earth, vW/E , the boat relative to the water, vB/W , and
!
!
!
the boat relative to the earth, vB/E . vB/E is due east, vW/E is due south and has magnitude 2.0 m/s. vB/W = 4.2 m/s .
!
!
!
vB/E = vB/W + vW/E . The velocity addition diagram is given in Figure 3.42. Motion in Two or Three Dimensions 3-17 EXECUTE: !
v
2.0 m/s
(a) Find the direction of vB/W . sin θ = W/E =
. θ = 28.4° , north of east.
vB/W 4.2 m/s 2
2
(b) vB/E = vB/W − vW/E = (4.2 m/s) 2 − (2.0 m/s) 2 = 3.7 m/s 800 m 800 m
=
= 216 s .
vB/E
3.7 m/s
EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.41. In the direction straight
across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s.
(c) t = Figure 3.42
3.43. IDENTIFY: Relative velocity problem in two dimensions.
!
!
(a) SET UP: vP/A is the velocity of the plane relative to the air. The problem states that vP/A has magnitude
35 m/s and direction south.
!
!
vA/E is the velocity of the air relative to the earth. The problem states that vA/E is to the southwest ( 45° S of W)
and has magnitude 10 m/s.
!
!
!
The relative velocity equation is vP/E = vP/A + vA/E . Figure 3.43a
EXECUTE: (b) (vP/A ) x = 0, (vP/A ) y = −35 m/s (vA/E ) x = −(10 m/s)cos 45° = −7.07 m/s,
(vA/E ) y = −(10 m/s)sin 45° = −7.07 m/s
(vP/E ) x = (vP/A ) x + (vA/E ) x = 0 − 7.07 m/s = −7.1 m/s
(vP/E ) y = (vP/A ) y + (vA/E ) y = −35 m/s − 7.07 m/s = −42 m/s
(c) vP/E = (vP/E ) 2 + (vP/E ) 2
x
y
vP/E = (−7.1 m/s) 2 + (−42 m/s) 2 = 43 m/s
tan φ = (vP/E ) x −7.1
=
= 0.169
(vP/E ) y −42 φ = 9.6°; ( 9.6° west of south)
Figure 3.43b
EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition must be
done using components. The wind adds both southward and westward components to the velocity of the plane
relative to the ground. 3-18 3.44. Chapter 3 IDENTIFY: Use Eqs.(2.17) and (2.18).
SET UP: At the maximum height v y = 0 .
EXECUTE: (a) vx = v0 x + α γ α β γ t 3 , v y = v0 y + β t − t 2 , and x = v0 xt + t 4 , y = v0 y t + t 2 − t 3 .
3
2
12
2
6 γ (b) Setting v y = 0 yields a quadratic in t , 0 = v0 y + β t − t 2 , which has as the positive solution
2
1⎡
t = β + β 2 + 2v0γ ⎤ = 13.59 s . Using this time in the expression for y(t) gives a maximum height of 341 m.
⎦
γ⎣
(c) The path of the rocket is sketched in Figure 3.44.
(d) y = 0 gives 0 = v0 yt + β 2 γ γ β t 2 − t 3 and t 2 − t − v0 y = 0 . The positive solution is t = 20.73 s . For this t,
6
6
2 x = 3.85 × 104 m .
EVALUATE: The graph in part (c) shows the path is not symmetric about the highest point and the time to return
to the ground is less than twice the time to the maximum height. Figure 3.44
3.45. IDENTIFY: !
!
! dr
! dv
and a =
v=
dt
dt dn
(t ) = nt n −1 . At t = 1.00 s , ax = 4.00 m/s 2 and a y = 3.00 m/s 2 . At t = 0 , x = 0 and y = 50.0 m .
dt
dx
dv
EXECUTE: (a) vx =
= 2 Bt . ax = x = 2 B , which is independent of t. ax = 4.00 m/s 2 gives B = 2.00 m/s 2 .
dt
dt
dv y
dy
= 3Dt 2 . a y =
= 6 Dt . a y = 3.00 m/s 2 gives D = 0.500 m/s 2 . x = 0 at t = 0 gives A = 0 . y = 50.0 m at
vy =
dt
dt
t = 0 gives C = 50.0 m .
!
!
ˆ
(b) At t = 0 , v = 0 and v = 0 , so v = 0 . At t = 0 , a = 2 B = 4.00 m/s 2 and a = 0 , so a = (4.00 m/s 2 )i .
SET UP: x x y y (c) At t = 10.0 s , vx = 2(2.00 m/s )(10.0 s) = 40.0 m/s and v y = 3(0.500 m/s )(10.0 s) 2 = 150 m/s .
2 3 2
2
v = vx + v y = 155 m/s . 3.46. !
ˆ
(d) x = (2.00 m/s 2 )(10.0 s) 2 = 200 m , y = 50.0 m + (0.500 m/s 3 )(10.0 s)3 = 550 m . r = (200 m)i + (550 m) ˆ .
j
EVALUATE: The velocity and acceleration vectors as functions of time are
!
!
ˆ
ˆ
v (t ) = (2 Bt )i + (3Dt 2 ) ˆ and a (t ) = (2 B) i + (6 Dt ) ˆ . The acceleration is not constant.
j
j
!
t!
!!
dv
IDENTIFY: r = r0 + ∫ v (t )dt and a =
.
0
dt
SET UP: At t = 0 , x0 = 0 and y0 = 0 .
!
!
βˆγ
ˆ
(a) Integrating, r = (α t − t 3 )i + ( t 2 ) ˆ . Differentiating, a = (−2 β t ) i + γ ˆ .
j
j
3
2
(b) The positive time at which x = 0 is given by t 2 = 3α β . At this time, the y-coordinate is
EXECUTE: γ
3αγ 3(2.4 m/s)(4.0 m/s 2 )
=
= 9.0 m .
y = t2 =
2
2β
2(1.6 m/s3 )
EVALUATE: The acceleration is not constant. Motion in Two or Three Dimensions 3-19 3.47. IDENTIFY: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the incline
determines the initial velocity and initial position for the projectile motion.
2
2
SET UP: For motion along the incline let + x be directed up the incline. vx = v0 x + 2ax ( x − x0 ) gives vx = 2(1.25 m/s 2 )(200 m) = 22.36 m/s . When the projectile motion begins the rocket has v0 = 22.36 m/s at
35.0° above the horizontal and is at a vertical height of (200.0 m)sin 35.0° = 114.7 m . For the projectile motion
let + x be horizontal to the right and let + y be upward. Let y = 0 at the ground. Then y0 = 114.7 m ,
v0 x = v0 cos35.0° = 18.32 m/s , v0 y = v0 sin 35.0° = 12.83 m/s , ax = 0 , a y = −9.80 m/s 2 . Let x = 0 at point A, so
x0 = (200.0 m)cos35.0° = 163.8 m .
EXECUTE: y − y0 = 2
2
(a) At the maximum height v y = 0 . v y = v0 y + 2a y ( y − y0 ) gives 2
2
v y − v0 y 2a y = 0 − (12.83 m/s) 2
= 8.40 m and y = 114.7 m + 8.40 m = 123 m . The maximum height above
2( −9.80 m/s 2 ) ground is 123 m.
(b) The time in the air can be calculated from the vertical component of the projectile motion: y − y0 = −114.7 m ,
v0 y = 12.83 m/s , a y = −9.80 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives (4.90 m/s 2 )t 2 − (12.83 m/s)t − 114.7 m . The
2 3.48. ( ) 1
12.83 ± (12.83) 2 + 4(4.90)(114.7) s . The positive root is t = 6.32 s . Then
9.80
x − x0 = v0 xt + 1 axt 2 = (18.32 m/s)(6.32 s) = 115.8 m and x = 163.8 m + 115.8 m = 280 m . The horizontal range of
2
the rocket is 280 m.
EVALUATE: The expressions for h and R derived in Example 3.8 do not apply here. They are only for a projectile
fired on level ground.
IDENTIFY: The person moves in projectile motion. Use the results in Example 3.8 to determine how T, h and D
depend on g and set up a ratio.
2v sin α 0
v 2 sin 2 α 0
, the maximum height is h = 0
and the
SET UP: From Example 3.8, the time in the air is t = 0
g
2g
quadratic formula gives t = horizontal range (called D in the problem) is D = 2
v0 sin 2α 0
. The person has the same v0 and α 0 on Mars as on
g the earth.
EXECUTE: hg = ⎛g ⎞
⎛ gE ⎞
tg = 2v0 sin α 0 , which is constant, so tE g E = tM g M . tM = ⎜ E ⎟ tE = ⎜
⎟ tE = 2.64tE .
gM ⎠
⎝
⎝ 0.379 g E ⎠ 2
⎛g ⎞
v0 sin 2 α 0
2
, which is constant, so hE g E = hM g M . hM = ⎜ E ⎟ hE = 2.64hE . Dg = v0 sin 2α 0 , which is constant,
gM ⎠
2
⎝ ⎛g ⎞
so DE g E = DM g M . DM = ⎜ E ⎟ DE = 2.64 DE .
⎝ gM ⎠
EVALUATE: All three quantities are proportional to 1/ g so all increase by the same factor of g E / g M = 2.64 .
3.49. IDENTIFY:
SET UP:
EXECUTE: 3.50. The range for a projectile that lands at the same height from which it was launched is R = 2
v0 sin 2α
.
g The maximum range is for α = 45° .
Assuming α = 45° , and R = 50 m , v0 = gR = 22 m/s . EVALUATE: We have assumed that debris was launched at all angles, including the angle of 45° that gives
maximum range.
IDENTIFY: The velocity has a horizontal tangential component and a vertical component. The vertical component
v2
of acceleration is zero and the horizontal component is arad = x
R
SET UP: Let + y be upward and + x be in the direction of the tangential velocity at the instant we are
considering. 3-20 Chapter 3 EXECUTE: (a) The bird’s tangential velocity can be found from vx = circumference 2 (8.00 m) 50.27 m
=
=
= 10.05 m/s
time of rotation
5.00 s
5.00 s Thus its velocity consists of the components vx = 10.05 m/s and v y = 3.00 m/s . The speed relative to the ground is
2
2
then v = vx + v y = 10.5 m/s . 3.51. (b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the horizontal direction, toward
v 2 (10.05 m/s) 2
the center of its spiral path–and has magnitude arad = x =
= 12.6 m/s 2 .
r
8.00 m
3.00 m/s
(c) Using the vertical and horizontal velocity components θ = tan −1
= 16.6° .
10.05 m/s
EVALUATE: The angle between the bird’s velocity and the horizontal remains constant as the bird rises.
IDENTIFY: Take + y to be downward. Both objects have the same vertical motion, with v0 y and a y = + g . Use constant acceleration equations for the x and y components of the motion.
SET UP: Use the vertical motion to find the time in the air:
v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 25 m, t = ?
EXECUTE: 3.52. y − y0 = v0 yt + 1 a yt 2 gives t = 2.259 s
2 During this time the dart must travel 90 m, so the horizontal component of its velocity must be
x − x0 90 m
v0 x =
=
= 40 m/s
t
2.25 s
EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle velocity
greater than 40 m/s.
IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air.
SET UP: Take + y upward. v0 x = 15.0 m/s , v0 y = +10.0 m/s , ax = 0 , a y = −9.80 m/s 2 .
EXECUTE: (a) Use the vertical motion to find the time in the air: y − y0 = v0 yt + 1 a yt 2 with y − y0 = −30.0 m
2 gives −30.0 m = (10.0 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives
t= ( ) 1
+10.0 ± ( −10.0) 2 − 4(4.9)( −30) s . The positive solution is t = 3.70 s . During this time she travels a
2(4.9) horizontal distance x − x0 = v0 xt + 1 axt 2 = (15.0 m/s)(3.70 s) = 55.5 m . She will land 55.5 m south of the point
2
where she drops from the helicopter and this is where the mats should have been placed.
(b) The x-t, y-t, vx -t and v y -t graphs are sketched in Figure 3.52.
2(30.0 m)
= 2.47 s .
9.80 m/s 2
She is in the air longer than this because she has an initial vertical component of velocity that is upward.
EVALUATE: If she had dropped from rest at a height of 30.0 m it would have taken her t = Figure 3.52
3.53. IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant acceleration
equations for the x and y components of motion. Motion in Two or Three Dimensions 3-21 SET UP: Take the origin of
coordinates at the point
where the canister is
released. Take + y to be
upward. The initial
velocity of the canister is
the velocity of the plane,
64.0 m/s in the
+ x-direction.
Figure 3.53 Use the vertical motion to find the time of fall:
t = ?, v0 y = 0, a y = −9.80 m/s 2 , y − y0 = −90.0 m (When the canister reaches the ground it is 90.0 m below the
origin.)
y − y0 = v0 yt + 1 a yt 2
2
EXECUTE: Since v0 y = 0, t = 2( y − y0 )
2(−90.0 m)
=
= 4.286 s.
−9.80 m/s 2
ay SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this time:
x − x0 = ?, ax − 0, v0 x = 64.0 m/s, 3.54. EXECUTE: x − x0 = v0t + 1 at 2 = (64.0 m/s)(4.286 s) + 0 = 274 m.
2
EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at
constant speed.
IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment
plus the distance the ship moves while the equipment is in the air.
SET UP: For the motion of the equipment take + x to be to the right and + y to be upwards. Then ax = 0 , a y = −9.80 m/s 2 , v0 x = v0 cos α 0 = 7.50 m/s and v0 y = v0 sin α 0 = 13.0 m/s . When the equipment lands in the front
of the ship, y − y0 = −8.75 m .
EXECUTE: Use the vertical motion of the equipment to find its time in the air: y − y0 = v0 yt + 1 a yt 2 gives
2 ) ( 1
13.0 ± (−13.0) 2 + 4(4.90)(8.75) s . The positive root is t = 3.21 s . The horizontal range of the
9.80
equipment is x − x0 = v0 xt + 1 axt 2 = (7.50 m/s)(3.21 s) = 24.1 m . In 3.21 s the ship moves a horizontal distance
2
(0.450 m/s)(3.21 s) = 1.44 m , so D = 24.1 m + 1.44 m = 25.5 m .
t= 2
v0 sin 2α 0
from Example 3.8 can't be used because the starting and ending points
g
of the projectile motion are at different heights.
IDENTIFY: Projectile motion problem.
Take the origin of
coordinates at the point
where the ball leaves the
bat, and take + y to be
upward.
v0 x = v0 cos α 0 EVALUATE:
3.55. The equation R = v0 y = v0 sin α 0 ,
but we don’t know v0 .
Figure 3.55 Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding
equation for the vertical displacement. The time t is the same for both components, so this will give us two
equations in two unknowns ( v0 and t). 3-22 Chapter 3 (a) SET UP: y-component:
a y = −9.80 m/s 2 , y − y0 = −0.9 m, v0 y = v0 sin 45° y − y0 = v0 y + t 1 a yt 2
2
EXECUTE: −0.9 m = (v0 sin 45°)t + 1 (−9.80 m/s 2 )t 2
2
SET UP: x-component:
ax = 0, x − x0 = 188 m, v0 x = v0 cos 45° x − x0 = v0 xt + 1 axt 2
2
EXECUTE: t= x − x0
188 m
=
v0 x
v0 cos 45° Put the expression for t from the x-component motion into the y-component equation and solve for v0 . (Note that
sin 45° = cos 45°. ) ⎛ 188 m ⎞
2 ⎛ 188 m ⎞
−0.9 m = (v0 sin 45°) ⎜
⎟ − (4.90 m/s ) ⎜
⎟
⎝ v0 cos 45° ⎠
⎝ v0 cos 45° ⎠ 2 2 ⎛ 188 m ⎞
4.90 m/s ⎜
⎟ = 188 m + 0.9 m = 188.9 m
⎝ v0 cos 45° ⎠
2 2 2
2
⎛ 188 m ⎞ 4.90 m/s
⎛ v0 cos 45° ⎞ 4.90 m/s
= 42.8 m/s
, v0 = ⎜
⎟
⎜
⎟=
188.9 m
⎝ cos 45° ⎠ 188.9 m
⎝ 188 m ⎠ (b) Use the horizontal motion to find the time it takes the ball to reach the fence:
SET UP: x-component:
x − x0 = 116 m, ax = 0, v0 x = v0 cos 45° = (42.8 m/s)cos 45° = 30.3 m/s, t = ? x − x0 = v0 xt + 1 axt 2
2
x − x0
116 m
=
= 3.83 s
v0 x
30.3 m/s
SET UP: Find the vertical displacement of the ball at this t:
y-component:
y − y0 = ?, a y = −9.80 m/s 2 , v0 y = v0 sin 45° = 30.3 m/s, t = 3.83 s
EXECUTE: t= y − y0 = v0 yt + 1 a yt 2
2
EXECUTE: y − y0 = (30.3 s)(3.83 s) + 1 (−9.80 m/s 2 )(3.83 s) 2
2 y − y0 = 116.0 m − 71.9 m = +44.1 m, above the point where the ball was hit. The height of the ball above the
ground is 44.1 m + 0.90 m = 45.0 m. It’s height then above the top of the fence is 45.0 m − 3.0 m = 42.0 m.
EVALUATE: With v0 = 42.8 m/s, v0 y = 30.3 m/s and it takes the ball 6.18 s to return to the height where it was 3.56. hit and only slightly longer to reach a point 0.9 m below this height. t = (188 m) /(v0 cos 45°) gives t = 6.21 s,
which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 = 94 m from
home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence
is reasonable.
IDENTIFY: The water moves in projectile motion.
SET UP: Let x0 = y0 = 0 and take + y to be positive. ax = 0 , a y = − g .
EXECUTE: The equations of motions are y = (v0 sin α )t − 1 gt 2 and x = (v0 cos α )t . When the water goes in the
2
tank for the minimum velocity, y = 2 D and x = 6 D . When the water goes in the tank for the maximum velocity, y = 2 D and x = 7 D . In both cases, sin α = cos α = 2 / 2.
To reach the minimum distance: 6 D = 2
2
v0t , and 2 D =
v0t − 1 gt 2 . Solving the first equation for t gives
2
2
2
2 ⎛ 6D 2 ⎞
6D 2
. Substituting this into the second equation gives 2 D = 6 D − 1 g ⎜
t=
2
⎜ v ⎟ . Solving this for v0 gives
⎟
v0
0
⎝
⎠
v0 = 3 gD . Motion in Two or Three Dimensions 3-23 To reach the maximum distance: 7 D = 2
2
v0t , and 2 D =
v0t − 1 gt 2 . Solving the first equation for t gives
2
2
2
2 t= ⎛ 7D 2 ⎞
7D 2
. Substituting this into the second equation gives 2 D = 7 D − 1 g ⎜
2
⎜ v ⎟ . Solving this for v0 gives
⎟
v0
0
⎝
⎠ v0 = 49 gD / 5 = 3.13 gD , which, as expected, is larger than the previous result. 3.57. EVALUATE: A launch speed of v0 = 6 gD = 2.45 gD is required for a horizontal range of 6D. The minimum
speed required is greater than this, because the water must be at a height of at least 2D when it reaches the front of
the tank.
IDENTIFY: The equations for h and R from Example 3.8 can be used.
v 2 sin 2 α 0
v 2 sin 2α 0
SET UP: h = 0
and R = 0
. If the projectile is launched straight up, α 0 = 90° .
2g
g
EXECUTE: (a) h = 2
v0
and v0 = 2 gh .
2g (b) Calculate α 0 that gives a maximum height of h when v0 = 2 2 gh . h = 8 gh sin 2 α 0
= 4h sin 2 α 0 . sin α 0 = 1 and
2
2g α 0 = 30.0° . (2
(c) R =
EVALUATE: 3.58. ) 2 2 gh sin 60.0°
g = 6.93h . 2
v0
2h
2h sin(2α 0 )
=
so R =
. For a given α 0 , R increases when h increases. For α 0 = 90° ,
2
g sin α 0
sin 2 α 0 R = 0 and for α 0 = 0° , h = 0 and R = 0 . For α 0 = 45° , R = 4h .
IDENTIFY: To clear the bar the ball must have a height of 10.0 ft when it has a horizontal displacement of 36.0 ft.
The ball moves as a projectile. When v0 is very large, the ball reaches the goal posts in a very short time and the
acceleration due to gravity causes negligible downward displacement.
SET UP: 36.0 ft = 10.97 m ; 10.0 ft = 3.048 m . Let + x be to the right and + y be upward, so ax = 0 , a y = − g ,
v0 x = v0 cos α 0 and v0 y = v0 sin α 0
EXECUTE: (a) The ball cannot be aimed lower than directly at the bar. tan α 0 = (b) x − x0 = v0 xt + 1 axt 2 gives t =
2 10.0 ft
and α 0 = 15.5° .
36.0 ft x − x0
x − x0
=
. Then y − y0 = v0 yt + 1 a yt 2 gives
2
v0 x
v0 cos α 0 ⎛ x − x0 ⎞ 1 ( x − x0 ) 2
1 ( x − x0 ) 2
.
y − y0 = (v0 sin α 0 ) ⎜
= ( x − x0 ) tan α 0 − g 2
⎟− g 2
2
2 v0 cos 2 α 0
⎝ v0 cos α 0 ⎠ 2 v0 cos α 0
v0 = ( x − x0 )
g
10.97 m
9.80 m/s 2
=
= 12.2 m/s
cos α 0 2[( x − x0 ) tan α 0 − ( y − y0 )] cos 45.0° 2[10.97 m − 3.048 m] EVALUATE: 3.59. With the v0 in part (b) the horizontal range of the ball is R = 2
v0 sin 2α 0
= 15.2 m = 49.9 ft . The ball
g reaches the highest point in its trajectory when x − x0 = R / 2 , so when it reaches the goal posts it is on its way
down.
IDENTIFY: Apply Eq.(3.27) and solve for x.
SET UP: The change in height is y = − h .
EXECUTE: (a) We get a quadratic equation in x, the solution to which is
x= 2
v0 cos α 0 ⎡ 2
2 gh ⎤ v0 cos α 0 ⎡
2
v sin α 0 + v0 sin 2 α 0 + 2 gh ⎤ .
⎢ tan α 0 + 2
⎥=
⎣0
⎦
g
v0 cos α 0 ⎦
g
⎣ If h = 0 , the square root reduces to v0 sin α 0 , and x = R . 3-24 Chapter 3 (b) The expression for x becomes x = (10.2 m)cos α 0 + [sin 2 α 0 + sin 2 α 0 + 0.98] . The graph of x as a function of α 0 is sketched in Figure 3.59. The angle α 0 = 90° corresponds to the projectile being launched straight up, and
there is no horizontal motion. If α 0 = 0 , the projectile moves horizontally until it has fallen the distance h.
(d) The graph shows that the maximum horizontal distance is for an angle less than 45° .
EVALUATE: For α 0 = 45° the x and y components of the initial velocity are equal. For α 0 < 45° the x component
of the initial velocity is less than the y component. Height comes from the initial position and less vertical
component of initial velocity is needed for the maximum range. Figure 3.59
3.60. IDENTIFY: The snowball moves in projectile motion. In part (a) the vertical motion determines the time in the air.
In part (c), find the height of the snowball above the ground after it has traveled horizontally 4.0 m.
SET UP: Let + y be downward. ax = 0 , a y = +9.80 m/s 2 . v0 x = v0 cosθ 0 = 5.36 m/s , v0 y = v0 sin θ 0 = 4.50 m/s .
EXECUTE: (a) Use the vertical motion to find the time in the air: y − y0 = v0 yt + 1 a yt 2 with y − y0 = 14.0 m gives
2 14.0 m = (4.50 m/s)t + (4.9 m/s 2 )t 2 . The quadratic formula gives t = ( ) 1
−4.50 ± (4.50) 2 − 4(4.9)( −14.0) s .
2(4.9) The positive root is t = 1.29 s . Then x − x0 = v0 xt + 1 axt 2 = (5.36 m/s)(1.29 s) = 6.91 m .
2
(b) The x-t, y-t, vx -t and v y -t graphs are sketched in Figure 3.60.
(c) x − x0 = v0 xt + 1 axt 2 gives t =
2 x − x0
4.0 m
=
= 0.746 s . In this time the snowball travels downward a
v0 x
5.36 m/s distance y − y0 = v0 yt + 1 a y t 2 = 6.08 m and is therefore 14.0 m − 6.08 m = 7.9 m above the ground. The snowball
2
passes well above the man and doesn’t hit him.
EVALUATE: If the snowball had been released from rest at a height of 14.0 m it would have reached the ground
2(14.0 m)
in t =
= 1.69 s . The snowball reaches the ground in a shorter time than this because of its initial
9.80 m/s 2
downward component of velocity. Figure 3.60
3.61. (a) IDENTIFY and SET UP: Use the equation derived in Example 3.8:
⎛ 2v sin α 0 ⎞
R = (v0 cos α 0 ) ⎜ 0
⎟
g
⎝
⎠ Motion in Two or Three Dimensions 3-25 Call the range R1 when the angle is α 0 and R2 when the angle is 90° − α .
⎛ 2v sin α 0 ⎞
R1 = (v0 cos α 0 ) ⎜ 0
⎟
g
⎝
⎠
⎛ 2v sin(90° − α 0 ) ⎞
R2 = (v0 cos(90° − α 0 )) ⎜ 0
⎟
g
⎝
⎠
The problem asks us to show that R1 = R2 .
EXECUTE: We can use the trig identities in Appendix B to show:
cos(90° − α 0 ) = cos(α 0 − 90°) = sin α 0
sin(90° − α 0 ) = − sin(α 0 − 90°) = −( − cos α 0 ) = + cos α 0
⎛ 2v cos α 0 ⎞
⎛ 2v0 sin α 0 ⎞
Thus R2 = (v0 sin α 0 ) ⎜ 0
⎟ = (v0 cos α 0 ) ⎜
⎟ = R1.
g
g
⎝
⎠
⎝
⎠
2
v0 sin 2α 0
Rg (0.25 m)(9.80 m/s 2 )
.
so sin 2α 0 = 2 =
g
v0
(2.2 m/s) 2
This gives α = 15° or 75°.
2
EVALUATE: R = (v0 sin 2α 0 ) / g , so the result in part (a) requires that sin 2 (2α 0 ) = sin 2 (180° − 2α 0 ), which is true. (b) R = 3.62. (Try some values of α 0 and see!)
IDENTIFY: Mary Belle moves in projectile motion.
SET UP: Let + y be upward. ax = 0 , a y = − g .
EXECUTE: (a) Eq.(3.27) with x = 8.2 m , y = 6.1 m and α 0 = 53° gives v0 = 13.8 m/s . (b) When she reached Joe Bob, t = 8.2 m
= 0.9874 s . vx = v0 x = 8.31 m/s and v y = v0 y + a yt = +1.34 m/s .
v0 cos53° v = 8.4 m/s , at an angle of 9.16° .
(c) The graph of vx (t ) is a horizontal line. The other graphs are sketched in Figure 3.62.
(d) Use Eq. (3.27), which becomes y = (1.327) x − (0.071115 m −1 ) x 2 . Setting y = −8.6 m gives x = 23.8 m as the
positive solution. Figure 3.62
3.63. (a) IDENTIFY: Projectile motion.
Take the origin of coordinates at the top
of the ramp and take + y to be upward.
The problem specifies that the object is
displaced 40.0 m to the right when it is
15.0 m below the origin.
Figure 3.63 We don’t know t, the time in the air, and we don’t know v0 . Write down the equations for the horizontal and
vertical displacements. Combine these two equations to eliminate one unknown.
SET UP: y-component:
y − y0 = −15.0 m, a y = −9.80 m/s 2 , v0 y = v0 sin 53.0°
y − y0 = v0 yt + 1 a yt 2
2
EXECUTE: −15.0 m = (v0 sin 53.0°)t − (4.90 m/s 2 )t 2 3-26 Chapter 3 SET UP: x-component:
x − x0 = 40.0 m, ax = 0, v0 x = v0 cos53.0° x − x0 = v0 xt + 1 axt 2
2
EXECUTE: 40.0 m = (v0t )cos53.0° 40.0 m
= 66.47 m.
cos53.0°
Use this to replace v0t in the first equation:
The second equation says v0t = −15.0 m = (66.47 m)sin 53° − (4.90 m/s 2 )t 2
t= (66.46 m)sin 53° + 15.0 m
68.08 m
=
= 3.727 s.
2
4.90 m/s
4.90 m/s 2 Now that we have t we can use the x-component equation to solve for v0 :
v0 =
EVALUATE: 40.0 m
40.0 m
=
= 17.8 m/s.
t cos53.0° (3.727 s)cos53.0° Using these values of v0 and t in the y = y0 = v0 y + 1 a y t 2 equation verifies that y − y0 = −15.0 m.
2 (b) IDENTIFY: v0 = (17.8 m/s) / 2 = 8.9 m/s
This is less than the speed required to make it to the other side, so he lands in the river.
Use the vertical motion to find the time it takes him to reach the water:
SET UP: y − y0 = −100 m; v0 y = +v0 sin 53.0° = 7.11 m/s; a y = −9.80 m/s 2 y − y0 = v0 yt + 1 a yt 2 gives −100 = 7.11t − 4.90t 2
2
EXECUTE: ( 1
4.90t 2 − 7.11t − 100 = 0 and t = 9.80 7.11 ± (7.11) 2 − 4(4.90)(−100) ) t = 0.726 s ± 4.57 s so t = 5.30 s.
The horizontal distance he travels in this time is
x − x0 = v0 xt = (v0 cos53.0°)t = (5.36 m/s)(5.30 s) = 28.4 m. 3.64. He lands in the river a horizontal distance of 28.4 m from his launch point.
EVALUATE: He has half the minimum speed and makes it only about halfway across.
IDENTIFY: The rock moves in projectile motion.
SET UP: Let + y be upward. ax = 0 , a y = − g . Eqs.(3.22) and (3.23) give vx and v y .
EXECUTE: Combining equations 3.25, 3.22 and 3.23 gives
2
2
v 2 = v0 cos 2 α 0 + (v0 sin α 0 − gt ) 2 = v0 (sin 2 α 0 + cos 2 α 0 ) − 2v0 sin α 0 gt + ( gt ) 2 . 1
2
2
v 2 = v0 − 2 g (v0 sin α 0t − gt 2 ) = v0 − 2 gy , where Eq.(3.21) has been used to eliminate t in favor of y. For the case
2
of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y = − h
2
into the above expression, yielding v = v0 + 2 gh , which is independent of α 0 . EVALUATE: This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid
2
for any y, positive, negative or zero, as long as v0 − 2 gy > 0 .
3.65. IDENTIFY and SET UP: Take + y to be upward. The rocket moves with projectile motion, with v0 y = +40.0 m/s and v0 x = 30.0 m/s relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity.
EXECUTE: (a) v y = 0 (at maximum height), v0 y = +40.0 m/s, a y = −9.80 m/s 2 , y − y0 = ? 2
2
v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 81.6 m (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal
distance while the rocket is in the air and the rocket lands in the cart.
(c) Use the vertical motion of the rocket to find the time it is in the air.
v0 y = 40 m/s, a y = −9.80 m/s 2 , v y = −40 m/s, t = ? v y = v0 y + a yt gives t = 8.164 s
Then x − x0 = v0 xt = (30.0 m/s)(8.164 s) = 245 m. Motion in Two or Three Dimensions 3-27 (d) Relative to the ground the rocket has initial velocity components v0 x = 30.0 m/s and v0 y = 40.0 m/s, so it is traveling at 53.1° above the horizontal.
(e) (i) Figure 3.65a Relative to the cart, the rocket travels straight up and then straight down
(ii) Figure 3.65b 3.66. Relative to the ground the rocket travels in a parabola.
EVALUATE: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart.
IDENTIFY: The ball moves in projectile motion.
SET UP: The woman and ball travel for the same time and must travel the same horizontal distance, so for the
ball v0 x = 6.00 m/s .
v0 x 6.00 m/s
=
and θ 0 = 72.5° .
v0 20.0 m/s
(b) Relative to the ground the ball moves in a parabola. The ball and the runner have the same horizontal
component of velocity, so relative to the runner the ball has only vertical motion. The trajectories as seen by each
observer are sketched in Figure 3.66.
EVALUATE: The ball could be thrown with a different speed, so long as the angle at which it was thrown was
adjusted to keep v0 x = 6.00 m/s .
EXECUTE: (a) v0 x = v0cosθ 0 . cosθ 0 = Figure 3.66
3.67. IDENTIFY: The boulder moves in projectile motion.
SET UP: Take + y downward. v0 x = v0 , v0 y = 0 . ax = 0 , a y = +9.80 m/s 2 .
(a) Use the vertical motion to find the time for the boulder to reach the level of the lake:
2( y − y0 )
2(20 m)
y − y0 = v0 yt + 1 a yt 2 with y − y0 = +20 m gives t =
=
= 2.02 s . The rock must travel
2
ay
9.80 m/s 2 EXECUTE: x − x0 100 m
=
= 49.5 m/s
t
2.02 s
(b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of y − y0 = 45 m . horizontally 100 m during this time. x − x0 = v0 xt + 1 axt 2 gives v0 = v0 x =
2 t= 3.68. 2( y − y0 )
2(45 m)
=
= 3.03 s and x − x0 = v0 xt = (49.5 m/s)(3.03 s) = 150 m . The rock lands
ay
9.80 m/s 2 150 m − 100 m = 50 m beyond the foot of the dam.
EVALUATE: The boulder passes over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s
before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake.
IDENTIFY: The bagels move in projectile motion. Find Henrietta’s location when the bagels reach the ground,
and require the bagels to have this horizontal range. 3-28 Chapter 3 SET UP: Let + y be downward and let x0 = y0 = 0 . ax = 0 , a y = + g . When the bagels reach the ground, y = 43.9 m .
EXECUTE: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels to
1
1
fall 43.9 m from rest. Get the time to fall: y = gt 2 , 43.9 m = (9.80 m/s 2 )t 2 and t = 2.99 s . So, she has been
2
2
jogging for 9.00 s + 2.99 s = 12.0 s . During this time she has gone x = vt = (3.05 m/s)(12.0 s) = 36.6 m . Bruce 3.69. must throw the bagels so they travel 36.6 m horizontally in 2.99 s. This gives x = vt . 36.6 m = v(2.99 s) and
v = 12.2 m/s .
(b) 36.6 m from the building.
EVALUATE: If v > 12.2 m/s the bagels land in front of her and if v < 12.2 m/s they land behind her. There is a
range of velocities greater than 12.2 m/s for which she would catch the bagels in the air, at some height above the
sidewalk.
IDENTIFY: The shell moves in projectile motion. To find the horizontal distance between the tanks we must find
the horizontal velocity of one tank relative to the other. Take + y to be upward.
(a) SET UP: The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the vertical
motion of the shell to find the time the shell is in the air:
v0 y = v0 sin α = 43.4 m/s, a y = −9.80 m/s 2 , y − y0 = 0 (returns to initial height), t = ?
EXECUTE: 3.70. y − y0 = v0 yt + 1 a yt 2 gives t = 8.86 s
2 SET UP: Consider the motion of one tank relative to the other.
EXECUTE: Relative to tank #1 the shell has a constant horizontal velocity v0 cos α = 246.2 m/s. Relative to the
ground the horizontal velocity component is 246.2 m/s + 15.0 m/s = 261.2 m/s. Relative to tank #2 the shell has
horizontal velocity component 261.2 m/s − 35.0 m/s = 226.2 m/s. The distance between the tanks when the shell
was fired is the (226.2 m/s)(8.86 s) = 2000 m that the shell travels relative to tank #2 during the 8.86 s that the
shell is in the air.
(b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels 310 m, in the same
direction. Therefore, their separation increases by 310 m − 133 m = 177 m. So, the separation becomes 2180 m
(rounding to 3 significant figures).
EVALUATE: The retreating tank has greater speed than the approaching tank, so they move farther apart while the
shell is in the air. We can also calculate the separation in part (b) as the relative speed of the tanks times the time
the shell is in the air: (35.0 m/s − 15.0 m/s)(8.86 s) = 177 m.
IDENTIFY: The object moves with constant acceleration in both the horizontal and vertical directions.
SET UP: Let + y be downward and let + x be the direction in which the firecracker is thrown.
EXECUTE: The firecracker’s falling time can be found from the vertical motion: t = 2h
g . The firecracker’s horizontal position at any time t (taking the student’s position as x = 0 ) is x = vt − 1 at 2 .
2
x = 0 when cracker hits the ground, so t = 2v / a . Combining this with the expression for the falling time gives 3.71. 2v
2h
2v 2 g
=
and h = 2 .
a
g
a
EVALUATE: When h is smaller, the time in the air is smaller and either v must be smaller or a must be larger.
!
!
IDENTIFY: The velocity vT/G of the tank relative to the ground is related to the velocity vR/G of the rocket relative
!
!
!
!
to the ground and the velocity vT/R of the tank relative to the rocket by vT/G = vT/R + vR/G .
SET UP: Let + y be upward and take y = 0 at the ground. Let + x be in the direction of the horizontal component
of the tank's motion. Once the tank is released it has ax = 0 , a y = −9.80 m/s 2 , relative to the ground.
EXECUTE: (a) For the rocket v y = v0 y + a yt = (1.75m/s 2 )(22.0 s) = 38.5 m/s and vx = 0 . The rocket has speed 38.5 m/s at the instant when the fuel tank is released.
!
(b) (i) The rocket's path is vertical, so relative to the crew member vT/R-x = +25.0 m/s and vT/R-y = 0 . (ii) vR/G is
!
vertical and vT/R is horizontal, so vT/G-x = +25.0 m/s and vT/G − y = +38.5 m/s .
(c) (i) The tank initially moves horizontally, at an angle of zero. (ii) tan α 0 = vT/G-y
vT/G-x = 38.5 m/s
and α 0 = 57.0° .
25.0 m/s Motion in Two or Three Dimensions 3-29 (d) Consider the motion of the tank, in the reference frame of the technician on the ground. At the instant the tank
is released the rocket at a height y − y0 = v0 yt + 1 a yt 2 = 1 (1.75 m/s 2 )(22.0 s) 2 = 423.5 m . So, for the tank
2
2
2
2
y0 = 423.5 m , v0 y = 38.5 m/s and a y = −9.80 m/s 2 . v y = 0 at the maximum height. v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 3.72. 2
2
v y − v0 y 2a y = 0 − (38.5 m/s) 2
= 75.6 m . y = 423.5 m + 75.6 m = 499 m . The tank reaches a height of 499 m
2(9.80 m/s 2 ) above the launch pad.
EVALUATE: Relative to the crew member in the rocket the jettisoned tank has an acceleration of
1.75 m/s 2 + 9.80 m/s 2 = 11.5 m/s 2 , downward. Relative to the rocket the tank follows a parabolic path, but with
zero initial vertical velocity and with a downward acceleration that has magnitude greater than g.
!
!
IDENTIFY: The velocity vR/G of the rocket relative to the ground is related to the velocity vS/G of the secondary
!
rocket relative to the ground and the velocity vS/R of the secondary rocket relative to the rocket by
!
!
!
vS/G = vS/R + vR/G .
SET UP: Let + y be upward and let y = 0 at the ground. Let + x be in the direction of the horizontal component
of the secondary rocket's motion. After it is launched the secondary rocket has ax = 0 and a y = −9.80 m/s 2 , relative
to the ground.
EXECUTE: (a) (i) vS/R-x = (12.0 m/s)cos53.0° = 7.22 m/s and vS/R-y = (12.0 m/s)sin 53.0° = 9.58 m/s .
(ii) vR/G-x = 0 and vR/G-y = 8.50 m/s . vS/G-x = vS/R-x + vR/G-x = 7.22 m/s and vS/G-y = vS/R-y + vR/G-y =
9.58 m/s + 8.50 m/s = 18.1 m/s .
(b) vS/G = (vS/G-x ) 2 + (vS/G-y ) 2 = 19.5 m/s . tan α 0 = vS/G-y
vS/G-x = 18.1 m/s
and α 0 = 68.3° .
7.22 m/s (c) Relative to the ground the secondary rocket has y0 = 145 m , v0 y = +18.1 m/s , a y = −9.80 m/s 2 and v y = 0 (at
2
2
the maximum height). v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 2
2
v y − v0 y 2a y = 0 − (18.1 m/s) 2
= 16.7 m .
2( −9.80 m/s 2 ) y = 145 m + 16.7 m = 162 m .
EVALUATE: 3.73. The secondary rocket reaches its maximum height in time t = v y − v0 y
ay = −18.1 m/s
= 1.85 s after it
−9.80 m/s 2 is launched. At this time the primary rocket has height 145 m + (8.50 m/s)(1.85 s) = 161 m , so is at nearly the same
height as the secondary rocket. The secondary rocket first moves upward from the primary rocket but then loses
vertical velocity due to the acceleration of gravity.
IDENTIFY: The original firecracker moves as a projectile. At its maximum height it's velocity is horizontal. The
!
!
velocity vA/G of fragment A relative to the ground is related to the velocity vF/G of the original firecracker relative to
!
!
!
!
the ground and the velocity vA/F of the fragment relative to the original firecracker by vA/G = vA/F + vF/G . Fragment
B obeys a similar equation.
SET UP: Let + x be along the direction of the horizontal motion of the firecracker before it explodes and let + y
be upward. Fragment A moves at 53.0° above the + x direction and fragment B moves at 53.0° below the + x
direction. Before it explodes the firecracker has ax = 0 and a y = −9.80 m/s 2
EXECUTE: The horizontal component of the firecracker's velocity relative to the ground is constant (since
ax = 0 ), so vF/G-x = (25.0 m/s)cos30.0° = 21.65 m/s . At the time of the explosion, vF/G-y = 0 . For fragment A, vA/F-x = (20.0 m/s)cos53.0° = 12.0 m/s and vA/F-y = (20.0 m/s)sin 53.0° = 16.0 m/s .
vA/G-x = vA/F-x + vF/G-x = 12.0 m/s + 21.65 m/s = 33.7 m/s . vA/G-y = vA/F-y + vF/G-y = 16.0 m/s .
tan α 0 = 3.74. vA/G-y
vA/G-x = 16.0 m/s
and α 0 = 25.4° . The calculation for fragment B is the same, except vA/F-y = −16.0 m/s .
33.7 m/s The fragments move at 25.4° above and 25.4° below the horizontal.
EVALUATE: As the initial velocity of the firecracker increases the angle with the horizontal for the fragments, as
measured from the ground, decreases.
IDENTIFY: The grenade moves in projectile motion. 110 km/h = 30.6 m/s . The horizontal range R of the grenade
must be 15.8 m plus the distance d that the enemy's car travels while the grenade is in the air. 3-30 Chapter 3 For the grenade take + y upward, so ax = 0 , a y = − g . Let v0 be the magnitude of the velocity of the SET UP: grenade relative to the hero. v0 x = v0 cos 45° , v0 y = v0 sin 45° . 90 km/h = 25 m/s ; The enemy’s car is traveling
away from the hero’s car with a relative velocity of vrel = 30.6 m/s − 25 m/s = 5.6 m/s .
EXECUTE: y − y0 = v0 yt + 1 a yt 2 with y − y0 = 0 gives t = −
2 R = v0 xt = v0 (cos 45°)t = 2v0 y
ay = 2v0 sin 45°
2v0vrel
. d = vrelt =
.
g
g 2
2
2v0 sin 45° cos 45° v0
v2
2vrel
v0 + 15.8 m .
= . R = d + 15.8 m gives that 0 =
g
g
g
g 2
2
v0 − 2vrelv0 − (15.8 m)g = 0 . v0 − 7.92v0 − 154.8 = 0 . The quadratic formula gives v0 = 17.0 m/s = 61.2 km/h . The
grenade has velocity of magnitude 61.2 km/h relative to the hero. Relative to the hero the velocity of the grenade
has components v0 x = v0 cos 45° = 43.3 km/h and v0 y = v0 sin 45° = 43.3 km/h . Relative to the earth the velocity of the grenade has components vEx = 43.3 km/h + 90 km/h = 133.3 km/h and vEy = 43.3 km/h . The magnitude of the
2
2
velocity relative to the earth is vE = vEx + vEy = 140 km/h . EVALUATE: The time the grenade is in the air is t = 2v0 sin 45° 2(17.0 m/s)sin 45°
=
= 2.45 s . During this time
g
9.80 m/s 2 the grenade travels a horizontal distance x − x0 = (133.3 km/h)(2.45 s)(1 h / 3600 s) = 90.7 m , relative to the earth, 3.75. and the enemy’s car travels a horizontal distance x − x0 = (110 km/h)(2.45 s)(1 h / 3600 s) = 74.9 m , relative to the
earth. The grenade has traveled 15.8 m farther.
IDENTIFY and SET UP: Use Eqs. (3.4) and (3.12) to get the velocity and acceleration components from the
position components.
EXECUTE: x = R cos ωt , y = R sin ωt
(a) r = x 2 + y 2 = R 2 cos 2 ωt + R 2 sin 2 ωt = R 2 (sin 2 ωt + cos 2 ωt ) = R 2 = R, since sin 2 ωt + cos 2 ωt = 1.
dx
dy
(b) vx =
= − Rω sin ωt , v y =
= Rω cos ωt
dt
dt
!!
v ⋅ r = vx x + v y y = (− Rω sin ωt )( R cos ωt ) + ( Rω cos ωt )( R sin ωt )
!!
!
!
v ⋅ r = R 2ω (− sin ωt cos ωt + sin ωt cos ωt ) = 0, so v is perpendicular to r .
(c) ax = ay = dv y
dt dvx
= − Rω 2 cos ωt = −ω 2 x
dt
= − Rω 2 sin ωt = −ω 2 y 2
2
a = ax + a y = ω 4 x 2 + ω 4 y 2 = ω 2 x 2 + y 2 = Rω 2 .
!
!
ˆ
ˆj
a = ax i + a y ˆ = −ω 2 ( xi + yˆ) = −ω 2 r .
j !
!
Since ω 2 is positive this means that the direction of a is opposite to the direction of r .
2
2
(d) v = vx + v y = R 2ω 2 sin 2 ωt + R 2ω 2 cos 2 ωt = R 2ω 2 (sin 2 ωt + cos 2 ωt ). v = R 2ω 2 = Rω. 3.76. (e) a = Rω 2 , ω = v / R, so a = R(v 2 / R 2 ) = v 2 / R.
EVALUATE: The rock moves in uniform circular motion. The position vector is radial, the velocity is tangential,
and the acceleration is radially inward.
IDENTIFY: All velocities are constant, so the distance traveled is d = vB/Et , where vB/E is the magnitude of the
!
!
velocity of the boat relative to the earth. The relative velocities vB/E , vS/W (boat relative to the water)
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!
!
!
and v W/E (water relative to the earth) are related by vB/E = vB/W + v W/E . Let + x be east and let + y be north. vW/E-x = +30.0 m/min and vW/E-y = 0 . vB/W = 100.0 m/min . The
!
direction of vB/W is the direction in which the boat is pointed or aimed.
SET UP: EXECUTE: (a) vB/W-y = +100.0 m/min and vB/W-x = 0 . vB/E-x = vB/W-x + vW/E-x = 30.0 m/min and vB/E-y = vB/W-y + vW/E-y = 100.0 m/min . The time to cross the river is t = y − y0
400.0 m
=
= 4.00 min .
vB/E-y 100.0 m/min Motion in Two or Three Dimensions 3-31 x − x0 = (30.0 m/min)(4.00 min) = 120.0 m . You will land 120.0 m east of point B, which is 45.0 m east of
point C. The distance you will have traveled is (400.0 m) 2 + (120.0 m) 2 = 418 m . !
75.0 m
(b) vB/W is directed at angle φ east of north, where tan φ =
and φ = 10.6° .
400.0 m
vB/W-x = (100.0 m/min)sin10.6° = 18.4 m/min and vB/W-y = (100.0 m/min)cos10.6° = 98.3 m/min . vB/E-x = vB/W-x + vW/E-x = 18.4 m/min + 30.0 m/min = 48.4 m/min . vB/E-y = vB/W-y + vW/E-y = 98.3 m/min .
t= y − y0
400.0 m
=
= 4.07 min . x − x0 = (48.4 m/min)(4.07 min) = 197 m . You will land 197 m downstream
vB/E-y
98.3 m/min from B, so 122 m downstream from C.
!
(c) (i) If you reach point C, then vB/E is directed at 10.6° east of north, which is 79.4° north of east. We don't know
!
!
the magnitude of vB/E and the direction of vB/W . In part (a) we found that if we aim the boat due north we will land
!
east of C, so to land at C we must aim the boat west of north. Let vB/W be at an angle φ of north of west. The
relative velocity addition diagram is sketched in Figure 3.76. The law of sines says sin θ sin 79.4°
=
.
vW/E
vB/W ⎛ 30.0 m/min ⎞
sin θ = ⎜
⎟ sin 79.4° and θ = 17.15° . Then φ = 180° − 79.4° − 17.15° = 83.5° . The boat will head
⎝ 100.0 m/min ⎠
83.5° north of west, so 6.5° west of north.
vB/E-x = vB/W-x + vW/E-x = −(100.0 m/min)cos83.5° + 30.0 m/min = 18.7 m/min . vB/E-y = vB/W-y + vW/E-y = −(100.0 m/min)sin83.5° = 99.4 m/min . Note that these two components do give the
!
direction of vB/E to be 79.4° north of east, as required. (ii) The time to cross the river is
t= y − y0
400.0 m
=
= 4.02 min . (iii) You travel from A to C, a distance of
vB/E-y
99.4 m/min (400.0 m)2 + (75.0 m) 2 = 407 m . (iv) vB/E = (vB/E-x ) 2 + (vB/E-y ) 2 = 101 m/min . Note that vB/Et = 406 m , the distance traveled (apart from a small
difference due to rounding).
EVALUATE: You cross the river in the shortest time when you head toward point B, as in part (a), even though
you travel farther than in part (c). Figure 3.76
3.77. IDENTIFY: vx = dx / dt , v y = dy / dt , ax = dvx / dt and a y = dv y / dt . d (sin ωt )
d (cos ωt )
= ω cos(ωt ) and
= −ω sin(ωt ) .
dt
dt
EXECUTE: (a) The path is sketched in Figure 3.77.
(b) To find the velocity components, take the derivative of x and y with respect to time: vx = Rω (1 − cos ωt ), and
SET UP: v y = Rω sin ωt. To find the acceleration components, take the derivative of vx and v y with respect to time:
ax = Rω 2 sin ωt, and a y = Rω 2 cos ωt.
(c) The particle is at rest (v y = vx = 0) every period, namely at t = 0, 2π / ω, 4π / ω,.... At that time, x = 0 , 2πR, 4πR,...; and y = 0. The acceleration is a = Rω 2 in the + y - direction. 3-32 Chapter 3
1/ 2 2
2
(d) No, since a = ⎡( Rω 2 sin ωt ) + ( Rω 2 cos ωt ) ⎤ = Rω 2 . The magnitude of the acceleration is the same as for
⎢
⎥
⎣
⎦
uniform circular motion.
EVALUATE: The velocity is tangent to the path. v0 x is always positive; v y changes sign during the motion. Figure 3.77
3.78. IDENTIFY: At the highest point in the trajectory the velocity of the projectile relative to the earth is horizontal.
!
!
The velocity vP/E of the projectile relative to the earth, the velocity vF/P of a fragment relative to the projectile, and
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!
the velocity vF/E of a fragment relative to the earth are related by vF/E = vF/P + vP/E .
SET UP: Let + x be along the horizontal component of the projectile motion. Let the speed of each fragment
relative to the projectile be v. Call the fragments 1 and 2, where fragment 1 travels in the + x direction and
fragment 2 is in the − x-direction , and let the speeds just after the explosion of the two fragments relative to the
earth be v1 and v2 . Let vp be the speed of the projectile just before the explosion. vF/E-x = vF/P-x + vP/E-x gives v1 = vp + v and −v2 = vp − v . Both fragments start from the same height with EXECUTE: zero vertical component of velocity relative to the earth, so they both fall for the same time t, and this is also the
same time as it took for the projectile to travel a horizontal distance D, so vpt = D . Since fragment 2 lands at A it
travels a horizontal distance D as it falls and v2t = D . −v2 = +vp − v gives v = vp + v2 and vt = vpt + v2t = 2 D . Then
v1t = vpt + vt = 3D . This fragment lands a horizontal distance 3D from the point of explosion and hence 4D from A. 3.79. EVALUATE: Fragment 1, that is ejected in the direction of the motion of the projectile travels with greater speed
relative to the earth than the fragment that travels in the opposite direction.
v 2 4π 2 R
IDENTIFY: arad = =
. All points on the centrifuge have the same period T.
R
T2
60 s/min
SET UP: The period T in seconds is related to n, the number of revolutions per minute, by n =
.
T
a
a
a
4π 2
EXECUTE: (a) rad = 2 , which is constant. rad,1 = rad,2 . Let R1 = R , so arad,1 = 5.00 g and let R2 = R / 2 .
R
T
R1
R2 ⎛R ⎞
arad,2 = arad,1 ⎜ 2 ⎟ = (5.00 g )(1/ 2) = 2.50 g .
⎝ R1 ⎠
4π 2 R
a
4π 2 R
⎛ 60 s/min ⎞
(b) T = ⎜
and arad =
gives arad = 4π 2 Rn 2 /(60 s/min) 2 . rad =
, which is constant.
⎟
n
n 2 (60 s/min) 2
T2
⎝
⎠ arad,1
n12 = n2 = n1 arad,2
2
n2 . Let arad,1 = 5.00 g , so n1 = n and arad,2 = 5 g Mercury = 5(0.378) g . Then arad,2
arad,1 EVALUATE: =n 5(0.378) g
= 0.615n .
5.00 g The radial acceleration is less for points closer to the rotation axis. Since g Mercury < g , a smaller rotation rate is required to produce 5 g Mercury than to produce 5 g .
3.80. IDENTIFY: Use the relation that relates the relative velocities.
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!
SET UP: The relative velocities are the raindrop relative to the earth, vR/E , the raindrop relative to the train, vR/T ,
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and the train relative to the earth, vT/E . vR/E = vR/T + vT/E . vT/E is due east and has magnitude 12.0 m/s. vR/T is
!
30.0° west of vertical. vR/E is vertical. The relative velocity addition diagram is given in Figure 3.80.
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!
!
EXECUTE: (a) vR/E is vertical and has zero horizontal component. The horizontal component of vR/T is −vT/E , so
is 12.0 m/s westward.
vT/E
12.0 m/s
vT/E
12.0 m/s
(b) vR/E =
=
= 20.8 m/s . vR/T =
=
= 24.0 m/s .
tan 30.0° tan30.0°
sin 30.0° sin30.0° Motion in Two or Three Dimensions 3-33 EVALUATE: The speed of the raindrop relative to the train is greater than its speed relative to the earth, because
of the motion of the train. Figure 3.80
3.81. IDENTIFY: Relative velocity problem. The plane’s motion relative to the earth is determined by its velocity
relative to the earth.
SET UP: Select a coordinate system where + y is north and + x is east.
The velocity vectors in the problem are:
!
vP/E , the velocity of the plane relative to the earth.
!
vP/A , the velocity of the plane relative to the air (the magnitude vP/A is the air speed of the plane and the direction
!
of vP/A is the compass course set by the pilot).
!
vA/E , the velocity of the air relative to the earth (the wind velocity).
!
!
!
The rule for combining relative velocities gives vP/E = vP/A + vA/E .
(a) We are given the following information about the relative velocities:
!
vP/A has magnitude 220 km/h and its direction is west. In our coordinates is has components (vP/A ) x = −220 km/h and (vP/A ) y = 0. !
From the displacement of the plane relative to the earth after 0.500 h, we find that vP/E has components in our
coordinate system of (vP/E ) x = − 120 km
= −240 km/h (west)
0.500 h 20 km
= −40 km/h (south)
0.500 h
With this information the diagram corresponding to the velocity addition equation is shown in Figure 3.81a.
(vP/E ) y = − Figure 3.81a
!
We are asked to find vA/E , so solve for this vector:
!
!
!
!
!
!
vP/E = vP/A + vA/E gives vA/E = vP/E − vP/A .
EXECUTE: The x-component of this equation gives
(vA/E ) x = (vP/E ) x − (vP/A ) x = −240 km/h − (−220 km/h) = −20 km/h.
The y-component of this equation gives
(vA/E ) y = (vP/E ) y − (vP/A ) y = −40 km/h. 3-34 Chapter 3 !
Now that we have the components of vA/E we can find its magnitude and direction. vA/E = (vA/E ) 2 + (vA/E ) 2
x
y
vA/E = ( −20 km/h) 2 + (−40 km/h) 2 = 44.7 km/h
40 km/h
= 2.00; φ = 63.4°
20 km/h
The direction of the wind velocity is 63.4° S of W, or
26.6° W of S.
tan φ = Figure 3.81b
EVALUATE: The plane heads west. It goes farther west than it would without wind and also travels south, so the
wind velocity has components west and south.
!
!
!
(b) SET UP: The rule for combining the relative velocities is still vP/E = vP/A + vA/E , but some of these velocities
have different values than in part (a).
!
vP/A has magnitude 220 km/h but its direction is to be found.
!
vA/E has magnitude 40 km/h and its direction is due south.
!
The direction of vP/E is west; its magnitude is not given.
!
!
!
The vector diagram for vP/E = vP/A + vA/E and the specified directions for the vectors is shown in Figure 3.81c. Figure 3.81c 3.82. The vector addition diagram forms a right triangle.
v
40 km/h
EXECUTE: sin φ = A/E =
= 0.1818; φ = 10.5°.
vP/A 220 km/h
The pilot should set her course 10.5° north of west.
EVALUATE: The velocity of the plane relative to the air must have a northward component to counteract the wind
and a westward component in order to travel west.
IDENTIFY: Both the bolt and the elevator move vertically with constant acceleration.
SET UP: Let + y be upward and let y = 0 at the initial position of the floor of the elevator, so y0 for the bolt is
3.00 m.
EXECUTE: (a) The position of the bolt is 3.00 m + (2.50 m/s)t − (1/ 2)(9.80 m/s 2 )t 2 and the position of the floor
is (2.50 m/s)t. Equating the two, 3.00 m = (4.90 m/s 2 )t 2 . Therefore, t = 0.782 s .
(b) The velocity of the bolt is 2.50 m/s − (9.80 m/s 2 )(0.782 s) = −5.17 m/s relative to Earth, therefore, relative to
an observer in the elevator v = −5.17 m/s − 2.50 m/s = −7.67 m/s.
(c) As calculated in part (b), the speed relative to Earth is 5.17 m/s.
(d) Relative to Earth, the distance the bolt traveled is
(2.50 m/s)t − (1/ 2)(9.80 m/s 2 )t 2 = (2.50 m/s)(0.782 s) − (4.90 m/s 2 )(0.782 s) 2 = −1.04 m .
EVALUATE: As viewed by an observer in the elevator, the bolt has v0 y = 0 and a y = −9.80 m/s 2 , so in 0.782 s it falls − 1 (9.80 m/s 2 )(0.782 s) 2 = −3.00 m .
2
3.83. IDENTIFY: In an earth frame the elevator accelerates upward at 4.00 m/s 2 and the bolt accelerates downward at
9.80 m/s 2 . Relative to the elevator the bolt has a downward acceleration of 4.00 m/s 2 + 9.80 m/s 2 = 13.80 m/s 2 . In
either frame, that of the earth or that of the elevator, the bolt has constant acceleration and the constant acceleration
equations can be used.
SET UP: Let + y be upward. The bolt travels 3.00 m downward relative to the elevator.
EXECUTE: (a) In the frame of the elevator, v0 y = 0 , y − y0 = −3.00 m , a y = −13.8 m/s 2 . y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2( −3.00 m)
=
= 0.659 s .
ay
−13.8 m/s 2 Motion in Two or Three Dimensions 3-35 (b) v y = v0 y + a y t . v0 y = 0 and t = 0.659 s . (i) a y = −13.8 m/s 2 and v y = −9.09 m/s . The bolt has speed 9.09 m/s when it reaches the floor of the elevator. (ii) a y = −9.80 m/s 2 and v y = −6.46 m/s . In this frame the bolt has speed
6.46 m/s when it reaches the floor of the elevator.
(c) y − y0 = v0 yt + 1 a yt 2 . v0 y = 0 and t = 0.659 s . (i) a y = −13.8 m/s 2 and
2
y − y0 = 1 (−13.8 m/s 2 )(0.659 s) 2 = −3.00 m . The bolt falls 3.00 m, which is correctly the distance between the
2
floor and roof of the elevator. (ii) a y = −9.80 m/s 2 and y − y0 = 1 (−9.80 m/s 2 )(0.659 s) 2 = −2.13 m . The bolt falls
2 3.84. 2.13 m.
EVALUATE: In the earth's frame the bolt falls 2.13 m and the elevator rises
1
(4.00 m/s 2 )(0.659 s) 2 = 0.87 m during the time that the bolt travels from the ceiling to the floor of the elevator.
2
!
!
IDENTIFY: The velocity vP/E of the plane relative to the earth is related to the velocity vP/A of the plane relative to
!
!
!
!
the air and the velocity vA/E of the air relative to the earth (the wind velocity) by vP/E = vP/A + vA/E .
SET UP: Let + x be to the east. With no wind vP/A = vP/E = 5550 km
= 840.9 km/h . vA/E-x = +225 km/h . The
6.60 h distance between A and B is 2775 km.
EXECUTE: vP/E-x = vP/A-x + vA/E-x . For the trip A to B, vP/A-x = +840.9 km/h and
2775 km
= 2.60 h . For the trip B to
1065.9 km/h
= −840.9 km/h + 225 km/h = −615.9 km/h and the travel time is vP/E-x = 840.9 km/h + 225 km/h = 1065.9 km/h and the travel time is t AB =
A, vP/A-x = −840.9 km/h and vP/E-x −2775 km
= 4.51 h . The total time for the round trip will be t = t AB + t BA = 7.11 h .
−615.9 km/h
EVALUATE: The round trip takes longer when the wind blows, even though the plane travels with the wind for
1065.9 km/h + 615.9 km/h
= 840.9 km/h ,
one leg of the trip. The arithmetic average of the speeds for each leg is
2
the same speed when there is no wind. But the plane spends more time traveling at the slower speed relative to the
ground and the average speed is less than the arithmetic average of the speeds for each half of the trip.
IDENTIFY: Relative velocity problem.
SET UP: The three relative velocities are:
!
vJ/G , Juan relative to the ground. This velocity is due north and has magnitude vJ/G = 8.00 m/s.
!
vB/G , the ball relative to the ground. This vector is 37.0° east of north and has magnitude vB/G = 12.00 m/s.
!
vB/J , the ball relative to Juan. We are asked to find the magnitude and direction of this vector.
!
!
!
!
!
!
The relative velocity addition equation is vB/G = vB/J + vJ/G , so vB/J = vB/G − v J/G .
The relative velocity addition diagram does not form a right triangle so we must do the vector addition using
components.
Take + y to be north and + x to be east. t BA = 3.85. EXECUTE: vB/Jx = + vB/G sin 37.0° = 7.222 m/s vB/Jy = + vB/G cos37.0° − vJ/G = 1.584 m/s 3.86. These two components give vB/J = 7.39 m/s at 12.4° north of east.
EVALUATE: Since Juan is running due north, the ball’s eastward component of velocity relative to him is the
same as its eastward component relative to the earth. The northward component of velocity for Juan and the ball
are in the same direction, so the component for the ball relative to Juan is the difference in their components of
velocity relative to the ground.
IDENTIFY: (a) The ball moves in projectile motion. When it is moving horizontally, v y = 0 .
SET UP:
EXECUTE: Let + x be to the right and let + y be upward. ax = 0 , a y = − g .
(a) v0 y = 2 gh = 2(9.80 m/s 2 )(4.90 m) = 9.80 m/s. (b) v0 y / g = 1.00 s .
(c) The horizontal component of the velocity of the ball relative to the man is (10.8 m/s) 2 − (9.80 m/s) 2 = 4.54 m/s , the horizontal component of the velocity relative to the hoop is
4.54 m/s + 9.10 m/s = 13.6 m/s , and the man must be 13.6 m in front of the hoop at release. 3-36 Chapter 3 ⎛ 9.80 m/s ⎞
(d) Relative to the flat car, the ball is projected at an angle θ = tan −1 ⎜
⎟ = 65°. Relative to the ground the
⎝ 4.54 m/s ⎠
⎛
⎞
9.80 m/s
angle is θ = tan −1 ⎜
⎟ = 35.7° .
4.54 m/s + 9.10 m/s ⎠
⎝
EVALUATE: In both frames of reference the ball moves in a parabolic path with ax = 0 and a y = − g . The only 3.87. difference between the description of the motion in the two frames is the horizontal component of the ball’s
velocity.
IDENTIFY: The pellets move in projectile motion. The vertical motion determines their time in the air.
SET UP: v0 x = v0 cos1.0° , v0 y = v0 sin1.0° .
⎛ 2v sin1.0° ⎞
. x − x0 = v0 xt gives x − x0 = (v0 cos1.0°)⎜ 0
⎟ = 80 m .
g
⎝
⎠
(b) The probability is 1000 times the ratio of the area of the top of the person’s head to the area of the circle in
⎛ π (10 × 10−2 m) 2 ⎞
−3
which the pellets land. (1000) ⎜
⎟ = 1.6 × 10 .
2
⎝ π (80 m)
⎠
(c) The slower rise will tend to reduce the time in the air and hence reduce the radius. The slower horizontal
velocity will also reduce the radius. The lower speed would tend to increase the time of descent, hence increasing
the radius. As the bullets fall, the friction effect is smaller than when they were rising, and the overall effect is to
decrease the radius.
EVALUATE: The small angle of deviation from the vertical still causes the pellets to spread over a large area
because their time in the air is large.
IDENTIFY: Write an expression for the square of the distance ( D 2 ) from the origin to the particle, expressed as a
EXECUTE: 3.88. (a) t = 2v0 y
g function of time. Then take the derivative of D 2 with respect to t, and solve for the value of t when this derivative
is zero. If the discriminant is zero or negative, the distance D will never decrease.
SET UP: D 2 = x 2 + y 2 , with x (t ) and y (t ) given by Eqs.(3.20) and (3.21). 3.89. EXECUTE: Following this process, sin −1 8/ 9 = 70.5°.
EVALUATE: We know that if the object is thrown straight up it moves away from P and then returns, so we are
not surprised that the projectile angle must be less than some maximum value for the distance to always increase
with time.
IDENTIFY: The baseball moves in projectile motion.
SET UP: Use coordinates where the x-axis is horizontal and the y-axis is vertical.
EXECUTE: (a) The trajectory of the projectile is given by Eq. (3.27), with α 0 = θ + φ, and the equation describing
the incline is y = x tan θ. Setting these equal and factoring out the x = 0 root (where the projectile is on the incline) gives a value for x0 ; the range measured along the incline is
⎡ 2v 2 ⎤
⎡ cos 2 (θ + φ) ⎤
x / cos θ = ⎢ 0 ⎥ [tan(θ + φ) − tan θ ] ⎢
⎥.
⎣g⎦
⎣ cos θ ⎦
(b) Of the many ways to approach this problem, a convenient way is to use the same sort of substitution, involving
double angles, as was used to derive the expression for the range along a horizontal incline. Specifically, write the
above in terms of α = θ + φ, as
2
⎡ 2v0 ⎤
2
R=⎢
⎥ [sin α cos α cosθ − cos α sin θ ] .
2
⎣ g cos θ ⎦ The dependence on α and hence φ is in the second term. Using the identities
sin α cos α = (1/ 2)sin 2α and cos 2 α = (1/ 2)(1 + cos 2α ), this term becomes
(1/ 2)[cos θ sin 2α − sin θ cos 2α − sin θ ] = (1/ 2)[sin(2α − θ ) − sin θ ] . 3.90. This will be a maximum when sin(2α − θ ) is a maximum, at 2α − θ = 2φ + θ = 90°, or φ = 45° − θ / 2.
EVALUATE: Note that the result reduces to the expected forms when θ = 0 (a flat incline, φ = 45° and when
θ = −90° (a vertical cliff), when a horizontal launch gives the greatest distance).
IDENTIFY: The arrow moves in projectile motion.
SET UP: Use coordinates that for which the axes are horizontal and vertical. Let θ be the angle of the slope and
let φ be the angle of projection relative to the sloping ground. Motion in Two or Three Dimensions 3-37 EXECUTE: The horizontal distance x in terms of the angles is
⎛ gx ⎞
1
tan θ = tan(θ + φ ) − ⎜ 2 ⎟
.
2
⎝ 2v0 ⎠ cos (θ + φ ) 2
Denote the dimensionless quantity gx / 2v0 by β ; in this case β= (9.80 m/s 2 )(60.0 m)cos30.0°
= 0.2486.
2(32.0 m/s) 2 The above relation can then be written, on multiplying both sides by the product cosθ cos(θ + φ ),
sin θ cos(θ + φ ) = sin(θ + φ )cosθ − β cosθ
,
cos(θ + φ ) β cosθ
. The term on the left is sin((θ + φ ) − θ ) = sin φ , so the result
cos(θ + φ )
of this combination is sin φ cos(θ + φ ) = β cosθ .
and so sin(θ + φ )cosθ − cos(θ + φ )sin θ = Although this can be done numerically (by iteration, trial-and-error, or other methods), the expansion
sin a cos b = 1 (sin( a + b) + sin( a − b)) allows the angle φ to be isolated; specifically, then
2
1
(sin(2φ + θ ) + sin(−θ )) = β cosθ , with the net result that sin(2φ + θ ) = 2 β cosθ + sin θ .
2
(a) For θ = 30°, and β as found above, φ = 19.3° and the angle above the horizontal is θ + φ = 49.3°. For level
ground, using β = 0.2871, gives φ = 17.5°.
(b) For θ = −30°, the same β as with θ = 30° may be used (cos30° = cos(−30°)) , giving φ = 13.0° and
φ + θ = −17.0°.
EVALUATE: 2
For θ = 0 the result becomes sin(2φ ) = 2 β = gx / v0 . This is equivalent to the expression 2
v0 sin(2α 0 )
derived in Example 3.8.
g
!
IDENTIFY: Find Δv and use this to calculate the magnitude and direction of the average acceleration.
v Δt
SET UP: In a time Δt, the velocity vector has moved through an angle (in radians) Δφ =
(see Figure 3.28 in
R
!
the textbook). By considering the isosceles triangle formed by the two velocity vectors, the magnitude Δv is seen R= 3.91. to be 2v sin(φ / 2) . EXECUTE: !
Δv
v
#
⎛ vΔt ⎞ 10 m/s
sin([1.0 / s]Δt)
aav =
= 2 sin ⎜
⎟=
Δt
Δt ⎝ 2 R ⎠
Δt Using the given values gives magnitudes of 9.59 m/s 2 , 9.98 m/s 2 and 10.0 m/s 2 . The changes in direction of the
vΔt
and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle of
R
π + Δθ
= π / 2 + 1/ 2 rad(or 118.6°),
the average acceleration vector with the original velocity vector is
2
π / 2 + 0.1 rad(or 95.7°) , and π / 2 + 0.05 rad(or 92.9°). velocity vectors are given by Δθ = 3.92. EVALUATE: The instantaneous acceleration magnitude, v 2 / R = (5.00 m/s) 2 /(2.50 m ) = 10.0 m/s 2 is indeed
#
approached in the limit at Δt → 0. Also, the direction of aav approaches the radially inward direction as Δt → 0 .
IDENTIFY: The rocket has two periods of constant acceleration motion.
SET UP: Let + y be upward. During the free-fall phase, ax = 0 and a y = − g . After the engines turn on,
ax = (3.00 g )cos30.0° and a y = (3.00 g )sin 30.0° . Let t be the total time since the rocket was dropped and let T be
the time the rocket falls before the engine starts.
EXECUTE: (i) The diagram is given in Figure 3.92a.
(ii) The x-position of the plane is (236 m/s)t and the x-position of the rocket is
(236 m/s)t + (1/ 2)(3.00)(9.80 m/s 2 )cos30°(t − T ) 2 . The graphs of these two equations are sketched in Figure
3.92b. 3-38 Chapter 3 (iii) If we take y = 0 to be the altitude of the airliner, then
y (t ) = −1/ 2 gT 2 − gT (t − T ) + 1/ 2(3.00)(9.80 m/s 2 )(sin 30°)(t − T ) 2 for the rocket. The airliner has constant y. The
graphs are sketched in Figure 3.92b.
In each of the Figures 3.92a-c, the rocket is dropped at t = 0 and the time T when the motor is turned on is
indicated.
By setting y = 0 for the rocket, we can solve for t in terms of T:
0 = −(4.90 m/s 2 )T 2 − (9.80 m/s 2 )T (t − T ) + (7.35 m/s 2 )(t − T ) 2 . Using the quadratic formula for the
variable x = t − T we find x = t − T = (9.80 m/s 2 )T + (9.80 m/s 2T ) 2 + (4)(7.35 m/s 2 )(4.9)T 2
, or t = 2.72 T . Now,
2(7.35 m/s 2 ) using the condition that xrocket − xplane = 1000 m , we find (236 m/s)t + (12.7 m/s 2 )(t − T )2 − (236 m/s)t = 1000 m, or
(1.72T ) 2 = 78.6 s 2 . Therefore T = 5.15 s.
EVALUATE: During the free-fall phase the rocket and airliner have the same x coordinate but the rocket moves
downward from the airliner. After the engines fire, the rocket starts to move upward and its horizontal component
of velocity starts to exceed that of the airliner. Figure 3.92
3.93. IDENTIFY: Apply the relative velocity relation.
SET UP: Let vC/W be the speed of the canoe relative to water and vW/G be the speed of the water relative to the
ground.
EXECUTE: (a) Taking all units to be in km and h, we have three equations. We know that heading upstream
vC/W − vW/G = 2 . We know that heading downstream for a time t , (vC/W + vW/G )t = 5. We also know that for the
bottle vW/G (t + 1) = 3. Solving these three equations for vW/G = x, vC/W = 2 + x, therefore (2 + x + x)t = 5 or
⎛3 ⎞
(2 + 2 x )t = 5. Also t = 3/ x − 1, so (2 + 2 x) ⎜ − 1⎟ = 5 or 2 x 2 + x − 6 = 0. The positive solution is
⎝x ⎠
x = vW/G = 1.5 km/h. (b) vC/W = 2 km/h + vW/G = 3.5 km/h.
EVALUATE: When they head upstream, their speed relative to the ground is 3.5 km/h − 1.5 km/h = 2.0 km/h .
When they head downstream, their speed relative to the ground is 3.5 km/h + 1.5 km/h = 5.0 km/h . The bottle is
moving downstream at 1.5 km/s relative to the earth, so they are able to overtake it. 4 NEWTON’S LAWS OF MOTION 4.1. IDENTIFY: Consider the vector sum in each case.
!
!
!
!
SET UP: Call the two forces F1 and F2 . Let F1 be to the right. In each case select the direction of F2 such that
!!!
F = F1 + F2 has the desired magnitude.
EXECUTE: (a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the
angle between them is zero. The two vectors and their sum are sketched in Figure 4.1a.
(b) The forces form the sides of a right isosceles triangle, and the angle between them is 90° . The two vectors and
their sum are sketched in Figure 4.1b.
(c) For the sum to have zero magnitude, the forces must be antiparallel, and the angle between them is 180° . The
two vectors are sketched in Figure 4.1c.
EVALUATE: The maximum magnitude of the sum of the two vectors is 2F, as in part (a). Figure 4.1
4.2. IDENTIFY: Add the three forces by adding their components.
SET UP: In the new coordinates, the 120-N force acts at an angle of 53° from the − x -axis, or 233° from
the + x -axis, and the 50-N force acts at an angle of 323° from the + x -axis.
EXECUTE: (a) The components of the net force are
Rx = (120 N)cos 233° + (50 N)cos323° = −32 N Ry = ( 250 N) + (120 N)sin 233° + (50 N)sin 323° = 124 N. 4.3. 4.4. ⎛ 124 ⎞
2
(b) R = Rx2 + Ry = 128 N, arctan ⎜
⎟ = 104° . The results have the same magnitude as in Example 4.1, and the
⎝ −32 ⎠
angle has been changed by the amount (37°) that the coordinates have been rotated.
EVALUATE: We can use any set of coordinate axes that we wish to and can therefore select axes for which the
analysis of the problem is the simplest.
IDENTIFY: Use right-triangle trigonometry to find the components of the force.
SET UP: Let + x be to the right and let + y be downward.
EXECUTE: The horizontal component of the force is (10 N)cos 45° = 7.1 N to the right and the vertical
component is (10 N)sin 45° = 7.1 N down.
EVALUATE: In our coordinates each component is positive; the signs of the components indicate the directions of
the component vectors.
IDENTIFY: Fx = F cosθ , Fy = F sin θ . SET UP: Let + x be parallel to the ramp and directed up the ramp. Let + y be perpendicular to the ramp and
directed away from it. Then θ = 30.0° . 4-1 4-2 Chapter 4 EXECUTE: (a) F = Fx
60.0 N
=
= 69.3 N.
cosθ cos30° (b) Fy = F sin θ = Fx tan θ = 34.6 N.
EVALUATE: We can verify that Fx2 + Fy2 = F 2 . The signs of Fx and Fy show their direction.
4.5. IDENTIFY: Vector addition.
!
SET UP: Use a coordinate system where the + x -axis is in the direction of FA , the force applied by dog A. The
forces are sketched in Figure 4.5.
EXECUTE: FAx = +270 N, FAy = 0
FBx = FB cos60.0° = (300 N)cos60.0° = +150 N
FBy = FB sin 60.0° = (300 N)sin 60.0° = +260 N Figure 4.5a
!!
!
R = FA + FB Rx = FAx + FBx = +270 N + 150 N = +420 N
Ry = FAy + FBy = 0 + 260 N = +260 N
2
R = Rx2 + Ry R = (420 N) 2 + (260 N) 2 = 494 N
tan θ = Ry Rx
θ = 31.8° = 0.619 Figure 4.5b 4.6. EVALUATE: The forces must be added as vectors. The magnitude of the resultant force is less than the sum of the
magnitudes of the two forces and depends on the angle between the two forces.
IDENTIFY: Add the two forces using components.
!
SET UP: Fx = F cosθ , Fy = F sin θ , where θ is the angle F makes with the + x axis.
EXECUTE: (a) F1x + F2 x = (9.00 N)cos120° + (6.00 N)cos(233.1°) = −8.10 N
F1 y + F2 y = (9.00 N)sin120° + (6.00 N)sin(233.1°) = +3.00 N. 4.7. 4.8. 4.9. 2
(b) R = Rx2 + Ry = (8.10 N) 2 + (3.00 N) 2 = 8.64 N.
!
EVALUATE: Since Fx < 0 and Fy > 0 , F is in the second quadrant.
!
!
IDENTIFY: Apply ∑ F = ma . SET UP: Let + x be in the direction of the force.
EXECUTE: ax = Fx / m = (132 N)/ (60 kg) = 2.2 m / s 2 .
EVALUATE: The acceleration is in the direction of the force.
!
!
IDENTIFY: Apply ∑ F = ma . SET UP: Let + x be in the direction of the acceleration.
EXECUTE: Fx = max = (135 kg)(1.40 m/s 2 ) = 189 N.
EVALUATE: The net force must be in the direction of the acceleration.
!
!
IDENTIFY: Apply ∑ F = ma to the box.
SET UP: Let + x be the direction of the force and acceleration. ∑F x = 48.0 N . Newton’s Laws of Motion 4.10. ∑F 48.0 N
x
=
= 16.0 kg .
3.00 m/s 2
ax
EVALUATE: The vertical forces sum to zero and there is no motion in that direction.
IDENTIFY: Use the information about the motion to find the acceleration and then use EXECUTE: ∑F x = max gives m = 4-3 calculate m.
SET UP: Let + x be the direction of the force. ∑F x ∑F x = max to = 80.0 N . EXECUTE: (a) x − x0 = 11.0 m , t = 5.00 s , v0 x = 0 . x − x0 = v0 xt + 1 axt 2 gives
2
ax = 2( x − x0 ) 2(11.0 m)
∑ Fx = 80.0 N = 90.9 kg .
=
= 0.880 m/s 2 . m =
2
2
t
(5.00 s)
0.880 m/s 2
ax (b) ax = 0 and vx is constant. After the first 5.0 s, vx = v0 x + axt = (0.880 m/s 2 )(5.00 s) = 4.40 m/s . 4.11. x − x0 = v0 xt + 1 axt 2 = (4.40 m/s)(5.00 s) = 22.0 m .
2
EVALUATE: The mass determines the amount of acceleration produced by a given force. The block moves farther
in the second 5.00 s than in the first 5.00 s.
IDENTIFY and SET UP: Use Newton’s second law in component form (Eq.4.8) to calculate the acceleration
produced by the force. Use constant acceleration equations to calculate the effect of the acceleration on the motion.
EXECUTE: (a) During this time interval the acceleration is constant and equal to
ax = Fx 0.250 N
=
= 1.562 m/s 2
m 0.160 kg We can use the constant acceleration kinematic equations from Chapter 2.
x − x0 = v0 xt + 1 axt 2 = 0 + 1 (1.562 m/s 2 )(2.00 s) 2 ,
2
2
so the puck is at x = 3.12 m.
vx = v0 x + axt = 0 + (1.562 m/s 2 )(2.00 s) = 3.13 m/s.
(b) In the time interval from t = 2.00 s to 5.00 s the force has been removed so the acceleration is zero. The speed
stays constant at vx = 3.12 m/s. The distance the puck travels is x − x0 = v0 xt = (3.12 m/s)(5.00 s − 2.00 s) = 9.36 m.
At the end of the interval it is at x = x0 + 9.36 m = 12.5 m.
In the time interval from t = 5.00 s to 7.00 s the acceleration is again ax = 1.562 m/s 2 . At the start of this interval v0 x = 3.12 m/s and x0 = 12.5 m.
x − x0 = v0 xt + 1 axt 2 = (3.12 m/s)(2.00 s) + 1 (1.562 m/s 2 )(2.00 s) 2 .
2
2
x − x0 = 6.24 m + 3.12 m = 9.36 m.
Therefore, at t = 7.00 s the puck is at x = x0 + 9.36 m = 12.5 m + 9.36 m = 21.9 m. vx = v0 x + axt = 3.12 m/s + (1.562 m/s 2 )(2.00 s) = 6.24 m/s 4.12. EVALUATE: The acceleration says the puck gains 1.56 m/s of velocity for every second the force acts. The force
acts a total of 4.00 s so the final velocity is (1.56 m/s)(4.0 s) = 6.24 m/s.
!
!
IDENTIFY: Apply ∑ F = ma . Then use a constant acceleration equation to relate the kinematic quantities. SET UP: Let + x be in the direction of the force.
EXECUTE: (a) ax = Fx / m = (140 N) /(32.5 kg) = 4.31 m/s 2 .
(b) x − x0 = v0 xt + 1 axt 2 . With v0 x = 0, x = 1 at 2 = 215 m .
2
2 4.13. (c) vx = v0 x + axt . With v0 x = 0, vx = axt = 2 x / t = 43.0 m/s .
EVALUATE: The acceleration connects the motion to the forces.
IDENTIFY: The force and acceleration are related by Newton’s second law.
SET UP: ∑ Fx = max , where ∑ Fx is the net force. m = 4.50 kg .
EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value.
∑ Fx = max = (4.50 kg)(10.0 m/s2 ) = 45.0 N . This maximum force occurs between 2.0 s and 4.0 s.
(b) The net force is constant when the acceleration is constant. This is between 2.0 s and 4.0 s.
(c) The net force is zero when the acceleration is zero. This is the case at t = 0 and t = 6.0 s .
EVALUATE: A graph of ∑ Fx versus t would have the same shape as the graph of ax versus t. 4-4 4.14. Chapter 4 IDENTIFY: The force and acceleration are related by Newton’s second law. ax = dvx
, so ax is the slope of the
dt graph of vx versus t. SET UP: The graph of vx versus t consists of straight-line segments. For t = 0 to t = 2.00 s , ax = 4.00 m/s 2 . For
t = 2.00 s to 6.00 s, ax = 0 . For t = 6.00 s to 10.0 s, ax = 1.00 m/s 2 . ∑F x = max , with m = 2.75 kg . ∑F x is the net force. EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value.
∑ Fx = max = (2.75 kg)(4.00 m/s2 ) = 11.0 N . This maximum occurs in the interval t = 0 to t = 2.00 s .
(b) The net force is zero when the acceleration is zero. This is between 2.00 s and 6.00 s.
(c) Between 6.00 s and 10.0 s, ax = 1.00 m/s 2 , so ∑ Fx = (2.75 kg)(1.00 m/s 2 ) = 2.75 N .
4.15. EVALUATE: The net force is largest when the velocity is changing most rapidly.
IDENTIFY: The net force and the acceleration are related by Newton’s second law. When the rocket is near the
!
surface of the earth the forces on it are the upward force F exerted on it because of the burning fuel and the
!
downward force Fgrav of gravity. Fgrav = mg .
SET UP: Let + y be upward. The weight of the rocket is Fgrav = (8.00 kg)(9.80 m/s 2 ) = 78.4 N .
EXECUTE: (a) At t = 0 , F = A = 100.0 N . At t = 2.00 s , F = A + (4.00 s 2 ) B = 150.0 N and
150.0 N − 100.0 N
= 12.5 N/s 2 .
4.00 s 2
(b) (i) At t = 0 , F = A = 100.0 N . The net force is
B= ∑F y ay = m ∑F y = y = F − Fgrav = 100.0 N − 78.4 N = 21.6 N . 21.6 N
= 2.70 m/s 2 . (ii) At t = 3.00 s, F = A + B(3.00 s) 2 = 212.5 N .
8.00 kg = 212.5 N − 78.4 N = 134.1 N . a y = ∑F y m = 134.1 N
= 16.8 m/s 2 .
8.00 kg 212.5 N
= 26.6 m/s 2 .
8.00 kg
EVALUATE: The acceleration increases as F increases.
IDENTIFY: Use constant acceleration equations to calculate ax and t. Then use (c) Now Fgrav = 0 and
4.16. ∑F ∑F y = F = 212.5 N . a y = ! ! ∑ F = ma to calculate the net force.
SET UP: Let + x be in the direction of motion of the electron.
2
2
EXECUTE: (a) v0 x = 0 , ( x − x0 ) = 1.80 × 10−2 m , vx = 3.00 × 106 m/s . vx = v0 x + 2ax ( x − x0 ) gives ax = 2
2
vx − v0 x
(3.00 × 106 m/s) 2 − 0
=
= 2.50 × 1014 m/s 2
2( x − x0 )
2(1.80 × 10−2 m) (b) vx = v0 x + axt gives t =
(c) 4.17. ∑F x = max = (9.11 × 10−31 kg)(2.50 × 1014 m/s 2 ) = 2.28 × 10−16 N . EVALUATE: The acceleration is in the direction of motion since the speed is increasing, and the net force is in the
direction of the acceleration.
IDENTIFY and SET UP: F = ma. We must use w = mg to find the mass of the boulder.
EXECUTE: 4.18. vx − v0 x 3.00 × 106 m/s − 0
=
= 1.2 × 10−8 s
2.50 × 1014 m/s 2
ax m= w
2400 N
=
= 244.9 kg
g 9.80 m/s 2 Then F = ma = (244.9 kg)(12.0 m/s 2 ) = 2940 N.
EVALUATE: We must use mass in Newton’s second law. Mass and weight are proportional.
!
!
IDENTIFY: Apply ∑ F = ma . SET UP: m = w / g = (71.2 N) /(9.80 m/s 2 ) = 7.27 kg . Fx 160 N
=
= 22.0 m/s 2
m 7.27 kg
EVALUATE: The weight of the ball is a vertical force and doesn’t affect the horizontal acceleration. However, the
weight is used to calculate the mass.
EXECUTE: ax = Newton’s Laws of Motion 4.19. 4.20. 4-5 IDENTIFY and SET UP: w = mg. The mass of the watermelon is constant, independent of its location. Its weight
differs on earth and Jupiter’s moon. Use the information about the watermelon’s weight on earth to calculate its
mass:
w
44.0 N
= 4.49 kg.
EXECUTE: w = mg gives that m = =
g 9.80 m/s 2
On Jupiter’s moon, m = 4.49 kg, the same as on earth. Thus the weight on Jupiter’s moon is w = mg = (4.49 kg)(1.81 m/s 2 ) = 8.13 N.
EVALUATE: The weight of the watermelon is less on Io, since g is smaller there.
IDENTIFY: Weight and mass are related by w = mg . The mass is constant but g and w depend on location.
SET UP: On earth, g = 9.80 m/s 2 .
EXECUTE: (a) 4.21. w
w
w
= m , which is constant, so E = A . wE = 17.5 N , g E = 9.80 m/s 2 , and wA = 3.24 N .
g
gE gA ⎛w ⎞
⎛ 3.24 N ⎞
2
2
gA = ⎜ A ⎟ gE = ⎜
⎟ (9.80 m/s ) = 1.81 m/s .
wE ⎠
⎝ 17.5 N ⎠
⎝
w
17.5 N
= 1.79 kg .
(b) m = E =
g E 9.80 m/s 2
EVALUATE: The weight at a location and the acceleration due to gravity at that location are directly proportional.
IDENTIFY: Apply ∑ Fx = max to find the resultant horizontal force. SET UP: Let the acceleration be in the + x direction.
EXECUTE: ∑ Fx = max = (55 kg)(15 m/s 2 ) = 825 N . The force is exerted by the blocks. The blocks push on the 4.22. 4.23. sprinter because the sprinter pushes on the blocks.
EVALUATE: The force the blocks exert on the sprinter has the same magnitude as the force the sprinter exerts on
the blocks. The harder the sprinter pushes, the greater the force on him.
!
!
IDENTIFY: ∑ F = ma refers to forces that all act on one object. The third law refers to forces that a pair of
objects exert on each other.
SET UP: An object is in equilibrium if the vector sum of all the forces on it is zero. A third law pair of forces
have the same magnitude regardless of the motion of either object.
EXECUTE: (a) the earth (gravity)
(b) 4 N; the book
(c) no, these two forces are exerted on the same object
(d) 4 N; the earth; the book; upward
(e) 4 N, the hand; the book; downward
(f) second (The two forces are exerted on the same object and this object has zero acceleration.)
(g) third (The forces are between a pair of objects.)
(h) No. There is a net upward force on the book equal to 1 N.
(i) No. The force exerted on the book by your hand is 5 N, upward. The force exerted on the book by the earth is
4 N, downward.
(j) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction.
(k) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction.
(l) One, only the gravity force.
(m) No. There is a net downward force of 5 N exerted on the book.
EVALUATE: Newton’s second and third laws give complementary information about the forces that act.
IDENTIFY: Identify the forces on the bottle.
SET UP: Classify forces as contact or noncontact forces. The noncontact force is gravity and the contact forces
come from things that touch the object. Gravity is always directed downward toward the center of the earth. Air
resistance is always directed opposite to the velocity of the object relative to the air.
EXECUTE: (a) The free-body diagram for the bottle is sketched in Figure 4.23a The only forces on the bottle are gravity
(downward) and air resistance (upward). Figure 4.23a
(b) 4-6 Chapter 4 w is the force of gravity that the earth exerts
on the bottle. The reaction to this force is w′,
force that the bottle exerts on the earth Figure 4.23b
Note that these two equal and opposite forces produce very different accelerations because the bottle and the earth
have very different masses.
Fair is the force that the air exerts on the bottle and is upward. The reaction to this force is a downward force Fa′ir 4.24. 4.25. that the bottle exerts on the air. These two forces have equal magnitudes and opposite directions.
EVALUATE: The only thing in contact with the bottle while it is falling is the air. Newton’s third law always
deals with forces on two different objects.
IDENTIFY: The reaction forces in Newton’s third law are always between a pair of objects. In Newton’s second
law all the forces act on a single object.
SET UP: Let + y be downward. m = w / g .
EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of
magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational
∑ Fy = a gives
force that the passenger exerts on the earth, upward and also of magnitude 650 N.
y
m
650 N − 620 N
ay =
= 0.452 m/s 2 . The passenger’s acceleration is 0.452 m / s 2 , downward.
(650 N)/(9.80 m/s 2 )
EVALUATE: There is a net downward force on the passenger and the passenger has a downward acceleration.
IDENTIFY: Apply Newton’s second law to the earth.
SET UP: The force of gravity that the earth exerts on her is her weight, w = mg = (45 kg)(9.8 m/s 2 ) = 441 N. By
Newton’s 3rd law, she exerts an equal and opposite force on the earth.
!
!
!
Apply ∑ F = ma to the earth, with ∑ F = w = 441 N, but must use the mass of the earth for m. EXECUTE: 4.26. a= w
441 N
=
= 7.4 × 10−23 m/s 2 .
m 6.0 × 1024 kg EVALUATE: This is much smaller than her acceleration of 9.8 m/s 2 . The force she exerts on the earth equals in
magnitude the force the earth exerts on her, but the acceleration the force produces depends on the mass of the
object and her mass is much less than the mass of the earth.
!
IDENTIFY and SET UP: The only force on the ball is the gravity force, Fgrav . This force is mg , downward and is
independent of the motion of the object.
EXECUTE: The free-body diagram is sketched in Figure 4.26. The free-body diagram is the same in all cases.
EVALUATE: Some forces, such as friction, depend on the motion of the object but the gravity force does not. Figure 4.26
4.27. IDENTIFY: Identify the forces on each object.
SET UP: In each case the forces are the noncontact force of gravity (the weight) and the forces applied by objects
!
that are in contact with each crate. Each crate touches the floor and the other crate, and some object applies F to
crate A.
EXECUTE: (a) The free-body diagrams for each crate are given in Figure 4.27.
FAB (the force on mA due to mB ) and FBA (the force on mB due to mA ) form an action-reaction pair.
(b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to
accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case
balanced by the upward force of the surface on them. Newton’s Laws of Motion 4-7 EVALUATE: Crate B is accelerated by FBA and crate A is accelerated by the net force F − FAB . The greater the
total weight of the two crates, the greater their total mass and the smaller will be their acceleration. 4.28. 4.29. 4.30. Figure 4.27
IDENTIFY: The surface of block B can exert both a friction force and a normal force on block A. The friction force is
directed so as to oppose relative motion between blocks B and A. Gravity exerts a downward force w on block A.
SET UP: The pull is a force on B not on A.
EXECUTE: (a) If the table is frictionless there is a net horizontal force on the combined object of the two blocks,
and block B accelerates in the direction of the pull. The friction force that B exerts on A is to the right, to try to
prevent A from slipping relative to B as B accelerates to the right. The free-body diagram is sketched in Figure
4.28a. f is the friction force that B exerts on A and n is the normal force that B exerts on A.
(b) The pull and the friction force exerted on B by the table cancel and the net force on the system of two blocks is
zero. The blocks move with the same constant speed and B exerts no friction force on A. The free-body diagram is
sketched in Figure 4.28b.
EVALUATE: If in part (b) the pull force is decreased, block B will slow down, with an acceleration directed to the
left. In this case the friction force on A would be to the left, to prevent relative motion between the two blocks by
giving A an acceleration equal to that of B. Figure 4.28
IDENTIFY: Since the observer in the train sees the ball hang motionless, the ball must have the same acceleration
as the train car. By Newton’s second law, there must be a net force on the ball in the same direction as its
acceleration.
!
SET UP: The forces on the ball are gravity, which is w, downward, and the tension T in the string, which is
directed along the string.
EXECUTE: (a) The acceleration of the train is zero, so the acceleration of the ball is zero. There is no net
horizontal force on the ball and the string must hang vertically. The free-body diagram is sketched in Figure 4.29a.
(b) The train has a constant acceleration directed east so the ball must have a constant eastward acceleration. There
must be a net horizontal force on the ball, directed to the east. This net force must come from an eastward
!
component of T and the ball hangs with the string displaced west of vertical. The free-body diagram is sketched in
Figure 4.29b.
EVALUATE: When the motion of an object is described in an inertial frame, there must be a net force in the
direction of the acceleration. Figure 4.29
IDENTIFY: Identify the forces for each object. Action-reaction pairs of forces act between two objects.
SET UP: Friction is parallel to the surfaces and is directly to oppose relative motion between the surfaces.
EXECUTE: The free-body diagram for the box is given in Figure 4.30a. The free body diagram for the truck is
given in Figure 4.30b. The box’s friction force on the truck bed and the truck bed’s friction force on the box form
an action-reaction pair. There would also be some small air-resistance force action to the left, presumably
negligible at this speed. 4-8 Chapter 4 EVALUATE: The friction force on the box, exerted by the bed of the truck, is in the direction of the truck's
acceleration. This friction force can't be large enough to give the box the same acceleration that the truck has and
the truck acquires a greater speed than the box. Figure 4.30
4.31. 4.32. IDENTIFY: Identify the forces on the chair. The floor exerts a normal force and a friction force.
SET UP: Let + y be upward and let + x be in the direction of the motion of the chair.
EXECUTE: (a) The free-body diagram for the chair is given in Figure 4.31.
(b) For the chair, a y = 0 so ∑ Fy = ma y gives n − mg − F sin 37° = 0 and n = 142 N .
!
EVALUATE: n is larger than the weight because F has a downward component. Figure 4.31
!
!
IDENTIFY: Identify the forces on the skier and apply ∑ F = ma . Constant speed means a = 0 .
SET UP: Use coordinates that are parallel and perpendicular to the slope.
EXECUTE: (a) The free-body diagram for the skier is given in Figure 4.32.
(b) ∑ Fx = max with ax = 0 gives T = mg sin θ = (65.0 kg)(9.80 m / s 2 )sin 26.0° = 279 N .
EVALUATE: T is less than the weight of the skier. It is equal to the component of the weight that is parallel to the
incline. 4.33. Figure 4.32
!
!
!
IDENTIFY: ∑ F = ma must be satisfied for each object. Newton’s third law says that the force FC on T that the car
!
exerts on the truck is equal in magnitude and opposite in direction to the force FT on C that the truck exerts on the
car.
!
SET UP: The only horizontal force on the car is the force FT on C exerted by the truck. The car exerts a force
!
!
FC on T on the truck. There is also a horizontal friction force f that the highway surface exerts on the truck. Assume
the system is accelerating to the right in the free-body diagrams.
EXECUTE: (a) The free-body diagram for the car is sketched in Figure 4.33a
(b) The free-body diagram for the truck is sketched in Figure 4.33b. Newton’s Laws of Motion 4-9 !
(c) The friction force f accelerates the system forward. The tires of the truck push backwards on the highway
surface as they rotate, so by Newton’s third law the roadway pushes forward on the tires.
!
EVALUATE: FT on C and FC on T each equal the tension T in the rope. Both objects have the same acceleration a .
T = mC a and f − T = mT a , so f = ( mC + mT ) a . The acceleration of the two objects is proportional to f. Figure 4.33
4.34. IDENTIFY: Use a constant acceleration equation to find the stopping time and acceleration. Then use
!
!
∑ F = ma to calculate the force.
!
SET UP: Let + x be in the direction the bullet is traveling. F is the force the wood exerts on the bullet.
⎛v +v ⎞
EXECUTE: (a) v0 x = 350 m/s , vx = 0 and ( x − x0 ) = 0.130 m . ( x − x0 ) = ⎜ 0 x x ⎟ t
⎝2⎠
gives t = 2( x − x0 ) 2(0.130 m)
=
= 7.43 × 10−4 s .
v0 x + vx
350 m/s 2
2
(b) vx = v0 x + 2ax ( x − x0 ) gives ax = ∑F x 4.35. 2
2
vx − v0 x
0 − (350 m/s) 2
=
= −4.71× 105 m/s 2
2( x − x0 )
2(0.130 m) = max gives − F = max and F = − max = −(1.80 × 10−3 kg)( − 4.71 × 105 m/s 2 ) = 848 N . EVALUATE: The acceleration and net force are opposite to the direction of motion of the bullet.
IDENTIFY: Vector addition problem. Write the vector addition equation in component form. We know one vector
and its resultant and are asked to solve for the other vector.
!
!
SET UP: Use coordinates with the + x -axis along F1 and the + y -axis along R; as shown in Figure 4.35a.
F1x = +1300 N, F1 y = 0
Rx = 0, Ry = +1300 N Figure 4.35a
!!
!
!
!!
F1 + F2 = R, so F2 = R − F1
EXECUTE: F2 x = Rx − F1x = 0 − 1300 N = −1300 N F2 y = Ry − F1 y = +1300 N − 0 = +1300 N
!
The components of F2 are sketched in Figure 4.35b.
F2 = F22x + F22y = (−1300 N) 2 + (1300 N) 2
F = 1840 N
F
+1300 N
tan θ = 2 y =
= −1.00
F2 x −1300 N
θ = 135° 4.36. Figure 4.35b
!
!
The magnitude of F2 is 1840 N and its direction is 135° counterclockwise from the direction of F1.
!
!
!
EVALUATE: F2 has a negative x-component to cancel F1 and a y-component to equal R.
IDENTIFY: Use the motion of the ball to calculate g, the acceleration of gravity on the planet. Then w = mg .
SET UP: Let + y be downward and take y0 = 0 . v0 y = 0 since the ball is released from rest. 4-10 Chapter 4 12
1
gt gives 10.0 m = g (2.2 s) 2 . g = 4.13 m / s 2 and then
2
2
wX = mg X = (0.100 kg)(4.03 m / s 2 ) = 0.41 N .
EVALUATE: g on Planet X is smaller than on earth and the object weighs less than it would on earth.
IDENTIFY: If the box moves in the + x -direction it must have a y = 0, so ∑ Fy = 0. EXECUTE: Get g on X: y = 4.37. The smallest force the child can exert and still
produce such motion is a force that makes the
y-components of all three forces sum to zero,
but that doesn’t have any x-component. Figure 4.37
!
!
!
F1 and F2 are sketched in Figure 4.37. Let F3 be the force exerted by the child. SET UP: ∑F y = ma y implies F1 y + F2 y + F3 y = 0, so F3 y = −( F1 y + F2 y ). EXECUTE: F1 y = + F1 sin 60° = (100 N)sin 60° = 86.6 N F2 y = + F2 sin( −30°) = − F2 sin 30° = −(140 N)sin 30° = −70.0 N
Then F3 y = −( F1 y + F2 y ) = −(86.6 N − 70.0 N) = −16.6 N; F3 x = 0
The smallest force the child can exert has magnitude 17 N and is directed at 90° clockwise from the + x -axis
shown in the figure.
(b) IDENTIFY and SET UP: Apply ∑ Fx = max . We know the forces and ax so can solve for m. The force exerted
by the child is in the − y -direction and has no x-component. EXECUTE: F1x = F1 cos 60° = 50 N F2 x = F2 cos30° = 121.2 N ∑F = F
∑F
m=
x 1x + F2 x = 50 N + 121.2 N = 171.2 N 171.2 N
x
=
= 85.6 kg
2.00 m/s 2
ax
Then w = mg = 840 N. EVALUATE: In part (b) we don’t need to consider the y-component of Newton’s second law. a y = 0 so the mass ∑ Fy = ma y equation.
!
!
Use ∑ F = ma to calculate the acceleration of the tanker and then use constant acceleration doesn’t appear in the 4.38. IDENTIFY: kinematic equations.
SET UP: Let + x be the direction the tanker is moving initially. Then ax = − F / m . EXECUTE: 2
2
vx = v0 x + 2ax ( x − x0 ) says that if the reef weren't there the ship would stop in a distance of x − x0 = − 2
2
v0 x
v0
mv 2 (3.6 × 107 kg)(1.5 m / s) 2
=
= 0=
= 506 m,
2ax 2( F / m) 2 F
2(8.0 × 104 N) 2
2
so the ship would hit the reef. The speed when the tanker hits the reef is found from vx = v0 x + 2ax ( x − x0 ) , so it is 2
v = v0 − (2 Fx / m) = (1.5 m/s) 2 − 4.39. 2(8.0 × 104 N)(500 m)
= 0.17 m/s,
(3.6 × 107 kg) and the oil should be safe.
EVALUATE: The force and acceleration are directed opposite to the initial motion of the tanker and the speed
decreases.
IDENTIFY: We can apply constant acceleration equations to relate the kinematic variables and we can use
Newton’s second law to relate the forces and acceleration.
(a) SET UP: First use the information given about the height of the jump to calculate the speed he has at the
instant his feet leave the ground. Use a coordinate system with the + y -axis upward and the origin at the position
when his feet leave the ground. Newton’s Laws of Motion 4-11 v y = 0 (at the maximum height), v0 y = ?, a y = −9.80 m/s 2 , y − y0 = +1.2 m
2
2
v y = v0 y + 2a y ( y − y0 ) EXECUTE: v0 y = −2a y ( y − y0 ) = −2( −9.80 m/s 2 )(1.2 m) = 4.85 m/s (b) SET UP: Now consider the acceleration phase, from when he starts to jump until when his feet leave the
ground. Use a coordinate system where the + y -axis is upward and the origin is at his position when he starts his
jump.
EXECUTE: Calculate the average acceleration:
v y − v0 y 4.89 m/s − 0
=
= 16.2 m/s 2
0.300 s
t
(c) SET UP: Finally, find the average upward force that the ground must exert on him to produce this average
upward acceleration. (Don’t forget about the downward force of gravity.) The forces are sketched in Figure 4.39.
EXECUTE:
890 N
m = w/ g =
= 90.8 kg
9.80 m/s 2
∑ Fy = ma y
(aav ) y = Fav − mg = m(aav ) y
Fav = m( g + (aav ) y )
Fav = 90.8 kg(9.80 m/s 2 + 16.2 m/s 2 )
Fav = 2360 N Figure 4.39 4.40. This is the average force exerted on him by the ground. But by Newton’s 3rd law, the average force he exerts on
the ground is equal and opposite, so is 2360 N, downward.
EVALUATE: In order for him to accelerate upward, the ground must exert an upward force greater than his
weight.
IDENTIFY: Use constant acceleration equations to calculate the acceleration ax that would be required. Then use ∑F x = max to find the necessary force. SET UP: Let + x be the direction of the initial motion of the auto.
2
v0 x
2
2
. The force F is directed opposite to the
EXECUTE: vx = v0 x + 2ax ( x − x0 ) with vx = 0 gives ax = −
2( x − x0 )
motion and ax = − F
. Equating these two expressions for ax gives
m
F =m 4.41. 2
v0 x
(12.5 m / s) 2
= (850 kg)
= 3.7 × 106 N.
2( x − x0 )
2(1.8 × 10−2 m) EVALUATE: A very large force is required to stop such a massive object in such a short distance.
IDENTIFY: Apply Newton’s second law to calculate a.
(a) SET UP: The free-body diagram for the bucket is sketched in Figure 4.41. The net force on the bucket
is T − mg , upward. Figure 4.41 4-12 Chapter 4 (b) EXECUTE: ∑F y a= 4.42. = ma y gives T − mg = ma
T − mg 75.0 N − (4.80 kg)(9.80 m/s 2 ) 75.0 N − 47.04 N
=
=
= 5.82 m/s 2 .
m
4.80 kg
4.80 kg EVALUATE: The weight of the bucket is 47.0 N. The upward force exerted by the cord is larger than this, so the
bucket accelerates upward.
!
!
IDENTIFY: Apply ∑ F = ma to the parachutist.
!
SET UP: Let + y be upward. Fair is the force of air resistance.
EXECUTE: (a) w = mg = (55.0 kg)(9.80 m/s 2 ) = 539 N
(b) The free-body diagram is given in Fig. 4.42. ∑F ∑F y = Fair − w = 620 N − 539 N = 81 N . The net force is upward. 81 N
y
=
= 1.5 m/s 2 , upward.
55.0 kg
m
EVALUATE: Both the net force and the acceleration are upward. Since her velocity is downward and her
acceleration is upward, her speed decreases. (c) a y = Figure 4.42
4.43. IDENTIFY: Use Newton’s 2nd law to relate the acceleration and forces for each crate.
(a) SET UP: Since the crates are connected by a rope, they both have the same acceleration, 2.50 m/s 2 .
(b) The forces on the 4.00 kg crate are shown in Figure 4.43a. EXECUTE:
∑ Fx = max
T = m1a = (4.00 kg)(2.50 m/s 2 ) = 10.0 N. Figure 4.43a
(c) SET UP: Forces on the 6.00 kg crate are shown in Figure 4.43b The crate accelerates to the right,
so the net force is to the right.
F must be larger than T. Figure 4.43b
(d) EXECUTE: ∑F x = max gives F − T = m2 a F = T + m2 a = 10.0 N + (6.00 kg)(2.50 m/s 2 ) = 10.0 N + 15.0 N = 25.0 N Newton’s Laws of Motion 4-13 EVALUATE: We can also consider the two crates and the rope connecting them as a single object of mass
m = m1 + m2 = 10.0 kg. The free-body diagram is sketched in Figure 4.43c. ∑F x = max F = ma = (10.0 kg)(2.50 m/s 2 ) = 25.0 N
This agrees with our answer in part (d). Figure 4.43c
4.44. 4.45. IDENTIFY: Apply Newton's second and third laws.
SET UP: Action-reaction forces act between a pair of objects. In the second law all the forces act on the same
object.
EXECUTE: (a) The force the astronaut exerts on the cable and the force that the cable exerts on the astronaut are
an action-reaction pair, so the cable exerts a force of 80.0 N on the astronaut.
(b) The cable is under tension.
F
80.0 N
= 0.762 m / s 2 .
(c) a = =
m 105.0 kg
(d) There is no net force on the massless cable, so the force that the shuttle exerts on the cable must be 80.0 N (this
is not an action-reaction pair). Thus, the force that the cable exerts on the shuttle must be 80.0 N.
F
80.0 N
(e) a = =
= 8.84 × 10−4 m / s 2 .
m 9.05 × 104 kg
EVALUATE: Since the cable is massless the net force on it is zero and the tension is the same at each end.
IDENTIFY and SET UP: Take derivatives of x (t ) to find vx and ax . Use Newton’s second law to relate the
acceleration to the net force on the object.
EXECUTE:
(a) x = (9.0 × 103 m/s 2 )t 2 − (8.0 × 104 m/s3 )t 3
x = 0 at t = 0
When t = 0.025 s, x = (9.0 × 103 m/s 2 )(0.025 s) 2 − (8.0 × 104 m/s3 )(0.025 s)3 = 4.4 m.
The length of the barrel must be 4.4 m.
dx
(b) vx =
= (18.0 × 103 m/s 2 )t − (24.0 × 104 m/s3 )t 2
dt
At t = 0, vx = 0 (object starts from rest).
At t = 0.025 s, when the object reaches the end of the barrel,
vx = (18.0 × 103 m/s 2 )(0.025 s) − (24.0 × 104 m/s3 )(0.025 s) 2 = 300 m/s (c) ∑F x = max , so must find ax . dvx
= 18.0 × 103 m/s 2 − (48.0 × 104 m/s3 )t
dt
(i) At t = 0, ax = 18.0 × 103 m/s 2 and ∑ Fx = (1.50 kg)(18.0 × 103 m/s 2 ) = 2.7 × 104 N.
ax = (ii) At t = 0.025 s, ax = 18 × 103 m/s 2 − (48.0 × 104 m/s3 )(0.025 s) = 6.0 × 103 m/s 2 and ∑F x 4.46. = (1.50 kg)(6.0 × 103 m/s 2 ) = 9.0 × 103 N. EVALUATE: The acceleration and net force decrease as the object moves along the barrel.
!
!
IDENTIFY: Apply ∑ F = ma and solve for the mass m of the spacecraft.
SET UP: w = mg . Let + y be upward.
EXECUTE: (a) The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward.
When it is speeding up, the acceleration is downward.
(b) In each case the net force is in the direction of the acceleration. Speeding up: w > F and the net force is
downward. Slowing down: w < F and the net force is upward. 4-14 Chapter 4 (c) Denote the y-component of the acceleration when the thrust is F1 by a1 and the y-component of the
acceleration when the thrust is F2 by a1. a y = +1.20 m/s 2 and a2 = −0.80 m/s 2 . The forces and accelerations are
then related by F1 − w = ma1 , F2 − w = ma2 . Dividing the first of these by the second to eliminate the mass gives
F1 − w a1
= , and solving for the weight w gives
F2 − w a2
w= a1F2 − a2 F1
. Substituting the given numbers, with + y upward, gives
a1 − a2
w= 4.47. (1.20 m / s 2 )(10.0 × 103 N) − (−0.80 m / s 2 )(25.0 × 103 N)
= 16.0 × 103 N.
1.20 m / s 2 − (−0.80 m / s 2 ) EVALUATE: The acceleration due to gravity at the surface of Mercury did not need to be found.
IDENTIFY: The ship and instrument have the same acceleration. The forces and acceleration are related by
Newton’s second law. We can use a constant acceleration equation to calculate the acceleration from the
information given about the motion.
!
SET UP: Let + y be upward. The forces on the instrument are the upward tension T exerted by the wire and the
!
downward force w of gravity. w = mg = (6.50 kg)(9.80 m/s 2 ) = 63.7 N
EXECUTE: (a) The free-body diagram is sketched in Figure 4.47. The acceleration is upward, so T > w .
2( y − y0 ) 2(276 m)
=
= 2.45 m/s 2 .
y − y0 = 276 m , t = 15.0 s , v0 y = 0 . y − y0 = v0 yt + 1 a yt 2 gives a y =
2
t2
(15.0 s) 2 ∑F y = ma y gives T − w = ma and T = w + ma = 63.7 N + (6.50 kg)(2.45 m/s 2 ) = 79.6 N . EVALUATE: There must be a net force in the direction of the acceleration. Figure 4.47
4.48.
4.49. If the rocket is moving downward and its speed is decreasing, its acceleration is upward, just as in Problem 4.47.
The solution is identical to that of Problem 4.47.
!
!
IDENTIFY: Apply ∑ F = ma to the gymnast. SET UP: The upward force on the gymnast gives the tension in the rope. The free-body diagram for the gymnast
is given in Figure 4.49.
EXECUTE: (a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so the tension must
equal the weight; T = mg .
(b) No motion is no acceleration, so the tension is again the gymnast’s weight.
!
!
(c) T − w = T − mg = ma = m a (the acceleration is upward, the same direction as the tension), so T = m( g + a ) .
!
(d) T − w = T − mg = ma = − m a (the acceleration is downward, the opposite direction to the tension), so
!
T = m( g − a ) .
EVALUATE: When she accelerates upward the tension is greater than her weight and when she accelerates
downward the tension is less than her weight. Figure 4.49 Newton’s Laws of Motion 4.50. 4-15 !
!
Apply ∑ F = ma to the elevator to relate the forces on it to the acceleration. IDENTIFY: The free-body diagram for the elevator is sketched in Figure 4.50. (a) SET UP: The net force is T − mg (upward). Figure 4.50 Take the + y -direction to be upward since that is the direction of the acceleration. The maximum upward
acceleration is obtained from the maximum possible tension in the cables.
EXECUTE: ∑ Fy = ma y gives T − mg = ma
T − mg 28,000 N − (2200 kg)(9.80 m/s 2 )
=
= 2.93 m/s 2 .
m
2200 kg
(b) What changes is the weight mg of the elevator.
T − mg 28,000 N − (2200 kg)(1.62 m/s 2 )
a=
=
= 11.1 m/s 2 .
m
2200 kg
EVALUATE: The cables can give the elevator a greater acceleration on the moon since the downward force of
gravity is less there and the same T then gives a greater net force.
IDENTIFY: He is in free-fall until he contacts the ground. Use the constant acceleration equations and
!
!
apply ∑ F = ma .
a= 4.51. SET UP: Take + y downward. While he is in the air, before he touches the ground, his acceleration is a y = 9.80 m/s 2 .
EXECUTE: 2
2
(a) v0 y = 0 , y − y0 = 3.10 m , and a y = 9.80 m/s 2 . v y = v0 y + 2a y ( y − y0 ) gives v y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(3.10 m) = 7.79 m/s
2
2
(b) v0 y = 7.79 m/s , v y = 0 , y − y0 = 0.60 m . v y = v0 y + 2a y ( y − y0 ) gives
2
2
v y − v0 y 0 − (7.79 m/s) 2
= −50.6 m/s 2 . The acceleration is upward.
2( y − y0 )
2(0.60 m)
!
(c) The free-body diagram is given in Fig. 4.51. F is the force the ground exerts on him.
∑ Fy = ma y gives mg − F = −ma . F = m( g + a) = (75.0 kg)(9.80 m/s2 + 50.6 m/s2 ) = 4.53 × 103 N , upward.
ay = = F
4.53 × 103 N
=
= 6.16 , so F = 6.16w .
w (75.0 kg)(9.80 m/s 2 ) !
By Newton's third law, the force his feet exert on the ground is − F .
EVALUATE: The force the ground exerts on him is about six times his weight. 4.52. IDENTIFY: Figure 4.51
!
!
Apply ∑ F = ma to the hammer head. Use a constant acceleration equation to relate the motion to the acceleration.
SET UP: Let + y be upward.
EXECUTE: (a) The free-body diagram for the hammer head is sketched in Figure 4.52.
2
2
(b) The acceleration of the hammer head is given by v y = v0 y + 2a y ( y − y0 ) with v y = 0 , v0 y = −3.2 m/s 2 and
2
y − y0 = −0.0045 m . a y = v0 y / 2( y − y0 ) = (3.2 m / s) 2 / 2(0.0045 cm) = 1.138 × 103 m / s 2 . The mass of the hammer 4-16 Chapter 4 head is its weight divided by g , (4.9 N) /(9.80 m / s 2 ) = 0.50 kg , and so the net force on the hammer head is
(0.50 kg)(1.138 × 103 m / s 2 ) = 570 N. This is the sum of the forces on the hammer head: the upward force that the
nail exerts, the downward weight and the downward 15-N force. The force that the nail exerts is then 590 N, and
this must be the magnitude of the force that the hammer head exerts on the nail.
(c) The distance the nail moves is 0.12 m, so the acceleration will be 4267 m / s 2 , and the net force on the hammer
head will be 2133 N. The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude
of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N.
EVALUATE: For the shorter stopping distance the acceleration has a larger magnitude and the force between the
nail and hammer head is larger. 4.53. IDENTIFY: Figure 4.52
!
!
Apply ∑ F = ma to some portion of the cable. SET UP: The free-body diagrams for the whole cable, the top half of the cable and the bottom half are sketched in
Figure 4.53. The cable is at rest, so in each diagram the net force is zero.
EXECUTE: (a) The net force on a point of the cable at the top is zero; the tension in the cable must be equal to the
weight w.
(b) The net force on the cable must be zero; the difference between the tensions at the top and bottom must be
equal to the weight w, and with the result of part (a), there is no tension at the bottom.
(c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be
half the weight, w / 2 . Equivalently, the net force on the upper half of the cable must be zero. From part (a) the
tension at the top is w, the weight of the top half is w / 2 and so the tension in the cable at the middle must
be w − w / 2 = w / 2 .
(d) A graph of T vs. distance will be a negatively sloped line.
EVALUATE: The tension decreases linearly from a value of w at the top to zero at the bottom of the cable. Figure 4.53
4.54. IDENTIFY: Note that in this problem the mass of the rope is given, and that it is not negligible compared to the
!
!
other masses. Apply ∑ F = ma to each object to relate the forces to the acceleration.
(a) SET UP: The free-body diagrams for each block and for the rope are given in Figure 4.54a. Figure 4.54a Newton’s Laws of Motion 4-17 Tt is the tension at the top of the rope and Tb is the tension at the bottom of the rope.
EXECUTE: (b) Treat the rope and the two blocks together as a single object, with mass
m = 6.00 kg + 4.00 kg + 5.00 kg = 15.0 kg. Take + y upward, since the acceleration is upward. The free-body
diagram is given in Figure 4.54b.
∑ Fy = ma y F − mg = ma
F − mg
m
200 N − (15.0 kg)(9.80 m/s 2 )
a=
= 3.53 m/s 2
15.0 kg a= Figure 4.54b
(c) Consider the forces on the top block (m = 6.00 kg), since the tension at the top of the rope (Tt ) will be one of
these forces. ∑F y = ma y F − mg − Tt = ma
Tt = F − m( g + a )
T = 200 N − (6.00 kg)(9.80 m/s 2 + 3.53 m/s 2 ) = 120 N
Figure 4.54c Alternatively, can consider the forces on the combined object rope plus bottom block (m = 9.00 kg): ∑F y = ma y Tt − mg = ma
Tt = m( g + a ) = 9.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 120 N,
which checks
Figure 4.54d
(d) One way to do this is to consider the forces on the top half of the rope (m = 2.00 kg). Let Tm be the tension at
the midpoint of the rope. ∑F y = ma y Tt − Tm − mg = ma
Tm = Tt − m( g + a ) = 120 N − 2.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 93.3 N
Figure 4.54e To check this answer we can alternatively consider the forces on the bottom half of the rope plus the lower block
taken together as a combined object (m = 2.00 kg + 5.00 kg = 7.00 kg): ∑F y = ma y Tm − mg = ma
Tm = m( g + a ) = 7.00 kg(9.80 m/s 2 + 3.53 m/s 2 ) = 93.3 N,
which checks
Figure 4.54f 4-18 4.55. Chapter 4 EVALUATE: The tension in the rope is not constant but increases from the bottom of the rope to the top. The
tension at the top of the rope must accelerate the rope as well the 5.00-kg block. The tension at the top of the rope
is less than F; there must be a net upward force on the 6.00-kg block.
!
!
IDENTIFY: Apply ∑ F = ma to the barbell and to the athlete. Use the motion of the barbell to calculate its acceleration.
SET UP: Let + y be upward.
EXECUTE: (a) The free-body diagrams for the baseball and for the athlete are sketched in Figure 4.55.
(b) The athlete’s weight is mg = (90.0 kg)(9.80 m / s 2 ) = 882 N . The upward acceleration of the barbell is found
from y − y0 = v0 yt + 1 a yt 2 . a y =
2 2( y − y0 ) 2(0.600 m)
=
= 0.469 m/s 2 . The force needed to lift the barbell is given
t2
(1.6 s) 2 by Flift − wbarbell = ma y . The barbell’s mass is (490 N) / (9.80 m/s 2 ) = 50.0 kg , so
Flift = wbarbell + ma = 490 N + (50.0 kg)(0.469 m / s 2 ) = 490 N + 23 N = 513 N .
The athlete is not accelerating, so Ffloor − Flift − wathlete = 0 . Ffloor = Flift + wathlete = 513 N + 882 N = 1395 N .
EVALUATE: Since the athlete pushes upward on the barbell with a force greater than its weight the barbell pushes
down on him and the normal force on the athlete is greater than the total weight, 1362 N, of the athlete plus
barbell. 4.56. IDENTIFY: Figure 4.55
!
!
Apply ∑ F = ma to the balloon and its passengers and cargo, both before and after objects are dropped overboard.
SET UP: When the acceleration is downward take + y to be downward and when the acceleration is upward take
+ y to be upward.
EXECUTE: (a) The free-body diagram for the descending balloon is given in Figure 4.56.
L is the lift force.
(b) ∑ Fy = ma y gives Mg − L = M ( g / 3) and L = 2Mg / 3 .
(c) Now + y is upward, so L − mg = m( g / 2) , where m is the mass remaining.
L = 2 Mg / 3 , so m = 4M / 9 . Mass 5M / 9 must be dropped overboard.
EVALUATE: In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the
total weight. Figure 4.56 Newton’s Laws of Motion 4.57. IDENTIFY:
SET UP: !
!
Apply ∑ F = ma to the entire chain and to each link. m = mass of one link. Let + y be upward. EXECUTE: (a) The free-body diagrams are sketched in Figure 4.57. Ftop is the force the top and middle links exert on each other. Fmiddle is the force the middle and bottom links exert on each other.
(b) (i) The weight of each link is mg = (0.300 kg)(9.80 m / s 2 ) = 2.94 N . Using the free-body diagram for the
whole chain: a= Fstudent − 3mg 12 N − 3(2.94 N)
3.18 N
=
=
= 3.53 m / s 2
3m
0.900 kg
0.900 kg (ii) The top link also accelerates at 3.53 m / s 2 , so Fstudent − Ftop − mg = ma .
Ftop = Fstudent − m( g + a ) = 12 N − (0.300 kg)(9.80 m/s 2 + 3.53 m/s 2 ) = 8.0 N .
EVALUATE: The force exerted by the middle link on the bottom link is given by Fmiddle − mg = ma and Fmiddle = m( g + a ) = 4.0 N . We can verify that with our results 4.58. y = ma y is satisfied for the middle link. Figure 4.57
!
!
!
!
!2
IDENTIFY: Calculate a from a = d 2 r / dt . Then Fnet = ma .
SET UP: w = mg
EXECUTE: Differentiating twice, the acceleration of the helicopter as a function of time is
!
!
ˆ
ˆ
ˆ
ˆ
a = (0.120 m / s3 )ti − (0.12 m / s 2 )k and at t = 5.0s , the acceleration is a = (0.60 m / s 2 )i − (0.12 m / s 2 )k .
The force is then
!
! w ! (2.75 × 105 N) ⎡
ˆ
ˆ
ˆ
ˆ
(0.60 m / s 2 )i − (0.12 m / s 2 )k ⎤ = (1.7 × 104 N)i − (3.4 × 103 N)k
F = ma = a =
⎦
g
(9.80 m / s 2 ) ⎣
EVALUATE: 4.59. ∑F IDENTIFY: The force and acceleration are in the same direction. They are both time dependent.
d 2x
Fx = max and ax = 2 .
dt dn
(t ) = nt n −1
dt
EXECUTE: The velocity as a function of time is vx (t ) = A − 3Bt 2 and the acceleration as a function of time is
SET UP: 4.60. ax (t ) = −6 Bt , and so the force as a function of time is Fx (t ) = ma (t ) = −6mBt .
EVALUATE: Since the acceleration is along the x-axis, the force is along the x-axis.
t!
!!
!!
IDENTIFY: a = F / m . v = v0 + ∫ a dt .
0 v0 = 0 since the object is initially at rest.
!
1 t!
1⎛ ˆ k
⎞
EXECUTE: v (t ) = ∫ F dt = ⎜ k1ti + 2 t 4 ˆ ⎟ .
j
0
m
m⎝
4
⎠
!
!
EVALUATE: F has both x and y components, so v develops x and y components.
IDENTIFY: Follow the steps specified in the problem.
dv dv dv dv
SET UP: The chain rule for differentiating says
=
= v.
dt dx dt dx
SET UP: 4.61. 4-19 4-20 Chapter 4 dv
cannot be integrated with respect to time, as the unknown
dt
C
dv
function v(t ) is part of the integrand. The equation must be separated before integration; that is, − dt = 2 and
m
v
Ct
11
− =− + ,
m
v v0 EXECUTE: (a) The equation of motion, −Cv 2 = m where v0 is the constant of integration that gives v = v0 at t = 0 . Note that this form shows that if v0 = 0 , there is
−1 dx ⎛ 1 Ct ⎞
no motion. This expression may be rewritten as v =
=⎜ + ⎟ ,
dt ⎝ v0 m ⎠
m ⎡ Ctv0 ⎤
which may be integrated to obtain x − x0 = ln ⎢1 +
.
C⎣
m⎥
⎦
To obtain x as a function of v, the time t must be eliminated in favor of v; from the expression obtained after the
m ⎛v ⎞
Ctv0 v0
first integration,
= − 1 , so x − x0 = ln ⎜ 0 ⎟ .
C ⎝v⎠
m
v
dv
dv
(b) Applying the chain rule, ∑ F = m = mv . Using the given expression for the net force,
dt
dx
⎛v⎞
C
m ⎛v ⎞
C
dv
⎛ dv ⎞
−Cv 2 = ⎜ v ⎟ m . − dx =
. Integrating gives − ( x − x0 ) = ln ⎜ ⎟ and x − x0 = ln ⎜ 0 ⎟ .
m
v0 ⎠
C ⎝v⎠
m
v
⎝ dx ⎠
⎝
EVALUATE: If C is positive, our expression for v (t ) shows it decreases from its value of v0 . As v decreases, so
does the acceleration and therefore the rate of decrease of v.
4.62. IDENTIFY: t t 0 0 x = ∫ vx dt and vx = ∫ ax dt , and similar equations apply to the y-component. SET UP: In this situation, the x-component of force depends explicitly on the y-component of position. As the ycomponent of force is given as an explicit function of time, v y and y can be found as functions of time and used in the expression for ax (t ) .
EXECUTE: a y = (k3 / m)t , so v y = (k3 / 2m)t 2 and y = ( k3 / 6m)t 3 , where the initial conditions v0 y = 0, y0 = 0 have been used. Then, the expressions for ax , vx and x are obtained as functions of time: ax = k1 k2 k3 3
+
t,
m 6m 2 k1
kk
k
kk
t + 2 32 t 4 and x = 1 t 2 + 2 3 2 t 5 .
m
24m
2m
120m
! ⎛ k1 2
k2 k3 5 ⎞ ˆ ⎛ k3 3 ⎞ ˆ
! ⎛k
kk
⎞ˆ ⎛ k
⎞
t+
t ⎟i + ⎜
t ⎟ j and v = ⎜ 1 t + 2 32 t 4 ⎟ i + ⎜ 3 t 2 ⎟ ˆ .
In vector form, r = ⎜
j
2
120m ⎠ ⎝ 6m ⎠
24m ⎠ ⎝ 2m ⎠
⎝m
⎝ 2m
EVALUATE: ax depends on time because it depends on y, and y is a function of time.
vx = 5 APPLYING NEWTON’S LAWS 5.1. IDENTIFY: a = 0 for each object. Apply ∑F y = ma y to each weight and to the pulley. SET UP: Take + y upward. The pulley has negligible mass. Let Tr be the tension in the rope and let Tc be the
tension in the chain.
EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a.
∑ Fy = ma y gives Tr = w = 25.0 N .
(b) The free-body diagram for the pulley is given in Figure 5.1b. Tc = 2Tr = 50.0 N .
EVALUATE: The tension is the same at all points along the rope. Figure 5.1a, b
5.2. 5.3. IDENTIFY: !
!
Apply ∑ F = ma to each weight. SET UP: Two forces act on each mass: w down and T ( = w) up.
EXECUTE: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w.
EVALUATE: The tension is the same in all three cases.
IDENTIFY: Both objects are at rest and a = 0 . Apply Newton’s first law to the appropriate object. The maximum
tension Tmax is at the top of the chain and the minimum tension is at the bottom of the chain.
SET UP: Let + y be upward. For the maximum tension take the object to be the chain plus the ball. For the
minimum tension take the object to be the ball. For the tension T three-fourths of the way up from the bottom of
the chain, take the chain below this point plus the ball to be the object. The free-body diagrams in each of these
three cases are sketched in Figures 5.3a, 5.3b and 5.3c. mb+c = 75.0 kg + 26.0 kg = 101.0 kg . mb = 75.0 kg . m is the mass of three-fourths of the chain: m = 3 (26.0 kg) = 19.5 kg .
4
EXECUTE: (a) From Figure 5.3a, ∑F y = 0 gives Tmax − mb+c g = 0 and Tmax = (101.0 kg)(9.80 m/s 2 ) = 990 N . ∑ F = 0 gives T − m g = 0 and T = (75.0 kg)(9.80 m/s ) = 735 N .
(b) From Figure 5.3c, ∑ F = 0 gives T − (m + m ) g = 0 and T = (19.5 kg + 75.0 kg)(9.80 m/s ) = 926 N .
From Figure 5.3b, 2 min y min b 2 y b 5-1 5-2 Chapter 5 EVALUATE: The tension in the chain increases linearly from the bottom to the top of the chain. Figure 5.3a–c
5.4. IDENTIFY: Apply Newton’s 1st law to the person. Each half of the rope exerts a force on him, directed along the
rope and equal to the tension T in the rope.
SET UP: (a) The force diagram for the person is given in Figure 5.4 T1 and T2 are the
tensions in each half of
the rope.
Figure 5.4
EXECUTE: ∑F x =0 T2 cosθ − T1 cosθ = 0
This says that T1 = T2 = T (The tension is the same on both sides of the person.) ∑F y =0 T1 sin θ + T2 sin θ − mg = 0
But T1 = T2 = T , so 2T sin θ = mg
T= mg
(90.0 kg)(9.80 m/s 2 )
=
= 2540 N
2sin θ
2sin10.0° (b) The relation 2T sin θ = mg still applies but now we are given that T = 2.50 × 104 N (the breaking strength) and
are asked to find θ . sin θ = 5.5. mg (90.0 kg)(9.80 m/s 2 )
=
= 0.01764, θ = 1.01°.
2T
2(2.50 × 104 N) EVALUATE: T = mg /(2sin θ ) says that T = mg / 2 when θ = 90° (rope is vertical).
T → ∞ when θ → 0 since the upward component of the tension becomes a smaller fraction of the tension.
!
!
IDENTIFY: Apply ∑ F = ma to the frame.
SET UP: Let w be the weight of the frame. Since the two wires make the same angle with the vertical, the tension
is the same in each wire. T = 0.75w .
EXECUTE: The vertical component of the force due to the tension in each wire must be half of the weight, and
w 3w
cosθ
this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical. =
2
4
and θ = arccos 2 = 48° .
3
EVALUATE: If θ = 0° , T = w / 2 and T → ∞ as θ → 90° . Therefore, there must be an angle where T = 3w / 4 . Applying Newton’s Laws 5.6. 5-3 IDENTIFY: Apply Newton’s 1st law to the car. The forces are the same as in Example 5.5.
SET UP: The free-body diagram is sketched in Figure 5.6. EXECUTE:
∑ Fx = max T cos α − n sin α = 0
T cos α = n sin α
∑ Fy = ma y n cos α + T sin α − w = 0
n cos α + T sin α = w Figure 5.6 ⎛ cos α ⎞
The first equation gives n = T ⎜
⎟.
⎝ sin α ⎠
Use this in the second equation to eliminate n:
⎛ cos α ⎞
⎜T
⎟ cos α + T sin α = w
⎝ sin α ⎠
Multiply this equation by sin α :
T (cos 2 α + sin 2 α ) = w sin α 5.7. T = w sin α (since cos 2 α + sin 2 α = 1 ).
⎛ cos α ⎞
⎛ cos α ⎞
Then n = T ⎜
⎟ = w sin α ⎜
⎟ = w cos α .
⎝ sin α ⎠
⎝ sin α ⎠
EVALUATE: These results are the same as obtained in Example 5.5. The choice of coordinate axes is up to us.
Some choices may make the calculation easier, but the results are the same for any choice of axes.
!
!
IDENTIFY: Apply ∑ F = ma to the car.
SET UP: Use coordinates with + x parallel to the surface of the street.
EXECUTE: ∑ Fx = 0 gives T = w sin α . F = mg sinθ = (1390 kg)(9.80 m/s 2 )sin17.5° = 4.10 × 103 N . 5.8. EVALUATE: The force required is less than the weight of the car by the factor sin α .
IDENTIFY: Apply Newton’s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the
cable.
SET UP: The force diagram for the wrecking ball is sketched in Figure 5.8. Figure 5.8
EXECUTE:
(a) ∑ Fy = ma y TB cos 40° − mg = 0
mg
(4090 kg)(9.80 m/s 2 )
=
= 5.23 × 104 N
cos 40°
cos 40°
(b) ∑ Fx = max
TB = TB sin 40° − TA = 0
TA = TB sin 40° = 3.36 × 104 N
EVALUATE: If the angle 40° is replaces by 0° (cable B is vertical), then TB = mg and TA = 0. 5-4 5.9. Chapter 5 Apply IDENTIFY: ! ! ∑ F = ma to the object and to the knot where the cords are joined. Let + y be upward and + x be to the right. SET UP:
EXECUTE: (a) TC = w, TA sin30° + TB sin 45° = TC = w, and TA cos30° − TB cos 45° = 0. Since sin 45° = cos 45°, adding the last two equations gives TA (cos30° + sin 30°) = w, and so TA =
TB = TA w
= 0.732 w. Then,
1.366 cos30°
= 0.897 w.
cos 45° (b) Similar to part (a), TC = w, − TA cos60° + TB sin 45° = w, and TA sin 60° − TB cos 45° = 0. Adding these two equations, TA =
EVALUATE:
5.10. w
sin 60°
= 2.73w, and TB = TA
= 3.35w.
cos 45°
(sin 60° − cos 60°) In part (a), TA + TB > w since only the vertical components of TA and TB hold the object against gravity. In part (b), since TA has a downward component TB is greater than w.
IDENTIFY: Apply Newton’s first law to the car.
SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp.
EXECUTE: (a) The free-body diagram for the car is given in Figure 5.10. The vertical weight w and the tension T
in the cable have each been replaced by their x and y components.
sin 25.0°
sin 25.0°
(b) ∑ Fx = 0 gives T cos31.0° − w sin 25.0° = 0 and T = w
= (1130 kg)(9.80 m/s 2 )
= 5460 N .
cos31.0°
cos31.0°
(c) ∑ Fy = 0 gives n + T sin 31.0° − w cos 25.0° = 0 and n = w cos 25.0°− T sin 31.0° = (1130 kg)(9.80 m/s 2 )cos 25.0° − (5460 N)sin 31.0° = 7220 N
EVALUATE: We could also use coordinates that are horizontal and vertical and would obtain the same values of n
and T. Figure 5.10
5.11. IDENTIFY: Since the velocity is constant, apply Newton’s first law to the piano. The push applied by the man
must oppose the component of gravity down the incline.
!
SET UP: The free-body diagrams for the two cases are shown in Figures 5.11a and b. F is the force applied by
the man. Use the coordinates shown in the figure.
EXECUTE: (a) ∑ Fx = 0 gives F − w sin11.0° = 0 and F = (180 kg)(9.80 m/s 2 )sin11.0° = 337 N .
(b) ∑F y = 0 gives n cos11.0° − w = 0 and n = w
.
cos11.0° w
⎛
⎞
F =⎜
⎟ sin11.0° = w tan11.0° = 343 N .
cos11.0° ⎠
⎝ ∑F x = 0 gives F − n sin11.0° = 0 and Applying Newton’s Laws 5-5 EVALUATE: A slightly greater force is required when the man pushes parallel to the floor. If the slope angle of
the incline were larger, sin α and tan α would differ more and there would be more difference in the force needed
in each case. Figure 5.11a, b
5.12. IDENTIFY: Apply Newton’s 1st law to the hanging weight and to each knot. The tension force at each end of a
string is the same.
(a) Let the tensions in the three strings be T, T ′, and T ′′, as shown in Figure 5.12a. Figure 5.12a
SET UP: The free-body diagram for the block is given in Figure 5.12b.
EXECUTE:
∑ Fy = 0 T′ − w = 0
T ′ = w = 60.0 N Figure 5.12b
SET UP: The free-body diagram for the lower knot is given in Figure 5.12c.
EXECUTE:
∑ Fy = 0 T sin 45° − T ′ = 0
T′
60.0 N
=
= 84.9 N
T=
sin 45° sin 45° Figure 5.12c 5-6 Chapter 5 (b) Apply ∑F x ∑F x = 0 to the force diagram for the lower knot: =0 F2 = T cos 45° = (84.9 N)cos 45° = 60.0 N
SET UP: The free-body diagram for the upper knot is given in Figure 5.12d.
EXECUTE:
∑ Fx = 0 T cos 45° − F1 = 0
F1 = (84.9 N)cos 45°
F1 = 60.0 N
Figure 5.12d Note that F1 = F2 .
EVALUATE: Applying ∑F y = 0 to the upper knot gives T ′′ = T sin 45° = 60.0 N = w. If we treat the whole system as a single object, the force diagram is given in Figure 5.12e. ∑F
∑F x = 0 gives F2 = F1 , which checks y = 0 gives T ′′ = w, which checks Figure 5.12e
5.13. IDENTIFY: Apply Newton’s first law to the ball. The force of the wall on the ball and the force of the ball on the
wall are related by Newton’s third law.
SET UP: The forces on the ball are its weight, the tension in the wire, and the normal force applied by the wall.
16.0 cm
To calculate the angle φ that the wire makes with the wall, use Figure 5.13a. sin φ =
and φ = 20.35°
46.0 cm
EXECUTE: (a) The free-body diagram is shown in Figure 5.13b. Use the x and y coordinates shown in the figure.
w
(45.0 kg)(9.80 m/s 2 )
= 470 N
∑ Fy = 0 gives T cosφ − w = 0 and T = cosφ =
cos 20.35°
(b) ∑F x = 0 gives T sin φ − n = 0 . n = (470 N)sin 20.35° = 163 N . By Newton’s third law, the force the ball exerts on the wall is 163 N, directed to the right.
⎛w⎞
EVALUATE: n = ⎜
⎟ sin φ = w tan φ . As the angle φ decreases (by increasing the length of the wire), T
⎝ cos φ ⎠
decreases and n decreases. 5.14. Figure 5.13a, b
!
!
IDENTIFY: Apply ∑ F = ma to each block. a = 0 .
SET UP: Take + y perpendicular to the incline and + x parallel to the incline. Applying Newton’s Laws 5-7 EXECUTE: The free-body diagrams for each block, A and B, are given in Figure 5.14.
(a) For B, ∑ Fx = max gives T1 − w sin α = 0 and T1 = w sin α .
(b) For block A,
(c) ∑F y ∑F x = max gives T1 − T2 − w sin α = 0 and T2 = 2w sin α . = ma y for each block gives nA = nB = w cos α . (d) For α → 0 , T1 = T2 → 0 and nA = nB → w . For α → 90° , T1 = w , T2 = 2w and nA = nB = 0 .
EVALUATE: The two tensions are different but the two normal forces are the same. Figure 5.14a, b
5.15. IDENTIFY: Apply Newton’s first law to the ball. Treat the ball as a particle.
SET UP: The forces on the ball are gravity, the tension in the wire and the normal force exerted by the surface.
The normal force is perpendicular to the surface of the ramp. Use x and y axes that are horizontal and vertical.
EXECUTE: (a) The free-body diagram for the ball is given in Figure 5.15. The normal force has been replaced by
its x and y components.
mg
(b) ∑ Fy = 0 gives n cos35.0° − w = 0 and n =
= 1.22mg .
cos35.0°
(c) ∑ Fx = 0 gives T − n sin 35.0° = 0 and T = (1.22mg )sin 35.0° = 0.700mg .
EVALUATE: Note that the normal force is greater than the weight, and increases without limit as the angle of the
ramp increases towards 90° . The tension in the wire is w tan φ , where φ is the angle of the ramp and T also
increases without limit as φ → 90° . Figure 5.15
5.16. IDENTIFY: Apply Newton’s second law to the rocket plus its contents and to the power supply. Both the rocket
and the power supply have the same acceleration.
SET UP: The free-body diagrams for the rocket and for the power supply are given in Figures 5.16a and b. Since
the highest altitude of the rocket is 120 m, it is near to the surface of the earth and there is a downward gravity
force on each object. Let + y be upward, since that is the direction of the acceleration. The power supply has mass mps = (15.5 N) /(9.80 m/s 2 ) = 1.58 kg 5-8 Chapter 5 EXECUTE: a=
(b) (a) ∑F y = ma y applied to the rocket gives F − mr g = mr a . F − mr g 1720 N − (125 kg)(9.80 m/s 2 )
=
= 3.96 m/s 2 .
125 kg
mr ∑F y = ma y applied to the power supply gives n − mps g = mps a . n = mps ( g + a) = (1.58 kg)(9.80 m/s 2 + 3.96 m/s 2 ) = 21.7 N .
EVALUATE: The acceleration is constant while the thrust is constant and the normal force is constant while the
acceleration is constant. The altitude of 120 m is not used in the calculation. Figure 5.16a, b
5.17. IDENTIFY: Use the kinematic information to find the acceleration of the capsule and the stopping time. Use
Newton’s second law to find the force F that the ground exerted on the capsule during the crash.
SET UP: Let + y be upward. 311 km/h = 86.4 m/s . The free-body diagram for the capsule is given in
Figure 15.17.
2
2
EXECUTE: y − y0 = −0.810 m , v0 y = −86.4 m/s , v y = 0 . v y = v0 y + 2a y ( y − y0 ) gives ay =
(b) 2
2
v y − v0 y 2( y − y0 ) ∑F y = 0 − (−86.4 m/s) 2
= 4610 m/s 2 = 470 g .
2( −0.810) m = ma y applied to the capsule gives F − mg = ma and F = m( g + a) = (210 kg)(9.80 m/s 2 + 4610 m/s 2 ) = 9.70 × 105 N = 471w.
⎛ v + vy ⎞
2( y − y0 )
2(−0.810 m)
=
= 0.0187 s
(c) y − y0 = ⎜ 0 y
⎟ t gives t =
v0 y + v y
−86.4 m/s 2 + 0
2⎠
⎝
EVALUATE: The upward force exerted by the ground is much larger than the weight of the capsule and stops the
capsule in a short amount of time. After the capsule has come to rest, the ground still exerts a force mg on the
capsule, but the large 9.00 × 105 N force is exerted only for 0.0187 s. Figure 5.17
5.18. IDENTIFY: Apply Newton’s second law to the three sleds taken together as a composite object and to each
individual sled. All three sleds have the same horizontal acceleration a.
SET UP: The free-body diagram for the three sleds taken as a composite object is given in Figure 5.18a and for
each individual sled in Figure 5.18b-d. Let + x be to the right, in the direction of the acceleration. mtot = 60.0 kg .
EXECUTE: a= (a) ∑F x = max for the three sleds as a composite object gives P = mtot a and P
125 N
=
= 2.08 m/s 2 .
mtot 60.0 kg Applying Newton’s Laws (b) ∑F x 5-9 = max applied to the 10.0 kg sled gives P − TA = m10 a and TA = P − m10 a = 125 N − (10.0 kg)(2.08 m/s 2 ) = 104 N . ∑F x = max applied to the 30.0 kg sled gives TB = m30 a = (30.0 kg)(2.08 m/s ) = 62.4 N .
2 EVALUATE: If we apply x = max to the 20.0 kg sled and calculate a from TA and TB found in part (b), we get TA − TB 104 N − 62.4 N
=
= 2.08 m/s 2 , which agrees with the value we calculated in part (a).
m20
20.0 kg TA − TB = m20 a . a = 5.19. ∑F Figure 5.18a–d
!
!
IDENTIFY: Apply ∑ F = ma to the load of bricks and to the counterweight. The tension is the same at each end of the rope. The rope pulls up with the same force (T ) on the bricks and on the counterweight. The counterweight
accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude.
(a) SET UP: The free-body diagrams for the bricks and counterweight are given in Figure 5.19. Figure 5.19 ∑F = ma y to each object. The acceleration magnitude is the same for the two objects.
!
For the bricks take + y to be upward since a for the bricks is upward. For the counterweight take + y to be
!
downward since a is downward.
bricks: ∑ Fy = ma y
(b) EXECUTE: Apply y T − m1 g = m1a
counterweight: ∑F y = ma y m2 g − T = m2 a
Add these two equations to eliminate T:
(m2 − m1 ) g = (m1 + m2 ) a
⎛ m − m1 ⎞
⎛ 28.0 kg − 15.0 kg ⎞
2
2
a =⎜ 2
⎟g = ⎜
⎟ (9.80 m/s ) = 2.96 m/s
⎝ 15.0 kg + 28.0 kg ⎠
⎝ m1 + m2 ⎠
(c) T − m1 g = m1a gives T = m1 (a + g ) = (15.0 kg)(2.96 m/s 2 + 9.80 m/s 2 ) = 191 N
As a check, calculate T using the other equation.
m2 g − T = m2 a gives T = m2 ( g − a) = 28.0 kg(9.80 m/s 2 − 2.96 m/s 2 ) = 191 N, which checks. 5-10 5.20. Chapter 5 EVALUATE: The tension is 1.30 times the weight of the bricks; this causes the bricks to accelerate upward. The
tension is 0.696 times the weight of the counterweight; this causes the counterweight to accelerate downward. If
m1 = m2 , a = 0 and T = m1 g = m2 g . In this special case the objects don’t move. If m1 = 0, a = g and T = 0; in
this special case the counterweight is in free-fall. Our general result is correct in these two special cases.
IDENTIFY: In part (a) use the kinematic information and the constant acceleration equations to calculate the
!
!
!
!
acceleration of the ice. Then apply ∑ F = ma . In part (b) use ∑ F = ma to find the acceleration and use this in the constant acceleration equations to find the final speed.
SET UP: Figures 5.20a and b give the free-body diagrams for the ice both with and without friction. Let + x be
directed down the ramp, so + y is perpendicular to the ramp surface. Let φ be the angle between the ramp and the
horizontal. The gravity force has been replaced by its x and y components.
2
2
EXECUTE: (a) x − x0 = 1.50 m , v0 x = 0 , vx = 2.50 m/s . vx = v0 x + 2ax ( x − x0 ) gives
ax = 2
2
vx − v0 x
(2.50 m/s) 2 − 0
=
= 2.08 m/s 2 .
2( x − x0 )
2(1.50 m) ∑F x = max gives mg sin φ = ma and sin φ = a 2.08 m/s 2
=
.
g 9.80 m/s 2 φ = 12.3° .
(b) ∑F a= mg sin φ − f (8.00 kg)(9.80 m/s 2 )sin12.3° − 10.0 N
=
= 0.838 m/s 2 .
m
8.00 kg x = max gives mg sin φ − f = ma and 2
2
Then x − x0 = 1.50 m , v0 x = 0 , ax = 0.838 m/s 2 and vx = v0 x + 2ax ( x − x0 ) gives vx = 2ax ( x − x0 ) = 2(0.838 m/s 2 )(1.50 m) = 1.59 m/s
EVALUATE: 5.21. IDENTIFY: With friction present the speed at the bottom of the ramp is less. Figure 5.20a, b
!
!
Apply ∑ F = ma to each block. Each block has the same magnitude of acceleration a. SET UP: Assume the pulley is to the right of the 4.00 kg block. There is no friction force on the 4.00 kg block,
the only force on it is the tension in the rope. The 4.00 kg block therefore accelerates to the right and the suspended
block accelerates downward. Let + x be to the right for the 4.00 kg block, so for it ax = a , and let + y be downward for the suspended block, so for it a y = a .
(a) The free-body diagrams for each block are given in Figures 5.21a and b.
T
10.0 N
=
= 2.50 m/s 2 .
(b) ∑ Fx = max applied to the 4.00 kg block gives T = (4.00 kg)a and a =
4.00 kg 4.00 kg
EXECUTE: (c) ∑F m= y = ma y applied to the suspended block gives mg − T = ma and T
10.0 N
=
= 1.37 kg .
g − a 9.80 m/s 2 − 2.50 m/s 2 (d) The weight of the hanging block is mg = (1.37 kg)(9.80 m/s 2 ) = 13.4 N . This is greater than the tension in the
rope; T = 0.75mg . Applying Newton’s Laws 5-11 EVALUATE: Since the hanging block accelerates downward, the net force on this block must be downward and
the weight of the hanging block must be greater than the tension in the rope. Note that the blocks accelerate no
matter how small m is. It is not necessary to have m > 4.00 kg , and in fact in this problem m is less than 4.00 kg. Figure 5.21a, b
5.22. (a) Consider both gliders together as a single object, apply IDENTIFY: ! ! ∑ F = ma , and solve for a. Use a in a constant acceleration equation to find the required runway length.
!
!
(b) Apply ∑ F = ma to the second glider and solve for the tension Tg in the towrope that connects the two
gliders.
SET UP: In part (a), set the tension Tt in the towrope between the plane and the first glider equal to its maximum value, Tt = 12,000 N .
EXECUTE: (a) The free-body diagram for both gliders as a single object of mass 2m = 1400 kg is given in Figure
5.22a. ∑F x = max gives Tt − 2 f = (2m)a and a = Tt − 2 f 12,000 N − 5000 N
=
= 5.00 m/s 2 . Then
2m
1400 kg 2
2
ax = 5.00 m/s 2 , v0 x = 0 and vx = 40 m/s in vx = v0 x + 2ax ( x − x0 ) gives ( x − x0 ) = 2
2
vx − v0 x
= 160 m .
2a x (b) The free-body diagram for the second glider is given in Figure 5.22b.
∑ Fx = max gives Tg − f = ma and T = f + ma = 2500 N + (700 kg)(5.00 m/s2 ) = 6000 N .
EVALUATE: We can verify that ∑F x = max is also satisfied for the first glider. Figure 5.22a, b
5.23. IDENTIFY: The maximum tension in the chain is at the top of the chain. Apply ! ! ∑ F = ma to the composite object of chain and boulder. Use the constant acceleration kinematic equations to relate the acceleration to the time.
SET UP: Let + y be upward. The free-body diagram for the composite object is given in Figure 5.23.
T = 2.50 wchain . mtot = mchain + mboulder = 1325 kg .
EXECUTE: (a) ∑F y = ma y gives T − mtot g = mtot a . a = ⎛ 2.50[575 kg] ⎞
a =⎜
− 1⎟ (9.80 m/s 2 ) = 0.832 m/s 2 .
⎝ 1325 kg
⎠ ⎞
T − mtot g 2.50mchain g − mtot g ⎛ 2.50mchain
=
=⎜
− 1⎟ g
mtot
mtot
⎝ mtot
⎠ 5-12 Chapter 5 (b) Assume the acceleration has its maximum value: a y = 0.832 m/s 2 , y − y0 = 125 m and v0 y = 0 . y − y0 = v0 yt + 1 a yt 2 gives t =
2 2( y − y0 )
2(125 m)
=
= 17.3 s
ay
0.832 m/s 2 EVALUATE: The tension in the chain is T = 1.41 × 104 N and the total weight is 1.30 × 104 N . The upward force
exceeds the downward force and the acceleration is upward. 5.24. Figure 5.23
!
!
Apply ∑ F = ma to the composite object of elevator plus student ( mtot = 850 kg ) and also to the IDENTIFY: student ( w = 550 N ). The elevator and the student have the same acceleration.
SET UP: Let + y be upward. The free-body diagrams for the composite object and for the student are given in
Figure 5.24a and b. T is the tension in the cable and n is the scale reading, the normal force the scale exerts on the
student. The mass of the student is m = w / g = 56.1 kg .
EXECUTE: ay = (a) ∑F y = ma y applied to the student gives n − mg = ma y . n − mg 450 N − 550 N
=
= −1.78 m/s 2 . The elevator has a downward acceleration of 1.78 m/s 2 .
m
56.1 kg
670 N − 550 N
= 2.14 m/s 2 .
56.1 kg (b) a y = (c) n = 0 means a y = − g . The student should worry; the elevator is in free-fall.
(d) ∑F y = ma y applied to the composite object gives T − mtot g = mtot a . T = mtot (a y + g ) . In part (a), T = (850 kg)(−1.78 m/s 2 + 9.80 m/s 2 ) = 6820 N . In part (c), a y = − g and T = 0 .
EVALUATE: In part (b), T = (850 kg)(2.14 m/s 2 + 9.80 m/s 2 ) = 10,150 N . The weight of the composite object is
8330 N. When the acceleration is upward the tension is greater than the weight and when the acceleration is
downward the tension is less than the weight. 5.25. Figure 5.24a, b
!
!
Apply ∑ F = ma to the puck. Use the information about the motion to calculate the acceleration. The IDENTIFY: table must slope downward to the right.
SET UP: Let α be the angle between the table surface and the horizontal. Let the + x -axis be to the right and
parallel to the surface of the table.
EXECUTE: ∑ Fx = max gives mg sin α = max . The time of travel for the puck is L / v0 , where L = 1.75 m and
v0 = 3.80 m/s . x − x0 = v0 xt + 1 axt 2 gives ax =
2 2
2
2 x 2 xv0
a
2 xv0
= 2 , where x = 0.0250 m . sin α = x =
.
2
2
t
L
g
gL ⎛ 2(2.50 × 10−2 m)(3.80 m / s) 2 ⎞
⎟ = 1.38° .
⎜ ( 9.80 m / s 2 ) (1.75 m) 2 ⎟
⎝
⎠ α = arcsin ⎜ Applying Newton’s Laws 5.26. 5-13 EVALUATE: The table is level in the direction along its length, since the velocity in that direction is constant. The
angle of slope to the right is small, so the acceleration and deflection in that direction are small.
!
dv
!
IDENTIFY: Acceleration and velocity are related by a y = y . Apply ∑ F = ma to the rocket.
dt
!
SET UP: Let + y be upward. The free-body diagram for the rocket is sketched in Figure 5.26. F is the thrust
force.
EXECUTE: (a) v y = At + Bt 2 . a y = A + 2 Bt . At t = 0 , a y = 1.50 m/s 2 so A = 1.50 m/s 2 . Then v y = 2.00 m/s at t = 1.00 s gives 2.00 m/s = (1.50 m/s 2 )(1.00 s) + B (1.00 s) 2 and B = 0.50 m/s3 .
(b) At t = 4.00 s , a y = 1.50 m/s 2 + 2(0.50 m/s3 )(4.00 s) = 5.50 m/s 2 .
(c) ∑F y = ma y applied to the rocket gives T − mg = ma and T = m(a + g ) = (2540 kg)(9.80 m/s 2 + 5.50 m/s 2 ) = 3.89 × 104 N . T = 1.56 w .
(d) When a = 1.50 m/s 2 , T = (2540 kg)(9.80 m/s 2 + 1.50 m/s 2 ) = 2.87 × 104 N
EVALUATE: During the time interval when v(t ) = At + Bt 2 applies the magnitude of the acceleration is increasing,
and the thrust is increasing. Figure 5.26
5.27. IDENTIFY: Consider the forces in each case. There is the force of gravity and the forces from objects that touch
the object in question.
SET UP: A surface exerts a normal force perpendicular to the surface, and a friction force, parallel to the surface.
EXECUTE: The free-body diagrams are sketched in Figure 5.27a-c.
EVALUATE: Friction opposes relative motion between the two surfaces. When one surface is stationary the
friction force on the other surface is directed opposite to its motion. Figure 5.27a–c
5.28. IDENTIFY: f s ≤ μs n and f k = μk n . The normal force n is determined by applying ! ! ∑ F = ma to the block. Normally, μ k ≤ μs . f s is only as large as it needs to be to prevent relative motion between the two surfaces.
SET UP: Since the table is horizontal, with only the block present n = 135 N . With the brick on the
block, n = 270 N .
EXECUTE: (a) The friction is static for P = 0 to P = 75.0 N . The friction is kinetic for P > 75.0 N .
(b) The maximum value of f s is μs n . From the graph the maximum f s is f s = 75.0 N , so
max fs 75.0 N
f
50.0 N
=
= 0.556 . f k = μk n . From the graph, f k = 50.0 N and μ k = k =
= 0.370 .
n
135 N
n 135 N
(c) When the block is moving the friction is kinetic and has the constant value f k = μk n , independent of P. This is μs = why the graph is horizontal for P > 75.0 N . When the block is at rest, f s = P since this prevents relative motion.
This is why the graph for P < 75.0 N has slope +1.
(d) max fs and f k would double. The values of f on the vertical axis would double but the shape of the graph
would be unchanged.
EVALUATE: The coefficients of friction are independent of the normal force. 5-14 5.29. Chapter 5 (a) IDENTIFY: Constant speed implies a = 0. Apply Newton’s 1st law to the box. The friction force is directed
opposite to the motion of the box.
!
SET UP: Consider the free-body diagram for the box, given in Figure 5.29a. Let F be the horizontal force
applied by the worker. The friction is kinetic friction since the box is sliding along the surface.
EXECUTE:
∑ Fy = ma y n − mg = 0
n = mg
So f k = μk n = μ k mg
Figure 5.29a ∑F x = max F − fk = 0
F = f k = μk mg = (0.20)(11.2 kg)(9.80 m/s 2 ) = 22 N
(b) IDENTIFY: Now the only horizontal force on the box is the kinetic friction force. Apply Newton’s 2nd law to
the box to calculate its acceleration. Once we have the acceleration, we can find the distance using a constant
acceleration equation. The friction force is f k = μ k mg , just as in part (a).
SET UP: The free-body diagram is sketched in Figure 5.29b.
EXECUTE:
∑ Fx = max − f k = max
− μk mg = max
ax = − μ k g = −(0.20)(9.80 m/s 2 ) = −1.96 m/s 2
Figure 5.29b Use the constant acceleration equations to find the distance the box travels:
vx = 0, v0 x = 3.50 m/s, ax = −1.96 m/s 2 , x − x0 = ?
2
2
vx = v0 x + 2ax ( x − x0 )
2
2
vx − v0 x 0 − (3.50 m/s) 2
=
= 3.1 m
2ax
2( −1.96 m/s 2 )
EVALUATE: The normal force is the component of force exerted by a surface perpendicular to the surface. Its
!
!
magnitude is determined by ∑ F = ma . In this case n and mg are the only vertical forces and a y = 0, so n = mg . x − x0 = 5.30. Also note that f k and n are proportional in magnitude but perpendicular in direction.
!
!
IDENTIFY: Apply ∑ F = ma to the box.
SET UP: Since the only vertical forces are n and w, the normal force on the box equals its weight. Static friction
is as large as it needs to be to prevent relative motion between the box and the surface, up to its maximum possible
value of f smax = μs n . If the box is sliding then the friction force is f k = μk n .
EXECUTE: (a) If there is no applied force, no friction force is needed to keep the box at rest.
(b) f smax = μs n = (0.40)(40.0 N) = 16.0 N . If a horizontal force of 6.0 N is applied to the box, then f s = 6.0 N in
the opposite direction.
(c) The monkey must apply a force equal to f smax , 16.0 N.
(d) Once the box has started moving, a force equal to f k = μk n = 8.0 N is required to keep it moving at constant
velocity.
EVALUATE: μ k < μs and less force must be applied to the box to maintain its motion than to start it moving. Applying Newton’s Laws 5.31. 5-15 !
!
Apply ∑ F = ma to the crate. f s ≤ μs n and f k = μk n . IDENTIFY: SET UP: Let + y be upward and let + x be in the direction of the push. Since the floor is horizontal and the push
is horizontal, the normal force equals the weight of the crate: n = mg = 441 N . The force it takes to start the crate moving equals max fs and the force required to keep it moving equals f k
313 N
208 N
= 0.710 . f k = 208 N , so μ k =
= 0.472 .
441 N
441 N
(b) The friction is kinetic. ∑ Fx = max gives F − f k = ma and F = f k + ma = 208 + (45.0 kg)(1.10 m/s 2 ) = 258 N .
max f s = 313 N , so μs = EXECUTE: (c) (i) The normal force now is mg = 72.9 N . To cause it to move, F = max fs = μs n = (0.710)(72.9 N) = 51.8 N . F − f k 258 N − (0.472)(72.9 N)
=
= 4.97 m/s 2
m
45.0 kg
EVALUATE: The kinetic friction force is independent of the speed of the object. On the moon, the mass of the
crate is the same as on earth, but the weight and normal force are less.
!
!
IDENTIFY: Apply ∑ F = ma to the box and calculate the normal and friction forces. The coefficient of kinetic
(ii) F = f k + ma and a = 5.32. fk
.
n
Let + x be in the direction of motion. ax = −0.90 m/s 2 . The box has mass 8.67 kg. friction is the ratio
SET UP:
EXECUTE: The normal force has magnitude 85 N + 25 N = 110 N. The friction force, from FH − f k = ma is
28 N
= 0.25.
110 N
The normal force is greater than the weight of the box, because of the downward component of the f k = FH − ma = 20 N − (8.67 kg)(−0.90 m/s 2 ) = 28 N . μ k = 5.33. EVALUATE:
push force.
IDENTIFY: Apply ! ! ∑ F = ma to the composite object consisting of the two boxes and to the top box. The friction the ramp exerts on the lower box is kinetic friction. The upper box doesn’t slip relative to the lower box, so the
friction between the two boxes is static. Since the speed is constant the acceleration is zero.
SET UP: Let + x be up the incline. The free-body diagrams for the composite object and for the upper box are
2.50 m
given in Figures 5.33a and b. The slope angle φ of the ramp is given by tan φ =
, so φ = 27.76° . Since the
4.75 m
boxes move down the ramp, the kinetic friction force exerted on the lower box by the ramp is directed up the
incline. To prevent slipping relative to the lower box the static friction force on the upper box is directed up the
incline. mtot = 32.0 kg + 48.0 kg = 80.0 kg .
EXECUTE: ∑F x (a) ∑F y = ma y applied to the composite object gives ntot = mtot g cos φ and f k = μk mtot g cos φ . = max gives f k + T − mtot g sin φ = 0 and T = (sin φ − μk cosφ )mtot g = (sin 27.76° − [0.444]cos 27.76°)(80.0 kg)(9.80 m/s 2 ) = 57.1 N .
The person must apply a force of 57.1 N, directed up the ramp.
(b) ∑ Fx = max applied to the upper box gives f s = mg sin φ = (32.0 kg)(9.80 m/s 2 )sin 27.76° = 146 N , directed up
the ramp.
EVALUATE: For each object the net force is zero. Figure 5.33a, b 5-16 5.34. Chapter 5 IDENTIFY: Use ! ! ∑ F = ma to find the acceleration that can be given to the car by the kinetic friction force. Then use a constant acceleration equation.
SET UP: Take + x in the direction the car is moving.
EXECUTE: (a) The free-body diagram for the car is shown in Figure 5.34. ∑F x ∑F y = ma y gives n = mg . 2
2
= max gives − μk n = max . − μk mg = max and ax = − μ k g . Then vx = 0 and vx = v0 x + 2ax ( x − x0 ) gives ( x − x0 ) = − 2
v0 x
v2
(29.1 m/s) 2
= + 0x =
= 54.0 m .
2a x
2 μ k g 2(0.80)(9.80 m/s 2 ) (b) v0 x = 2 μ k g ( x − x0 ) = 2(0.25)(9.80 m/s 2 )(54.0 m) = 16.3 m/s
EVALUATE: For constant stopping distance 2
v0 x μk is constant and v0 x is proportional to μk . The answer to part (b) can be calculated as (29.1 m/s) 0.25/ 0.80 = 16.3 m/s . Figure 5.34
5.35. 5.36. IDENTIFY: For a given initial speed, the distance traveled is inversely proportional to the coefficient of kinetic
friction.
SET UP: From Table 5.1 the coefficient of kinetic friction is 0.04 for Teflon on steel and 0.44 for brass on steel.
0.44
EXECUTE: The ratio of the distances is
= 11 .
0.04
EVALUATE: The smaller the coefficient of kinetic friction the smaller the retarding force of friction, and the
greater the stopping distance.
IDENTIFY: Constant speed means zero acceleration for each block. If the block is moving the friction force the
!
!
tabletop exerts on it is kinetic friction. Apply ∑ F = ma to each block.
SET UP: The free-body diagrams and choice of coordinates for each block are given by Figure 5.36.
m A = 4.59 kg and mB = 2.55 kg .
EXECUTE: (a) ∑F y = ma y with a y = 0 applied to block B gives mB g − T = 0 and T = 25.0 N . ax = 0 applied to block A gives T − f k = 0 and f k = 25.0 N . n A = mA g = 45.0 N and μ k = ∑F x = max with f k 25.0 N
=
= 0.556 .
nA 45.0 N (b) Now let A be block A plus the cat, so mA = 9.18 kg . n A = 90.0 N and f k = μk n = (0.556)(90.0 N) = 50.0 N . ∑F x = max for A gives T − f k = m Aax . ∑F y = ma y for block B gives mB g − T = mB a y . ax for A equals a y for B, so adding the two equations gives mB g − f k = ( mA + mB )a y and a y =
The acceleration is upward and block B slows down. mB g − f k
25.0 N − 50.0 N
=
= −2.13 m/s 2 .
mA + mB 9.18 kg + 2.55 kg Applying Newton’s Laws EVALUATE: 5-17 The equation mB g − f k = ( mA + mB )a y has a simple interpretation. If both blocks are considered together then there are two external forces: mB g that acts to move the system one way and f k that acts oppositely.
The net force of mB g − f k must accelerate a total mass of mA + mB . 5.37. IDENTIFY: Figure 5.36
!
!
Apply ∑ F = ma to each crate. The rope exerts force T to the right on crate A and force T to the left on crate B. The target variables are the forces T and F. Constant v implies a = 0.
SET UP: The free-body diagram for A is sketched in Figure 5.37a
EXECUTE:
∑ Fy = ma y nA − mA g = 0
nA = mA g
f kA = μ k n A = μ k m A g
Figure 5.37a ∑F x = max T − f kA = 0
T = μk mA g
SET UP: The free-body diagram for B is sketched in Figure 5.37b.
EXECUTE:
∑ Fy = ma y nB − mB g = 0
nB = mB g
f kB = μk nB = μ k mB g
Figure 5.37b ∑F x = max F − T − f kB = 0
F = T + μk mB g
Use the first equation to replace T in the second:
F = μ k mA g + μ k mB g .
(a) F = μ k (mA + mB ) g
(b) T = μk mA g
EVALUATE: We can also consider both crates together as a single object of mass (mA + mB ). this combined object gives F = f k = μk ( mA + mB ) g , in agreement with our answer in part (a). ∑F x = max for 5-18 5.38. Chapter 5 f = μr n . Apply IDENTIFY:
SET UP: ! ! ∑ F = ma to the tire. n = mg and f = ma . 2
v 2 − v0
, where L is the distance covered before the wheel’s speed is reduced to half its original
L
2
2
2
a v 2 − v 2 v0 − 1 v0 3 v0
4
=
=
speed and v = v0 / 2 . μ r = = 0
.
2 Lg
2 Lg
8 Lg
g ax = EXECUTE: 3
(3.50 m / s) 2
= 0.0259 .
8 (18.1 m)(9.80 m / s 2 ) Low pressure, L = 18.1 m and
High pressure, L = 92.9 m and
EVALUATE:
5.39. SET UP: μ r is inversely proportional to the distance L, so
Apply IDENTIFY: 3 (3.50 m / s) 2
= 0.00505 .
8 (3.50 m / s) 2 ! ! μr1 L2
=
.
μr2 L1 ∑ F = ma to the box. Use the information about sliding to calculate the mass of the box. f k = μk n , f r = μr n and n = mg .
Without the dolly: n = mg and F − μk n = 0 ( ax = 0 since speed is constant). EXECUTE: m= F μk g = 160 N
= 34.74 kg
(0.47) (9.80 m s 2 ) With the dolly: the total mass is 34.7 kg + 5.3 kg = 40.04 kg and friction now is rolling friction, f r = μr mg .
F − μr mg = ma . a =
EVALUATE:
5.40. F − μr mg
= 3.82 m s 2 .
m f k = μk mg = 160 N and f r = μr mg = 4.36 N , or, f r μr
=
. The rolling friction force is much less
fk μk than the kinetic friction force.
!
!
IDENTIFY: Apply ∑ F = ma to the truck. For constant speed, a = 0 and Fhoriz = f r .
SET UP:
EXECUTE: f r = μr n = μ r mg . Let m2 = 1.42m1 and μ r2 = 0.81μr1 .
Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and Fnet in the horizontal direction must be zero. Therefore f r = μr n = Fhoriz = 200 N before the weight and pressure changes
are made. After the changes, (0.81μr ) (1.42n) = Fhoriz , because the speed is still constant and Fnet = 0 . We can
(0.81μr )(1.42n)
F
= horiz and (0.81) (1.42) (200 N) = Fhoriz = 230 N .
μr n
200 N
EVALUATE: The increase in weight increases the normal force and hence the friction force, whereas the decrease
in μ r reduces it. The percentage increase in the weight is larger, so the net effect is an increase in the friction force.
!
!
IDENTIFY: Apply ∑ F = ma to each block. The target variables are the tension T in the cord and the simply divide the two equations: 5.41. acceleration a of the blocks. Then a can be used in a constant acceleration equation to find the speed of each block.
The magnitude of the acceleration is the same for both blocks.
SET UP: The system is sketched in Figure 5.41a. For each block take a positive
coordinate direction to be the
direction of the block’s acceleration. Figure 5.41a Applying Newton’s Laws 5-19 block on the table: The free-body is sketched in Figure 5.41b.
EXECUTE:
∑ Fy = ma y
n − mA g = 0
n = mA g
f k = μk n = μ k mA g
Figure 5.41b ∑F x = max T − f k = mAa
T − μk mA g = mAa
SET UP: hanging block: The free-body is sketched in Figure 5.41c.
EXECUTE:
∑ Fy = ma y mB g − T = mB a
T = mB g − mB a
Figure 5.41c
(a) Use the second equation in the first mB g − mB a − μk mA g = m Aa
(mA + mB ) a = (mB − μk mA ) g
a= ( mB − μ k mA ) g (1.30 kg − (0.45)(2.25 kg))(9.80 m/s 2 )
=
= 0.7937 m/s 2
mA + mB
2.25 kg + 1.30 kg SET UP: Now use the constant acceleration equations to find the final speed. Note that the blocks have the same
speeds. x − x0 = 0.0300 m, ax = 0.7937 m/s 2 , v0 x = 0, vx = ?
2
2
vx = v0 x + 2ax ( x − x0 ) EXECUTE: vx = 2ax ( x − x0 ) = 2(0.7937 m/s 2 )(0.0300 m) = 0.218 m/s = 21.8 cm/s. (b) T = mB g − mB a = mB ( g − a ) = 1.30 kg(9.80 m/s 2 − 0.7937 m/s 2 ) = 11.7 N Or, to check, T − μk m A g = m Aa
T = mA (a + μ k g ) = 2.25 kg(0.7937 m/s 2 + (0.45)(9.80 m/s 2 )) = 11.7 N, which checks.
EVALUATE: 5.42. The force T exerted by the cord has the same value for each block. T < mB g since the hanging block accelerates downward. Also, f k = μk mA g = 9.92 N. T > f k and the block on the table accelerates in the direction
of T.
!
!
IDENTIFY: Apply ∑ F = ma to the box. When the box is ready to slip the static friction force has its maximum
possible value, f s = μs n .
SET UP: Use coordinates parallel and perpendicular to the ramp.
EXECUTE: (a) The normal force will be wcos θ and the component of the gravitational force along the ramp
is wsin θ . The box begins to slip when w sin θ > μs w cos θ , or tan θ > μs = 0.35, so slipping occurs at
θ = arctan(0.35) = 19.3° .
(b) When moving, the friction force along the ramp is μ k w cos θ , the component of the gravitational force along
the ramp is w sin θ , so the acceleration is ( w sin θ − wμk cos θ ) m = g (sin θ − μk cos θ ) = 0.92 m s 2 .
(c) Since v0 x = 0 , 2ax = v 2 , so v = (2ax)1 2 , or v = [(2)(0.92m s 2 )(5 m)]1 2 = 3 m/s .
EVALUATE: When the box starts to move, friction changes from static to kinetic and the friction force becomes
smaller. 5-20 5.43. Chapter 5 (a) IDENTIFY: Apply ! ! ∑ F = ma to the crate. Constant v implies a = 0. Crate moving says that the friction is kinetic friction. The target variable is the magnitude of the force applied by the woman.
SET UP: The free-body diagram for the crate is sketched in Figure 5.43.
EXECUTE:
∑ Fy = ma y n − mg − F sin θ = 0
n = mg + F sin θ
f k = μk n = μ k mg + μ k F sin θ Figure 5.43 ∑F x = max F cosθ − f k = 0
F cosθ − μk mg − μk F sin θ = 0
F (cosθ − μk sin θ ) = μ k mg μk mg
cosθ − μk sin θ
(b) IDENTIFY and SET UP:
F= “start the crate moving” means the same force diagram as in part (a), except that
μs mg
μ k is replaced by μs . Thus F =
.
cosθ − μs sin θ cosθ
1
F → ∞ if cosθ − μs sin θ = 0. This gives μs =
=
.
sin θ tan θ
!
EVALUATE: F has a downward component so n > mg . If θ = 0 (woman pushes horizontally), n = mg and
EXECUTE: F = f k = μk mg .
5.44. IDENTIFY: Apply ! ! ∑ F = ma to the box. SET UP: Let + y be upward and + x be horizontal, in the direction of the acceleration. Constant speed means a = 0 .
EXECUTE: (a) There is no net force in the vertical direction, so n + F sin θ − w = 0, or
n = w − F sin θ = mg − F sin θ. The friction force is f k = μk n = μ k (mg − F sin θ ). The net horizontal force is F cos θ − f k = F cos θ − μk ( mg − F sin θ ) , and so at constant speed,
F=
(b) Using the given values, F =
EVALUATE:
5.45. IDENTIFY: μk mg
cos θ + μk sin θ (0.35)(90 kg)(9.80m s 2 )
= 290 N .
(cos 25° + (0.35)sin 25°) If θ = 0° , F = μ k mg .
!
!
Apply ∑ F = ma to each block. SET UP: For block B use coordinates parallel and perpendicular to the incline. Since they are connected by ropes,
blocks A and B also move with constant speed.
EXECUTE: (a) The free-body diagrams are sketched in Figure 5.45.
(b) The blocks move with constant speed, so there is no net force on block A; the tension in the rope connecting A
and B must be equal to the frictional force on block A, μ k = (0.35) (25.0 N) = 9 N.
(c) The weight of block C will be the tension in the rope connecting B and C; this is found by considering the
forces on block B. The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the
component of the weight along the ramp, the friction on block B and the tension in the second rope. Thus, the
weight of block C is
wC = 9 N + wB (sin 36.9° + μk cos36.9°) = 9 N + (25.0 N)(sin 36.9° + (0.35)cos 36.9°) = 31.0 N The intermediate calculation of the first tension may be avoided to obtain the answer in terms of the common
weight w of blocks A and B, wC = w( μk + (sin θ + μ k cos θ )), giving the same result.
(d) Applying Newton’s Second Law to the remaining masses (B and C) gives:
a = g ( wC − μ k wB cos θ − wB sin θ ) ( wB + wC ) = 1.54m s 2 . Applying Newton’s Laws 5-21 EVALUATE: Before the rope between A and B is cut the net external force on the system is zero. When the rope is
cut the friction force on A is removed from the system and there is a net force on the system of blocks B and C. Figure 5.45
5.46. IDENTIFY and SET UP: The derivative of v y gives a y as a function of time, and the integral of v y gives y as a function of time.
EXECUTE: Differentiating Eq. (5.10) with respect to time gives the acceleration
⎛k⎞
a = vt ⎜ ⎟ e − ( k m )t = ge −( k m )t , where Eq. (5.9), vt = mg k , has been used. Integrating Eq. (5.10) with respect to time
⎝m⎠
with y0 = 0 gives
t
⎡ ⎛m⎞
⎤
⎛m⎞
⎡m
⎤
y = ∫ vt [1 − e− ( k m ) t ] dt = vt ⎢t + ⎜ ⎟ e − ( k m )t ⎥ − vt ⎜ ⎟ = vt ⎢t − (1 − e− ( k m ) t ) ⎥ .
0
⎝k⎠
⎣k
⎦
⎣ ⎝k⎠
⎦ EVALUATE:
5.47. We can verify that dy / dt = v y . IDENTIFY and SET UP:
EXECUTE:
(b) vt = Apply Eq.(5.13). (a) Solving for D in terms of vt , D = mg (80 kg) (9.80 m s 2 )
=
= 0.44 kg m.
vt2
(42 m s) 2 mg
(45 kg)(9.80 m s 2 )
=
= 42 m s.
D
(0.25 kg m) EVALUATE: vt is less for the daughter since her mass is less.
!
!
Apply ∑ F = ma to the ball. At the terminal speed, f = mg . 5.48. IDENTIFY: 5.49. SET UP: The fluid resistance is directed opposite to the velocity of the object. At half the terminal speed, the
magnitude of the frictional force is one-fourth the weight.
EXECUTE: (a) If the ball is moving up, the frictional force is down, so the magnitude of the net force is (5/4)w
and the acceleration is (5/4)g, down.
(b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration
is (3/4)g, down.
EVALUATE: The frictional force is less than mg in each case and in each case the net force is downward and the
acceleration is downward.
!
!
IDENTIFY: Apply ∑ F = ma to one of the masses. The mass moves in a circular path, so has acceleration v2
, directed toward the center of the path.
R
SET UP: In each case, R = 0.200 m . In part (a), let + x be toward the center of the circle, so ax = arad . In part (b)
arad = let + y be toward the center of the circle, so a y = arad . + y is downward when the mass is at the top of the circle
!
and + y is upward when the mass is at the bottom of the circle. Since arad has its greatest possible value, F is in
!
the direction of arad at both positions.
EXECUTE: (a) ∑F x = max gives F = marad = m v2
. F = 75.0 N and v =
R FR
(75.0 N)(0.200 m)
=
= 3.61 m/s .
m
1.15 kg (b) The free-body diagrams for a mass at the top of the path and at the bottom of the path are given in figure 5.49.
At the top, ∑ Fy = ma y gives F = marad − mg and at the bottom it gives F = mg + marad . For a given rotation rate and hence value of arad , the value of F required is larger at the bottom of the path.
(c) F = mg + marad so v2 F
= − g and
Rm ⎛ 75.0 N
⎞
⎛F
⎞
v = R ⎜ − g ⎟ = (0.200 m) ⎜
− 9.80 m/s 2 ⎟ = 3.33 m/s
⎝m
⎠
⎝ 1.15 kg
⎠ 5-22 Chapter 5 !
EVALUATE: The maximum speed is less for the vertical circle. At the bottom of the vertical path F and the
weight are in opposite directions so F must exceed marad by an amount equal to mg. At the top of the vertical path
F and mg are in the same direction and together provide the required net force, so F must be larger at the bottom. Figure 5.49
5.50. IDENTIFY: Since the car travels in an arc of a circle, it has acceleration arad = v 2 / R , directed toward the center of
the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the
minimum coefficient of friction that is required, set the static friction force equal to its maximum value, f s = μs n .
Friction is static friction because the car is not sliding in the radial direction.
SET UP: The free-body diagram for the car is given in Figure 5.50. The diagram assumes the center of the curve
is to the left of the car.
v2
v2
EXECUTE: (a) ∑ Fy = ma y gives n = mg . ∑ Fx = max gives μs n = m . μs mg = m
and
R
R
v2
(25.0 m/s) 2
μs =
=
= 0.290
gR (9.80 m/s 2 )(220 m)
(b) v2 μs = Rg = constant , so v12 μs1 = 2
v2 μs2 . v2 = v1 μs2
μ /3
= (25.0 m/s) s1 = 14.4 m/s .
μs1
μs1 EVALUATE: A smaller coefficient of friction means a smaller maximum friction force, a smaller possible
acceleration and therefore a smaller speed. Figure 5.50
5.51. We can use the analysis done in Example 5.23. As in that example, we assume friction is negligible.
v2
SET UP: From Example 5.23, the banking angle β is given by tan β =
. Also, n = mg / cos β .
gR
65.0 mi/h = 29.1 m/s .
IDENTIFY: (29.1 m/s) 2
and β = 21.0° . The expression for tan β does not involve the mass
(9.80 m/s 2 )(225 m)
of the vehicle, so the truck and car should travel at the same speed.
(1125 kg)(9.80 m/s 2 )
= 1.18 × 104 N and ntruck = 2ncar = 2.36 × 104 N , since mtruck = 2mcar .
(b) For the car, ncar =
cos21.0°
EVALUATE: The vertical component of the normal force must equal the weight of the vehicle, so the normal
force is proportional to m.
IDENTIFY: The acceleration of the person is arad = v 2 / R , directed horizontally to the left in the figure in the
!
!
2π R
. Apply ∑ F = ma to the person.
problem. The time for one revolution is the period T =
v
EXECUTE: 5.52. (a) tan β = Applying Newton’s Laws 5-23 The person moves in a circle of radius R = 3.00 m + (5.00 m)sin 30.0° = 5.50 m . The free-body diagram
!
is given in Figure 5.52. F is the force applied to the seat by the rod.
v2
mg
EXECUTE: (a) ∑ Fy = ma y gives F cos30.0° = mg and F =
. ∑ Fx = max gives F sin 30.0° = m .
cos30.0°
R
SET UP: Combining these two equations gives v = Rg tan θ = (5.50 m)(9.80 m/s 2 ) tan 30.0° = 5.58 m/s . Then the period
2π R 2π (5.50 m)
=
= 6.19 s .
v
5.58 m/s
(b) The net force is proportional to m so in is T = ! ! ∑ F = ma the mass divides out and the angle for a given rate of rotation is independent of the mass of the passengers.
EVALUATE: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force
required for circular motion is produced by a component of the force exerted on the seat by the rod. 5.53. IDENTIFY: Figure 5.52
!
!
Apply ∑ F = ma to the composite object of the person plus seat. This object moves in a horizontal circle and has acceleration arad , directed toward the center of the circle.
SET UP: The free-body diagram for the composite object is given in Figure 5.53. Let + x be to the right, in the
!
direction of arad . Let + y be upward. The radius of the circular path is R = 7.50 m . The total mass is
(255 N + 825 N) /(9.80 m/s 2 ) = 110.2 kg . Since the rotation rate is 32.0 rev/min = 0.5333 rev/s , the period T is
1
= 1.875 s .
0.5333 rev/s
EXECUTE: ∑F x ∑F y = ma y gives TA cos 40.0° − mg = 0 and TA = mg
255 N + 825 N
=
= 1410 N .
cos 40.0°
cos 40.0° = max gives TA sin 40.0° + TB = marad and 4π 2 R
4π 2 (7.50 m)
− TA sin 40.0° = (110.2 kg)
− (1410 N)sin 40.0° = 8370 N .
2
T
(1.875 s) 2
The tension in the horizontal cable is 8370 N and the tension in the other cable is 1410 N.
EVALUATE: The weight of the composite object is 1080 N. The tension in cable A is larger than this since its
vertical component must equal the weight. marad = 9280 N . The tension in cable B is less than this because part of
the required inward force comes from a component of the tension in cable A.
TB = m Figure 5.53 5-24 5.54. Chapter 5 IDENTIFY: Apply ! ! ∑ F = ma to the button. The button moves in a circle, so it has acceleration a rad . The situation is equivalent to that of Example 5.22.
2
v2
2π R
EXECUTE: (a) μs =
. Expressing v in terms of the period T, v =
so μs = 4π2 R . A platform speed of
Tg
T
Rg
SET UP: 4π 2 (0.150 m)
= 0.269.
(1.50 s) 2 (9.80 m s 2 )
(b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period
(longer periods mean slower speeds, so the button may be moved further out) and so is inversely proportional to
2
⎛ 40.0 ⎞
the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m) ⎜
⎟ = 0.067 m .
⎝ 60.0 ⎠ 40.0 rev/min corresponds to a period of 1.50 s, so μs = 4π 2 R
. The maximum radial acceleration that friction can give is μs mg . At the faster rotation
T2
rate T is smaller so R must be smaller to keep arad the same. EVALUATE: 5.55. arad = 4π 2 R
.
T2
SET UP: R = 800 m . 1/ T is the number of revolutions per second.
EXECUTE: (a) Setting arad = g and solving for the period T gives
IDENTIFY: The acceleration due to circular motion is arad = T = 2π R
400 m
= 2π
= 40.1 s,
g
9.80 m s 2 so the number of revolutions per minute is (60 s min) (40.1 s) = 1.5 rev min .
(b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square
root of the ratio of the accelerations, T ′ = (1.5 rev min) × 3.70 9.8 = 0.92 rev min.
EVALUATE: In part (a) the tangential speed of a point at the rim is given by arad = v2
, so
R v = Rarad = Rg = 62.6 m/s ; the space station is rotating rapidly.
5.56. 2π R
. The apparent weight of a person is the normal force exerted on him by the seat he is sitting
v
on. His acceleration is arad = v 2 / R , directed toward the center of the circle.
SET UP: The period is T = 60.0 s. The passenger has mass m = w / g = 90.0 kg .
IDENTIFY: T= v 2 (5.24 m/s) 2
2π R 2π (50.0 m)
= 0.549 m/s 2 .
=
= 5.24 m/s . Note that arad = =
R
50.0 m
T
60.0 s
(b) The free-body diagram for the person at the top of his path is given in Figure 5.56a. The acceleration is
downward, so take + y downward. ∑ Fy = ma y gives mg − n = marad .
EXECUTE: (a) v = n = m( g − arad ) = (90.0 kg)(9.80 m/s 2 − 0.549 m/s 2 ) = 833 N .
The free-body diagram for the person at the bottom of his path is given in Figure 5.56b. The acceleration is
upward, so take + y upward. ∑ Fy = ma y gives n − mg = marad and n = m( g + arad ) = 931 N .
(c) Apparent weight = 0 means n = 0 and mg = marad . g = v2
and v = gR = 22.1 m/s . The time for one
R 2π R 2π (50.0 m)
=
= 14.2 s . Note that arad = g .
v
22.1 m/s
(d) n = m( g + arad ) = 2mg = 2(882 N) = 1760 N , twice his true weight.
revolution would be T = Applying Newton’s Laws 5-25 EVALUATE: At the top of his path his apparent weight is less than his true weight and at the bottom of his path
his apparent weight is greater than his true weight. 5.57. Figure 5.56a, b
!
!
IDENTIFY: Apply ∑ F = ma to the motion of the pilot. The pilot moves in a vertical circle. The apparent weight
!
is the normal force exerted on him. At each point arad is directed toward the center of the circular path.
(a) SET UP: “the pilot feels weightless” means that the vertical normal force n exerted on the pilot by the chair on
which the pilot sits is zero. The force diagram for the pilot at the top of the path is given in Figure 5.57a.
EXECUTE:
∑ Fy = ma y mg = marad
g= v2
R Figure 5.57a Thus v = gR = (9.80 m/s 2 )(150 m) = 38.34 m/s
⎛ 1 km ⎞⎛ 3600 s ⎞
v = (38.34 m/s) ⎜ 3 ⎟⎜
⎟ = 138 km/h
⎝ 10 m ⎠⎝ 1 h ⎠
(b) SET UP: The force diagram for the pilot at the bottom of the path is given in Figure 5.57b. Note that the
vertical normal force exerted on the pilot by the chair on which the pilot sits is now upward.
EXECUTE:
∑ Fy = ma y v2
R
v2
n = mg + m
R
This normal force is the pilot’s apparent weight.
n − mg = m Figure 5.57b w = 700 N, so m = w
= 71.43 kg
g 3
⎛ 1 h ⎞ ⎛ 10 m ⎞
v = (280 km/h) ⎜
⎟ = 77.78 m/s
⎟⎜
⎝ 3600 s ⎠ ⎝ 1 km ⎠ (77.78 m/s) 2
= 3580 N.
150 m
EVALUATE: In part (b), n > mg since the acceleration is upward. The pilot feels he is much heavier than when at Thus n = 700 N + 71.43 kg rest. The speed is not constant, but it is still true that arad = v 2 / R at each point of the motion.
5.58. !
IDENTIFY: arad = v 2 / R , directed toward the center of the circular path. At the bottom of the dive, arad is upward.
The apparent weight of the pilot is the normal force exerted on her by the seat on which she is sitting.
SET UP: The free-body diagram for the pilot is given in Figure 5.58. 5-26 Chapter 5 EXECUTE:
(b) ∑F y (a) arad = v2
v2
(95.0 m/s) 2
=
= 230 m .
gives R =
arad 4.00(9.80 m/s 2 )
R = ma y gives n − mg = marad . n = m( g + arad ) = m( g + 4.00 g ) = 5.00mg = (5.00)(50.0 kg)(9.80 m/s 2 ) = 2450 N
EVALUATE: Her apparent weight is five times her true weight, the force of gravity the earth exerts on her. 5.59. IDENTIFY: Figure 5.58
!
!
Apply ∑ F = ma to the water. The water moves in a vertical circle. The target variable is the speed v; we will calculate arad and then get v from arad = v 2 / R
SET UP: Consider the free-body diagram for the water when the pail is at the top of its circular path, as shown in
Figures 5.59a and b.
The radial acceleration is in toward the center
of the circle so at this point is downward.
n is the downward normal force exerted on
the water by the bottom of the pail.
Figure 5.59a
EXECUTE:
∑ Fy = ma y n + mg = m v2
R Figure 5.59b 5.60. At the minimum speed the water is just ready to lose contact with the bottom of the pail, so at this speed, n → 0.
(Note that the force n cannot be upward.)
v2
With n → 0 the equation becomes mg = m . v = gR = (9.80 m/s 2 )(0.600 m) = 2.42 m/s.
R
EVALUATE: At the minimum speed arad = g . If v is less than this minimum speed, gravity pulls the water (and
bucket) out of the circular path.
IDENTIFY: The ball has acceleration arad = v 2 / R , directed toward the center of the circular path. When the ball is
at the bottom of the swing, its acceleration is upward.
SET UP: Take + y upward, in the direction of the acceleration. The bowling ball has mass m = w / g = 7.27 kg .
v 2 (4.20 m/s) 2
=
= 4.64 m/s , upward.
R
3.80 m
(b) The free-body diagram is given in Figure 5.60. ∑ Fy = ma y gives T − mg = marad .
EXECUTE: (a) arad = T = m( g + arad ) = (7.27 kg)(9.80 m/s 2 + 4.64 m/s 2 ) = 105 N Applying Newton’s Laws EVALUATE: 5-27 The acceleration is upward, so the net force is upward and the tension is greater than the weight. Figure 5.60
5.61. IDENTIFY: !
!
Apply ∑ F = ma to the knot. SET UP: a = 0 . Use coordinates with axes that are horizontal and vertical.
EXECUTE: (a) The free-body diagram for the knot is sketched in Figure 5.61.
T1 is more vertical so supports more of the weight and is larger. You can also see this from ∑ Fx = max : T2 cos 40° − T1 cos60° = 0 . T2 cos 40° − T1 cos60° = 0 .
(b) T1 is larger so set T1 = 5000 N. Then T2 = T1 1.532 = 3263.5 N . ∑ Fy = ma y gives T1 sin 60° + T2 sin 40° = w and w = 6400 N .
EVALUATE: The sum of the vertical components of the two tensions equals the weight of the suspended object.
The sum of the tensions is greater than the weight. 5.62. IDENTIFY: Figure 5.61
!
!
Apply ∑ F = ma to each object . Constant speed means a = 0 . SET UP: The free-body diagrams are sketched in Figure 5.62. T1 is the tension in the lower chain, T2 is the
tension in the upper chain and T = F is the tension in the rope.
EXECUTE: The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have
no net force on it, so twice the tension in the rope must be equal to w and the tension in the rope, which equals F, is
w 2 . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force
w on the upper pulley, and the tension in the upper chain is also w.
EVALUATE: The pulley combination allows the worker to lift a weight w by applying a force of only w / 2 . Figure 5.62
5.63. IDENTIFY: !
!
Apply ∑ F = ma to the rope. SET UP: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the
force due to the tension in the rope are an action-reaction pair.
EXECUTE: (a) The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts
an upward vertical force of w 2 on the rope. Therefore, the downward force that the rope exerts at each end is Tend sin θ = w 2 , so Tend = w (2sin θ ) = Mg (2sin θ ). 5-28 Chapter 5 (b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that
each hook exerts, which is the same as the horizontal component of the force due to the tension at the end;
Tend cos θ = Tmiddle , so Tmiddle = Mg cos θ (2sin θ ) = Mg (2tan θ ). 5.64. (c) Mathematically speaking, θ ≠ 0 because this would cause a division by zero in the equation for Tend or Tmiddle .
Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight.
EVALUATE: The tension in the rope is not the same at all points along the rope.
!
!
!
!
IDENTIFY: Apply ∑ F = ma to the combined rope plus block to find a. Then apply ∑ F = ma to a section of the rope of length x. First note the limiting values of the tension. The system is sketched in Figure 5.64a. At the top of the rope T = F
At the bottom of the rope T = M ( g + a ) Figure 5.64a
SET UP: Consider the rope and block as one combined object, in order to calculate the acceleration: The freebody diagram is sketched in Figure 5.64b.
EXECUTE:
∑ Fy = ma y F − ( M + m) g = ( M + m) a
a= F
−g
M +m Figure 5.64b
SET UP: Now consider the forces on a section of the rope that extends a distance x < L below the top. The
tension at the bottom of this section is T ( x ) and the mass of this section is m( x / L). The free-body diagram is
sketched in Figure 5.64c.
EXECUTE:
∑ Fy = ma y F − T ( x ) − m( x / L) g = m( x / L) a
T ( x ) = F − m( x / L ) g − m ( x / L ) a
Figure 5.64c Using our expression for a and simplifying gives
⎛
⎞
mx
T ( x) = F ⎜1 −
⎟
L( M + m) ⎠
⎝ 5.65. EVALUATE: Important to check this result for the limiting cases:
x = 0 : The expression gives the correct value of T = F .
x = L : The expression gives T = F ( M /( M + m)). This should equal T = M ( g + a ), and when we use the
expression for a we see that it does.
!
!
IDENTIFY: Apply ∑ F = ma to each block.
SET UP: Constant speed means a = 0 . When the blocks are moving, the friction force is f k and when they are at rest, the friction force is f s .
EXECUTE: (a) The tension in the cord must be m2 g in order that the hanging block move at constant speed. This
tension must overcome friction and the component of the gravitational force along the incline, so
m2 g = ( m1 g sin α + μk m1 g cos α ) and m2 = m1 (sin α + μk cos α ) .
(b) In this case, the friction force acts in the same direction as the tension on the block of mass m1 , so m2 g = (m1 g sin α − μk m1 g cos α ) , or m2 = m1 (sin α − μk cos α ) . Applying Newton’s Laws 5-29 (c) Similar to the analysis of parts (a) and (b), the largest m2 could be is m1 (sin α + μs cos α ) and the smallest m2 could be is m1 (sin α − μs cos α ) .
EVALUATE: In parts (a) and (b) the friction force changes direction when the direction of the motion of
m1 changes. In part (c), for the largest m2 the static friction force on m1 is directed down the incline and for the
5.66. smallest m2 the static friction force on m1 is directed up the incline.
IDENTIFY: The system is in equilibrium. Apply Newton’s 1st law to block A, to the hanging weight and to the
knot where the cords meet. Target variables are the two forces.
(a) SET UP: The free-body diagram for the hanging block is given in Figure 5.66a.
EXECUTE:
∑ Fy = ma y T3 − w = 0
T3 = 12.0 N
Figure 5.66a
SET UP: The free-body diagram for the knot is given in Figure 5.66b.
EXECUTE:
∑ Fy = ma y T2 sin 45.0° − T3 = 0
T3
12.0 N
=
sin 45.0° sin 45.0°
T2 = 17.0 N T2 = Figure 5.66b ∑F x = max T2 cos 45.0° − T1 = 0
T1 = T2 cos 45.0° = 12.0 N
SET UP: The free-body diagram for block A is given in Figure 5.66c.
EXECUTE:
∑ Fx = max T1 − f s = 0
f s = T1 = 12.0 N
Figure 5.66c
EVALUATE: Also can apply ∑F y = ma y to this block:
n − wA = 0
n = wA = 60.0 N Then μs n = (0.25)(60.0 N) = 15.0 N; this is the maximum possible value for the static friction force. We see that
f s < μs n; for this value of w the static friction force can hold the blocks in place.
(b) SET UP: We have all the same free-body diagrams and force equations as in part (a) but now the static
friction force has its largest possible value, f s = μs n = 15.0 N. Then T1 = fs = 15.0 N.
EXECUTE: From the equations for the forces on the knot
15.0 N
T2 cos 45.0° − T1 = 0 implies T2 = T1 / cos 45.0° =
= 21.2 N
cos 45.0°
T2 sin 45.0° − T3 = 0 implies T3 = T2 sin 45.0° = (21.2 N)sin 45.0° = 15.0 N And finally T3 − w = 0 implies w = T3 = 15.0 N.
EVALUATE: Compared to part (a), the friction is larger in part (b) by a factor of (15.0/12.0) and w is larger by
this same ratio. 5-30 5.67. Chapter 5 IDENTIFY: Apply ! ! ∑ F = ma to each block. Use Newton’s 3rd law to relate forces on A and on B. SET UP: Constant speed means a = 0 .
EXECUTE: (a) Treat A and B as a single object of weight w = wA + wB = 4.80 N . The free-body diagram for this combined object is given in Figure 5.67a. ∑F y = ma y gives n = w = 4.80 N . f k = μk n = 1.44 N . ∑F x = max gives F = f k = 1.44 N
(b) The free-body force diagrams for blocks A and B are given in Figure 5.67b. n and f k are the normal and friction forces applied to block B by the tabletop and are the same as in part (a). f kB is the friction force that A
applies to B. It is to the right because the force from A opposes the motion of B. nB is the downward force that A
exerts on B. f kA is the friction force that B applies to A. It is to the left because block B wants A to move with it.
n A is the normal force that block B exerts on A. By Newton’s third law, f kB = f kA and these forces are in opposite directions. Also, n A = nB and these forces are in opposite directions. ∑F y = ma y for block A gives n A = wA = 1.20 N , so nB = 1.20 N . f kA = μ k nA = (0.300)(1.20 N) = 0.36 N , and f kB = 0.36 N. ∑F
∑F x = max for block A gives T = f kA = 0.36 N . x = max for block B gives F = f kB + f k = 0.36 N + 1.44 N = 1.80 N EVALUATE: In part (a) block A is at rest with respect to B and it has zero acceleration. There is no horizontal
force on A besides friction, and the friction force on A is zero. A larger force F is needed in part (b), because of the
friction force between the two blocks. 5.68. IDENTIFY: Figure 5.67a–c
!
!
Apply ∑ F = ma to the brush. Constant speed means a = 0. Target variables are two of the forces on the brush.
SET UP: Note that the normal force exerted by the wall is horizontal, since it is perpendicular to the wall. The
kinetic friction force exerted by the wall is parallel to the wall and opposes the motion, so it is vertically
downward. The free-body diagram is given in Figure 5.68.
EXECUTE:
∑ Fx = max n − F cos53.1° = 0
n = F cos53.1°
f k = μk n = μ k F cos53.1°
Figure 5.68 ∑F y = ma y F sin 53.1° − w − f k = 0
F sin 53.1° − w − μ k F cos53.1° = 0
F (sin 53.1° − μ k cos53.1°) = w
F= w
sin 53.1° − μ k cos53.1° Applying Newton’s Laws 120 N
= 16.9 N
sin 53.1° − μk cos53.1° sin 53.1° − (0.15)cos53.1°
(b) n = F cos53.1° = (16.9 N)cos53.1° = 10.1 N
EVALUATE: In the absence of friction w = F sin 53.1°, which agrees with our expression.
IDENTIFY: The net force at any time is Fnet = ma .
SET UP: At t = 0 , a = 62 g . The maximum acceleration is 140g at t = 1.2 ms .
(a) F = 5.69. 5-31 w = EXECUTE: (a) Fnet = ma = 62mg = 62(210 × 10−9 kg)(9.80 m/s 2 ) = 1.3 × 10−4 N . This force is 62 times the flea’s
weight.
(b) Fnet = 140mg = 2.9 × 10−4 N . 5.70. (c) Since the initial speed is zero, the maximum speed is the area under the ax -t graph. This gives 1.2 m/s.
EVALUATE: a is much larger than g and the net external force is much larger than the flea's weight.
!
!
IDENTIFY: Apply ∑ F = ma to the instrument and calculate the acceleration. Then use constant acceleration equations to describe the motion.
SET UP: The free-body diagram for the instrument is given in Figure 5.70. The instrument has mass
m = w g = 1.531 kg .
T − mg
= 13.07 m s 2 .
m
v0 y = 0, v y = 330 m s, a y = 13.07 m s 2 , t = ? Then v y = v0 y + a y t gives t = 25.3 s . Consider forces on the EXECUTE: (a) For on the instrument, ∑ Fy = ma y gives T − mg = ma and a = rocket; rocket has the same a y . Let F be the thrust of the rocket engines. F − mg = ma and
F = m( g + a ) = (25,000 kg) (9.80 m s 2 + 13.07 m s 2 ) = 5.72 × 105 N .
(b) y − y0 = v0 yt + 1 a yt 2 gives y − y0 = 4170 m.
2
EVALUATE: The rocket and instrument have the same acceleration. The tension in the wire is over twice the
weight of the instrument and the upward acceleration is greater than g. 5.71. IDENTIFY: Figure 5.70
!
!
a = dv / dt . Apply ∑ F = ma to yourself. SET UP: The reading of the scale is equal to the normal force the scale applies to you.
EXECUTE: The elevator’s acceleration is a= dv (t )
= 3.0 m s 2 + 2(0.20 m s3 )t = 3.0 m s 2 + (0.40 m s3 )t
dt At t = 4.0 s, a = 3.0 m s 2 + (0.40 m s3 )(4.0 s) = 4.6 m s 2 . From Newton’s Second Law, the net force on you is
Fnet = Fscale − w = ma and
Fscale = w + ma = (72 kg)(9.8 m s 2 ) + (72 kg)(4.6 m s 2 ) = 1040 N
5.72. EVALUATE: a increases with time, so the scale reading is increasing.
!
!
IDENTIFY: Apply ∑ F = ma to the passenger to find the maximum allowed acceleration. Then use a constant acceleration equation to find the maximum speed.
SET UP: The free-body diagram for the passenger is given in Figure 5.72.
EXECUTE: ∑ Fy = ma y gives n − mg = ma . n = 1.6mg , so a = 0.60 g = 5.88 m s 2 .
2 2
2
y − y0 = 3.0 m, a y = 5.88 m s , v0 y = 0 so v y = v0 y + 2a y ( y − y0 ) gives v y = 5.0 m s . 5-32 Chapter 5 EVALUATE: A larger final speed would require a larger value of a y , which would mean a larger normal force on the person. 5.73. IDENTIFY: Figure 5.72
!
!
Apply ∑ F = ma to the package. Calculate a and then use a constant acceleration equation to describe the motion.
SET UP: Let + x be directed up the ramp.
EXECUTE: (a) Fnet = − mg sin 37° − f k = − mg sin 37° − μ k mg cos37° = ma and
a = −(9.8 m s 2 )(0.602 + (0.30)(0.799)) = −8.25m s 2
2
2
Since we know the length of the slope, we can use vx = v0 x + 2ax ( x − x0 ) with x0 = 0 and vx = 0 at the top.
2
v0 = −2ax = 2( −8.25 m s 2 )(8.0 m) = 132 m 2 s 2 and v0 = 132 m 2 s 2 = 11.5 m s
(b) For the trip back down the slope, gravity and the friction force operate in opposite directions to each other.
Fnet = − mg sin 37° + μk mg cos37° = ma and a = g (− sin 37° + 0.30 cos37°) = (9.8 m s 2 )((−0.602) + (0.30)(0.799)) = −3.55 m s 2 .
2
Now we have v0 = 0, x0 = −8.0 m, x = 0 and v 2 = v0 + 2a ( x − x0 ) = 0 + 2(−3.55 m s 2 )( −8.0 m) = 56.8 m 2 s 2 , so 5.74. v = 56.8 m 2 s 2 = 7.54 m s .
EVALUATE: In both cases, moving up the incline and moving down the incline, the acceleration is directed down
the incline. The magnitude of a is greater when the package is going up the incline, because mg sin 37° and f k are
in the same direction whereas when the package is going down these two forces are in opposite directions.
!
!
IDENTIFY: Apply ∑ F = ma to the hammer. Since the hammer is at rest relative to the bus its acceleration equals that of the bus.
SET UP: The free-body diagram for the hammer is given in Figure 5.74.
EXECUTE: ∑ Fy = ma y gives T sin 74° − mg = 0 so T sin 74° = mg . ∑ Fx = max gives T cos74° = ma. Divide the
a
1
and a = 2.8 m s 2 .
=
g tan74°
EVALUATE: When the acceleration increases the angle between the rope and the ceiling of the bus decreases,
and the angle the rope makes with the vertical increases. second equation by the first: 5.75. IDENTIFY: Figure 5.74
!
!
Apply ∑ F = ma to the washer and to the crate. Since the washer is at rest relative to the crate, these two objects have the same acceleration.
SET UP: The free-body diagram for the washer is given in Figure 5.75.
EXECUTE: It’s interesting to look at the string’s angle measured from the perpendicular to the top of the crate.
This angle is θstring = 90° − angle measured from the top of the crate . The free-body diagram for the washer then
leads to the following equations, using Newton’s Second Law and taking the upslope direction as positive:
− mw g sin θslope + T sin θstring = mw a and T sin θstring = mw ( a + g sinθslope )
− mw g cosθslope + T cos θstring = 0 and T cosθstring = mw g cos θslope Applying Newton’s Laws Dividing the two equations: tanθstring = 5-33 a + g sin θslope
g cos θslope For the crate, the component of the weight along the slope is − mc g sin θslope and the normal force is mc g cos θslope .
Using Newton’s Second Law again: − mc g sin θslope + μk mc g cos θslope = mc a . μ k = a + g sin θslope
g cos θslope . This leads to the interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic
friction: μ k = tan θstring = tan(90° − 68°) = tan 22° = 0.40 .
EVALUATE: In the limit that μ k → 0 , θstring → 0 and the string is perpendicular to the top of the crate. As μ k increases, θstring increases. 5.76. IDENTIFY: Figure 5.75
!
!
Apply ∑ F = ma to yourself and calculate a. Then use constant acceleration equations to describe the motion.
SET UP: The free-body diagram is given in Figure 5.76.
EXECUTE: (a) ∑ Fy = ma y gives n = mg cos α . ∑ Fx = max gives mg sin α − f k = ma . Combining these two
equations, we have a = g (sin α − μk cos α ) = −3.094 m s 2 . Find your stopping distance:
2
2
vx = 0, ax = −3.094 m s 2 , v0 x = 20 m s . vx = v0 x + 2ax ( x − x0 ) gives x − x0 = 64.6 m, which is greater than 40 m.
You don’t stop before you reach the hole, so you fall into it.
2
2
(b) ax = −3.094 m s 2 , x − x0 = 40 m, vx = 0 . vx = v0 x + 2ax ( x − x0 ) gives v0 x = 16 m s.
EVALUATE: Your stopping distance is proportional to the square of your initial speed, so your initial speed is
proportional to the square root of your stopping distance. To stop in 40 m instead of 64.6 m your initial speed must
40 m
= 16 m/s .
be (20 m/s)
64.6 m 5.77. IDENTIFY: Figure 5.76
!
!
Apply ∑ F = ma to each block and to the rope. The key idea in solving this problem is to recognize that if the system is accelerating, the tension that block A exerts on the rope is different from the tension that block
B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn’t accelerate.)
SET UP: Take a positive coordinate direction for each object to be in the direction of the acceleration of that
object. All three objects have the same magnitude of acceleration.
EXECUTE: The Second Law equations for the three different parts of the system are:
Block A (The only horizontal forces on A are tension to the right, and friction to the left): − μk mA g + TA = m Aa.
Block B (The only vertical forces on B are gravity down, and tension up): mB g − TB = mB a.
Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the
portion of the rope that hangs vertically): mR d g + TB − TA = mR a.
L () 5-34 Chapter 5 To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations:
m + mR (d / L) − μ k mA
− μ k mA g + mB g + mR d g = (m A + mB + mR )a, or a = g B
.
L
( mA + mB + mR ) () (a) When μ k = 0, a = g mB + mR ( d / L)
. As the system moves, d will increase, approaching L as a limit, and thus
( mA + mB + mR ) the acceleration will approach a maximum value of a = g mB + mR
.
(m A + mB + mR ) (b) For the blocks to just begin moving, a > 0, so solve 0 = [mB + mR ( d / L) − μs m A ] for d. Note that we must use static friction to find d for when the block will begin to move. Solving for d, d = L
( μs mA − mB ) or
mR d = 1.0 m (0.25(2 kg) − 0.4 kg) = 0.63 m.
0.160 kg
(c) When mR = 0.04 kg, d = 1.0 m (0.25(2 kg) − 0.4 kg) = 2.50 m . This is not a physically possible situation
0.04 kg
since d > L. The blocks won’t move, no matter what portion of the rope hangs over the edge.
EVALUATE: For the blocks to move when released, the weight of B plus the weight of the rope that hangs
vertically must be greater than the maximum static friction force on A, which is μs n = 4.9 N .
5.78. Apply Newton’s 1st law to the rope. Let m1 be the mass of that part of the rope that is on the table, IDENTIFY: and let m2 be the mass of that part of the rope that is hanging over the edge. ( m1 + m2 = m, the total mass of the
rope). Since the mass of the rope is not being neglected, the tension in the rope varies along the length of the rope.
Let T be the tension in the rope at that point that is at the edge of the table.
SET UP: The free-body diagram for the hanging section of the rope is given in Figure 5.78a
EXECUTE:
∑ Fy = ma y T − m2 g = 0
T = m2 g
Figure 5.78a
SET UP: The free-body diagram for that part of the rope that is on the table is given in Figure 5.78b.
EXECUTE:
∑ Fy = ma y n − m1 g = 0
n = m1 g
Figure 5.78b When the maximum amount of rope hangs over the edge the static friction has its maximum value:
f s = μs n = μs m1 g ∑F x = max T − fs = 0
T = μs m1 g
Use the first equation to replace T:
m2 g = μs m1 g
m2 = μs m1
The fraction that hangs over is 5.79. m2
μs m1
μs
=
=
.
m m1 + μs m1 1 + μs EVALUATE: As μs → 0, the fraction goes to zero and as μs → ∞, the fraction goes to unity.
IDENTIFY: First calculate the maximum acceleration that the static friction force can give to the case. Apply
!
!
∑ F = ma to the case. Applying Newton’s Laws 5-35 (a) SET UP: The static friction force is to the right in Figure 5.79a (northward) since it tries to make the case
move with the truck. The maximum value it can have is f s = μs N .
EXECUTE:
∑ Fy = ma y n − mg = 0
n = mg
f s = μs n = μs mg
Figure 5.79a ∑F x = max f s = ma μs mg = ma
a = μs g = (0.30)(9.80 m/s 2 ) = 2.94 m/s 2
The truck’s acceleration is less than this so the case doesn’t slip relative to the truck; the case’s acceleration is
a = 2.20 m/s 2 (northward). Then f s = ma = (30.0 kg)(2.20 m/s 2 ) = 66 N, northward.
(b) IDENTIFY: Now the acceleration of the truck is greater than the acceleration that static friction can give the
case. Therefore, the case slips relative to the truck and the friction is kinetic friction. The friction force still tries to
keep the case moving with the truck, so the acceleration of the case and the friction force are both southward. The
free-body diagram is sketched in Figure 5.79b.
SET UP:
EXECUTE:
∑ Fy = ma y
n − mg = 0
n = mg
f k = μ k mg = (0.20)(30.0 kg)(9.80 m/s 2 )
f k = 59 N, southward
Figure 5.79b fk
59 N
=
= 2.0 m/s 2 . The magnitude of the acceleration of the case is less
m 30.0 kg
than that of the truck and the case slides toward the front of the truck. In both parts (a) and (b) the friction is in the
direction of the motion and accelerates the case. Friction opposes relative motion between two surfaces in contact.
!
!
IDENTIFY: Apply ∑ F = ma to the car to calculate its acceleration. Then use a constant acceleration equation to
f k = ma implies a = EVALUATE: 5.80. find the initial speed.
SET UP: Let + x be in the direction of the car’s initial velocity. The friction force f k is then in the − x -direction .
192 ft = 58.52 m .
EXECUTE: n = mg and f k = μk mg . ∑ Fx = max gives − μ k mg = max and
2
2
ax = − μ k g = −(0.750)(9.80 m/s 2 ) = −7.35 m/s 2 . vx = 0 (stops), x − x0 = 58.52 m . vx = v0 x + 2ax ( x − x0 ) gives v0 x = −2ax ( x − x0 ) = −2(−7.35 m/s 2 )(58.52 m) = 29.3 m/s = 65.5 mi/h . He was guilty.
x − x0 = EVALUATE: 2
2
vx − v0 x
v2
= − 0 x . If his initial speed had been 45 mi/h he would have stopped in
2a x
2ax 2 5.81. ⎛ 45 mi/h ⎞
⎜
⎟ (192 ft) = 91 ft .
⎝ 65.5 mi/h ⎠
!
!
IDENTIFY: Apply ∑ F = ma to the point where the three wires join and also to one of the balls. By symmetry
the tension in each of the 35.0 cm wires is the same. 5-36 Chapter 5 The geometry of the situation is sketched in Figure 5.81a. The angle φ that each wire makes with the
12.5 cm
and φ = 15.26° . Let TA be the tension in the vertical wire and let TB be the
vertical is given by sin φ =
47.5 cm
tension in each of the other two wires. Neglect the weight of the wires. The free-body diagram for the left-hand
ball is given in Figure 5.81b and for the point where the wires join in Figure 5.81c. n is the force one ball exerts on
the other.
EXECUTE: (a) ∑ Fy = ma y applied to the ball gives TB cos φ − mg = 0 .
SET UP: TB = mg
(15.0 kg)(9.80 m/s 2 )
=
= 152 N . Then
cos φ
cos15.26° ∑F y = ma y applied in Figure 5.81c gives TA − 2TB cos φ = 0 and TA = 2(152 N)cos φ = 294 N .
(b) ∑F x = max applied to the ball gives n − TB sin φ = 0 and n = (152 N)sin15.26° = 40.0 N . EVALUATE: 5.82. IDENTIFY: TA equals the total weight of the two balls. Figure 5.81a–c
!
!
Apply ∑ F = ma to the box. Compare the acceleration of the box to the acceleration of the truck and use constant acceleration equations to describe the motion.
SET UP: Both objects have acceleration in the same direction; take this to be the + x -direction.
EXECUTE: If the block were to remain at rest relative to the truck, the friction force would need to cause an
acceleration of 2.20 m s 2 ; however, the maximum acceleration possible due to static friction is
(0.19)(9.80 m s 2 ) = 1.86 m s 2 , and so the block will move relative to the truck; the acceleration of the box
would be μ k g = (0.15)(9.80 m s 2 ) = 1.47 m s 2 . The difference between the distance the truck moves and the
distance the box moves (i.e., the distance the box moves relative to the truck) will be 1.80 m after a time
t= 5.83. 2Δx
2(1.80 m)
=
= 2.221 s.
2
atruck − abox
(2.20 m s 2 − 1.47 m s ) In this time, the truck moves 1 atruck t 2 = 1 (2.20m s 2 ) (2.221 s) 2 = 5.43 m.
2
2
EVALUATE: To prevent the box from sliding off the truck the coefficient of static friction would have to be
μs = (2.20 m/s 2 ) / g = 0.224 .
!
!
IDENTIFY: Apply ∑ F = ma to each block. Forces between the blocks are related by Newton’s 3rd law. The
target variable is the force F. Block B is pulled to the left at constant speed, so block A moves to the right at
constant speed and a = 0 for each block.
SET UP: The free-body diagram for block A is given in Figure 5.83a. nBA is the normal force that B exerts on A.
f BA = μk nBA is the kinetic friction force that B exerts on A. Block A moves to the right relative to B, and f BA
opposes this motion, so f BA is to the left. Applying Newton’s Laws 5-37 Note also that F acts just on B, not on A.
EXECUTE:
∑ Fy = ma y nBA − wA = 0
nBA = 1.40 N
f BA = μk nBA = (0.30)(1.40 N) = 0.420 N
Figure 5.83a ∑F x = max T − f BA = 0
T = f BA = 0.420 N
SET UP: The free-body diagram for block B is given in Figure 5.83b. Figure 5.83b
EXECUTE: n AB is the normal force that block A exerts on block B. By Newton’s third law n AB and nBA are equal in magnitude and opposite in direction, so n AB = 1.40 N. f AB is the kinetic friction force that A exerts on B. Block
B moves to the left relative to A and f AB opposes this motion, so f AB is to the right.
f AB = μk n AB = (0.30)(1.40 N) = 0.420 N.
n and f k are the normal and friction force exerted by the floor on block B; f k = μk n. Note that block B moves to
the left relative to the floor and f k opposes this motion, so f k is to the right. ∑F y = ma y n − wB − n AB = 0
n = wB + n AB = 4.20 N + 1.40 N = 5.60 N
Then f k = μk n = (0.30)(5.60 N) = 1.68 N. ∑F x = max f AB + T + f k − F = 0
F = T + f AB + f k = 0.420 N + 0.420 N + 1.68 N = 2.52 N 5.84. EVALUATE: Note that f AB and f BA are a third law action-reaction pair, so they must be equal in magnitude and
opposite in direction and this is indeed what our calculation gives.
!
!
IDENTIFY: Apply ∑ F = ma to the person to find the acceleration the PAPS unit produces. Apply constant acceleration equations to her free-fall motion and to her motion after the PAPS fires.
SET UP: We take the upward direction as positive.
EXECUTE: The explorer’s vertical acceleration is −3.7 m s 2 for the first 20 s. Thus at the end of that time her
vertical velocity will be v y = a yt = (−3.7 m s 2 )(20 s) = −74 m s. She will have fallen a distance
⎛ −74 m s ⎞
d = vavt = ⎜
⎟ (20 s) = −740 m and will thus be 1200 m − 740 m = 460 m above the surface. Her vertical
2
⎝
⎠
velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as 5-38 Chapter 5
2
2
v y − v0 y 0 − (−74 m s) 2
= 5.95 m s 2 , which is the vertical
2( y − y0 )
−460 m
acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by
2
2
y0 = 0, v y = v0 y + 2a y ( y − y0 ) gives a y = t= v y − v0 y
ay = = 0 − (−74 m s)
= 12.4 s
5.95 m s 2 Using Newton’s Second Law for the vertical direction FPAPSv + mg = ma . This gives
FPAPSv = ma − mg = m( a + g ) = (150 kg)(5.95 − (−3.7)) m s 2 = 1450 N ,
which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in
12.4 seconds; the acceleration needed to do this is
ay = 5.85. v y − v0 y
t = 0 − 33 m s 2
= 2.66 m s 2
12.4 s and the force needed is FPAPSh = ma = (150 kg)(2.66 m s 2 ) = 400 N , since there are no other horizontal forces.
EVALUATE: The acceleration produced by the PAPS must bring to zero both her horizontal and vertical
components of velocity.
!
!
IDENTIFY: Apply ∑ F = ma to each block. Parts (a) and (b) will be done together. Figure 5.85a Note that each block has the same magnitude of acceleration, but in different directions. For each block let the
!
direction of a be a positive coordinate direction.
SET UP: The free-body diagram for block A is given in Figure 5.85b.
EXECUTE:
∑ Fy = ma y TAB − mA g = mA a
TAB = m A (a + g )
TAB = 4.00 kg(2.00 m/s 2 + 9.80 m/s 2 ) = 47.2 N
Figure 5.85b
SET UP: The free-body diagram for block B is given in Figure 5.85b.
EXECUTE:
∑ Fy = ma y n − mB g = 0
n = mB g
Figure 5.85c f k = μ k n = μ k mB g = (0.25)(12.0 kg)(9.80 m/s 2 ) = 29.4 N ∑F x = max TBC − TAB − f k = mB a
TBC = TAB + f k + mB a = 47.2 N + 29.4 N + (12.0 kg)(2.00 m/s 2 )
TBC = 100.6 N Applying Newton’s Laws SET UP: 5-39 The free-body diagram for block C is sketched in Figure 5.85d.
EXECUTE:
∑ Fy = ma y mC g − TBC = mC a
mC ( g − a ) = TBC
mC = TBC
100.6 N
=
= 12.9 kg
g − a 9.80 m/s 2 − 2.00 m/s 2 Figure 5.85d
EVALUATE: If all three blocks are considered together as a single object and ! ! ∑ F = ma is applied to this combined object, mC g − mA g − μk mB g = ( mA + mB + mC ) a. Using the values for μ k , mA and mB given in the
5.86. problem and the mass mC we calculated, this equation gives a = 2.00 m/s 2 , which checks.
!
!
IDENTIFY: Apply ∑ F = ma to each block. They have the same magnitude of acceleration, a.
SET UP: Consider positive accelerations to be to the right (up and to the right for the left-hand block, down and
to the right for the right-hand block).
EXECUTE: (a) The forces along the inclines and the accelerations are related by
T − (100 kg)g sin 30° = (100 kg)a and (50 kg)g sin 53° − T = (50 kg)a, where T is the tension in the cord and a the
mutual magnitude of acceleration. Adding these relations,
(50 kg sin 53° − 100 kg sin 30°) g = (50 kg + 100 kg)a, or a = −0.067 g. Since a comes out negative, the blocks will
slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive
accelerations were to the left, a would be +0.067 g .
2 5.87. (b) a = 0.067(9.80 m s 2 ) = 0.658 m s .
(c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the
above relations involving T yields 424 N.
EVALUATE: For part (a) we could have compared mg sin θ for each block to determine which direction the
system would move.
IDENTIFY: Let the tensions in the ropes be T1 and T2 . Figure 5.87a Consider the forces on each block. In each case take a positive coordinate direction in the direction of the
acceleration of that block.
SET UP: The free-body diagram for m1 is given in Figure 5.87b. EXECUTE:
∑ Fx = max T1 = m1a1 Figure 5.87b 5-40 Chapter 5 The free-body diagram for m2 is given in Figure 5.87c. SET UP: EXECUTE:
∑ Fy = ma y m2 g − T2 = m2 a2
Figure 5.87c This gives us two equations, but there are 4 unknowns ( T1 , T2 , a1 , and a2 ) so two more equations are required.
SET UP: The free-body diagram for the moveable pulley (mass m) is given in Figure 5.87d. EXECUTE:
∑ Fy = ma y mg + T2 − 2T1 = ma Figure 5.87d But our pulleys have negligible mass, so mg = ma = 0 and T2 = 2T1. Combine these three equations to eliminate T1
and T2 : m2 g − T2 = m2 a2 gives m2 g − 2T1 = m2 a2 . And then with T1 = m1a 1 we have m2 g − 2m1a1 = m2 a2 .
There are still two unknowns, a1 and a2 . But the accelerations a1 and a2 are related. In any time SET UP: interval, if m1 moves to the right a distance d, then in the same time m2 moves downward a distance d / 2. One of
the constant acceleration kinematic equations says x − x0 = v0 xt + 1 axt 2 , so if m2 moves half the distance it must
2
have half the acceleration of m1 : a2 = a1 / 2, or a1 = 2a2 .
EXECUTE: This is the additional equation we need. Use it in the previous equation and get
m2 g − 2m1 (2a2 ) = m2 a2 .
a2 (4m1 + m2 ) = m2 g
a2 = m2 g
2m2 g
and a1 = 2a2 =
.
4m1 + m2
4m1 + m2 EVALUATE:
5.88. IDENTIFY: If m2 → 0 or m1 → ∞, a1 = a2 = 0. If m2 >> m1 , a2 = g and a1 = 2 g .
!
!
Apply ∑ F = ma to block B, to block A and B as a composite object and to block C. If A and B slide together all three blocks have the same magnitude of acceleration.
SET UP: If A and B don’t slip the friction between them is static. The free-body diagrams for block B, for blocks
A and B, and for C are given in Figures 5.88a-c. Block C accelerates downward and A and B accelerate to the right.
In each case take a positive coordinate direction to be in the direction of the acceleration. Since block A moves to
the right, the friction force f s on block B is to the right, to prevent relative motion between the two blocks. When C
has its largest mass, f s has its largest value: f s = μs n .
EXECUTE: a = μs g . ∑F x ∑F x = max applied to the block B gives f s = mB a . n = mB g and f s = μs mB g . μs mB g = mB a and = max applied to blocks A + B gives T = mAB a = mAB μs g . mC g − T = mC a . mC g − mAB μs g = mC μs g . mC = ∑F y = ma y applied to block C gives mAB μs
⎛ 0.750 ⎞
= (5.00 kg + 8.00 kg) ⎜
⎟ = 39.0 kg .
1 − μs
⎝ 1 − 0.750 ⎠ Applying Newton’s Laws 5-41 EVALUATE: With no friction from the tabletop, the system accelerates no matter how small the mass of C is.
If mC is less than 39.0 kg, the friction force that A exerts on B is less than μs n . If mC is greater than 39.0 kg,
blocks C and A have a larger acceleration than friction can give to block B and A accelerates out from under B. Figure 5.88
5.89. IDENTIFY: Apply the method of Exercise 5.19 to calculate the acceleration of each object. Then apply constant
acceleration equations to the motion of the 2.00 kg object.
SET UP: After the 5.00 kg object reaches the floor, the 2.00 kg object is in free-fall, with downward acceleration g.
5.00 kg − 2.00 kg
EXECUTE: The 2.00-kg object will accelerate upward at g
= 3 g 7, and the 5.00-kg object will
5.00 kg + 2.00 kg accelerate downward at 3 g 7. Let the initial height above the ground be h0 . When the large object hits the
2
ground, the small object will be at a height 2h0 , and moving upward with a speed given by v0 = 2ah0 = 6 gh0 7.
2
The small object will continue to rise a distance v0 2 g = 3h0 7, and so the maximum height reached will be 5.90. 2h0 + 3h0 7 = 17h0 7 = 1.46 m above the floor , which is 0.860 m above its initial height.
EVALUATE: The small object is 1.20 m above the floor when the large object strikes the floor, and it rises an
additional 0.26 m after that.
!
!
IDENTIFY: Apply ∑ F = ma to the box.
SET UP: The box has an upward acceleration of a = 1.90 m/s 2 .
EXECUTE: The floor exerts an upward force n on the box, obtained from n − mg = ma, or n = m( a + g ). The
friction force that needs to be balanced is
2 μ k n = μk m(a + g ) = (0.32)(28.0 kg)(1.90 m s 2 + 9.80 m s ) = 105 N. 5.91. EVALUATE: If the elevator wasn't accelerating the normal force would be n = mg and the friction force that
would have to be overcome would be 87.8 N. The upward acceleration increases the normal force and that
increases the friction force.
!
!
IDENTIFY: Apply ∑ F = ma to the block. The cart and the block have the same acceleration. The normal force exerted by the cart on the block is perpendicular to the front of the cart, so is horizontal and to the right. The
friction force on the block is directed so as to hold the block up against the downward pull of gravity. We want to
calculate the minimum a required, so take static friction to have its maximum value, f s = μs n.
SET UP: The free-body diagram for the block is given in Figure 5.91.
EXECUTE:
∑ Fx = max
n = ma
f s = μs n = μs ma
Figure 5.91 ∑F y = ma y f s − mg = 0 μs ma = mg
a = g / μs
EVALUATE: An observer on the cart sees the block pinned there, with no reason for a horizontal force on it
because the block is at rest relative to the cart. Therefore, such an observer concludes that n = 0 and thus f s = 0,
and he doesn’t understand what holds the block up against the downward force of gravity. The reason for this 5-42 Chapter 5 difficulty is that 5.92. ! ! ∑ F = ma does not apply in a coordinate frame attached to the cart. This reference frame is accelerated, and hence not inertial. The smaller μs is, the larger a must be to keep the block pinned against the
front of the cart.
!
!
IDENTIFY: Apply ∑ F = ma to each block.
SET UP: Use coordinates where + x is directed down the incline.
EXECUTE: (a) Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be
pulled down the plane; i.e., the larger block will not move faster than the smaller block, and the blocks will have
the same acceleration. For the smaller block, (4.00 kg)g (sin30° − (0.25)cos 30°) − T = (4.00 kg)a, or
11.11 N − T = (4.00 kg)a, and similarly for the larger, 15.44 N + T = (8.00 kg)a . Adding these two relations, 26.55 N = (12.00 kg)a, a = 2.21 m s 2 .
(b) Substitution into either of the above relations gives T = 2.27 N.
(c) The string will be slack. The 4.00-kg block will have a = 2.78 m s 2 and the 8.00-kg block will have 5.93. a = 1.93 m s 2 , until the 4.00-kg block overtakes the 8.00-kg block and collides with it.
EVALUATE: If the string is cut the acceleration of each block will be independent of the mass of that block and
will depend only on the slope angle and the coefficient of kinetic friction. The 8.00-kg block would have a smaller
acceleration even though it has a larger mass, since it has a larger μ k .
!
!
IDENTIFY: Apply ∑ F = ma to the block and to the plank.
SET UP: Both objects have a = 0 .
EXECUTE: Let nB be the normal force between the plank and the block and n A be the normal force between the block and the incline. Then, nB = w cos θ and n A = nB + 3w cos θ = 4w cos θ . The net frictional force on the block is μ k (nA + nB ) = μ k 5w cosθ . To move at constant speed, this must balance the component of the block’s weight
3
3
along the incline, so 3w sin θ = μk 5w cos θ , and μ k = 5 tan θ = 5 tan 37° = 0.452. 5.94. EVALUATE: In the absence of the plank the block slides down at constant speed when the slope angle and
coefficient of friction are related by tan θ = μk . For θ = 36.9° , μ k = 0.75 . A smaller μ k is needed when the plank
is present because the plank provides an additional friction force.
!
!
IDENTIFY: Apply ∑ F = ma to the ball, to m1 and to m2
SET UP: The free-body diagrams for the ball, m1 and m2 are given in Figures 5.94a-c. All three objects have the
!
same magnitude of acceleration. In each case take the direction of a to be a positive coordinate direction.
EXECUTE: (a) ∑ Fy = ma y applied to the ball gives T cosθ = mg . ∑ Fx = max applied to the ball gives T sin θ = ma . Combining these two equations to eliminate T gives tan θ = a / g . (b) ∑F x = max applied to m2 gives T = m2 a . ∑F y = ma y applied to m1 gives m1 g − T = m1a . Combining these ⎛ m1 ⎞
m1
250 kg
=
and θ = 9.46° .
two equations gives a = ⎜
⎟ g . Then tan θ =
m1 + m2 ⎠
m1 + m2 1500 kg
⎝
(c) As m1 becomes much larger than m2 , a → g and tan θ → 1 , so θ → 45° .
EVALUATE: The device requires that the ball is at rest relative to the platform; any motion swinging back and
forth must be damped out. When m1 << m2 the system still accelerates, but with small a and θ → 0° . Figure 5.94a–c Applying Newton’s Laws 5.95. 5-43 !
!
Apply ∑ F = ma to the automobile. IDENTIFY: 2
v0
, as derived in Example 5.23.
gR
Use coordinates that are vertical and horizontal, since the acceleration is horizontal.
EXECUTE: For speeds larger than v0 , a frictional force is needed to keep the car from skidding. In this case, the SET UP: The "correct" banking angle is for zero friction and is given by tan β = inward force will consist of a part due to the normal force n and the friction force f ; n sinβ + f cos β = marad . The
normal and friction forces both have vertical components; since there is no vertical acceleration,
2
(1.5v0 ) 2
n cosβ − f sinβ = mg . Using f = μs n and arad = v =
= 2.25 g tan β , these two relations become
R
R
n sin β + μs n cos β = 2.25 mg tan β and n cos β − μs n sin β = mg . Dividing to cancel n gives
sin β + μs cos β
1.25 sinβ cosβ
. Using
= 2.25 tanβ . Solving for μs and simplifying yields μs =
cos β − μs sin β
1 + 1.25sin 2 β
⎛ ⎞
(20 m s) 2
⎟ = 18.79° gives μs = 0.34.
(9.80 m s 2 )(120 m) ⎠
⎝ β = arctan ⎜ 5.96. EVALUATE: If μs is insufficient, the car skids away from the center of curvature of the roadway, so the friction in
inward.
!
!
!!
IDENTIFY: Apply ∑ F = ma to the car. The car moves in the arc of a horizontal circle, so a = arad, directed toward the center of curvature of the roadway. The target variable is the speed of the car. arad will be calculated
from the forces and then v will be calculated from arad = v 2 / R.
(a) To keep the car from sliding up the banking the static friction force is directed down the incline. At maximum
speed the static friction force has its maximum value f s = μs n.
SET UP: The free-body diagram for the car is sketched in Figure 5.96a.
EXECUTE:
∑ Fy = ma y n cos β − fs sin β − mg = 0
But f s = μs n, so
n cos β − μs n sin β − mg = 0
n= mg
cos β − μs sin β Figure 5.96a ∑F x = max n sin β + μs n cos β = marad
n(sin β + μs cos β ) = marad
Use the ∑F y equation to replace n: ⎛
⎞
mg
⎜
⎟ (sin β + μs cos β ) = marad
⎝ cos β − μs sin β ⎠
⎛ sin β + μs cos β
arad = ⎜
⎝ cos β − μs sin β ⎞
⎛ sin 25° + (0.30)cos 25° ⎞
2
2
⎟g = ⎜
⎟ (9.80 m/s ) = 8.73 m/s
cos 25° − (0.30)sin 25° ⎠
⎝
⎠ arad = v 2 / R implies v = arad R = (8.73 m/s 2 )(50 m) = 21 m/s.
(b) IDENTIFY: To keep the car from sliding down the banking the static friction force is directed up the incline.
At the minimum speed the static friction force has its maximum value f s = μs n. 5-44 Chapter 5 SET UP: The free-body diagram for the car is sketched in Figure 5.96b. The free-body diagram is identical to that
in part (a) except that now the components
of f s have opposite directions. The force
equations are all the same except for the
opposite sign for terms containing μs .
Figure 5.96b
EXECUTE: ⎛ sin β − μs cos β ⎞
⎛ sin 25° − (0.30)cos 25° ⎞
2
2
arad = ⎜
⎟g = ⎜
⎟ (9.80 m/s ) = 1.43 m/s
⎝ cos 25° + (0.30)sin 25° ⎠
⎝ cos β + μs sin β ⎠ v = arad R = (1.43 m/s 2 )(50 m) = 8.5 m/s. 5.97. EVALUATE: For v between these maximum and minimum values, the car is held on the road at a constant height
by a static friction force that is less than μs n. When μs → 0, arad = g tan β . Our analysis agrees with the result of
Example 5.23 in this special case.
!
!
IDENTIFY: Apply ∑ F = ma to the car.
SET UP: 1 mi/h = 0.447 m/s . The acceleration of the car is arad = v 2 / r , directed toward the center of curvature
of the roadway.
EXECUTE: (a) 80 mi h = 35.7 m s . The centripetal force needed to keep the car on the road is provided by friction; thus μs mg = mv 2
v2
(35.7 m s) 2
and r =
=
= 171 m .
μs g (0.76)(9.8 m s 2 )
r (b) If μs = 0.20 , v = r μs g = (171 m) (0.20) (9.8 m/s 2 ) = 18.3 m s or about 41 mi h .
(c) If μs = 0.37 , v = (171 m) (0.37) (9.8 m/s 2 ) = 24.9 m s or about 56 mi h
The speed limit is evidently designed for these conditions.
EVALUATE: The maximum safe speed is proportional to
5.98. 0.20 / 0.76 = 0.51 , so the maximum safe speed for wet-ice conditions is about half what it is for a dry road.
IDENTIFY: The analysis of this problem is the same as that of Example 5.21.
a
v2
SET UP: From Example 5.21, tan β = rad =
.
g
rg
EXECUTE: 5.99. μs . Solving for v in terms of β and R, v = gR tan β = (9.80 m s 2 ) (50.0) tan 30.0° = 16.8 m s , about 60.6 km h.
EVALUATE: The greater the speed of the bus the larger will be the angle β , so T will have a larger horizontal,
inward component.
IDENTIFY and SET UP: The monkey and bananas have the same mass and the tension in the rope has the same
upward value at the bananas and at the monkey. Therefore, the monkey and bananas will have the same net force
and hence the same acceleration, in both magnitude and direction.
EXECUTE: (a) For the monkey to move up, T > mg . The bananas also move up.
(b) The bananas and monkey move with the same acceleration and the distance between them remains constant.
(c) Both the monkey and bananas are in free fall. They have the same initial velocity and as they fall the distance
between them doesn’t change.
(d) The bananas will slow down at the same rate as the monkey. If the monkey comes to a stop, so will the
bananas.
EVALUATE: None of these actions bring the monkey any closer to the bananas. Applying Newton’s Laws 5.100. SET UP: Follow the analysis that leads to Eq.(5.10), except now the initial speed is v0 y = 3mg / k = 3vt rather than zero.
EXECUTE: The separated equation of motion has a lower limit of 3vt instead of 0; specifically,
v dv ∫ v−v 3 vt 5.101. 5-45 !
!
IDENTIFY: Apply ∑ F = ma , with f = kv . = ln
t ⎛ v 1⎞
vt − v
k
⎡1
⎤
= ln ⎜
− ⎟ = − t , or v = 2vt ⎢ + e − ( k m )t ⎥ .
−2vt
m
⎣2
⎦
⎝ 2vt 2 ⎠ EVALUATE: As t → ∞ the speed approaches vt . The speed is always greater than vt and this limit is approached
from above.
!
!
IDENTIFY: Apply ∑ F = ma to the rock.
SET UP: Equations 5.9 through 5.13 apply, but with a0 rather than g as the initial acceleration.
EXECUTE: (a) The rock is released from rest, and so there is initially no resistive force and
a0 = (18.0 N) (3.00 kg) = 6.00 m s 2 .
(b) (18.0 N − (2.20 N ⋅ s m) (3.00 m s)) (3.00 kg) = 3.80 m s 2 .
(c) The net force must be 1.80 N, so kv = 16.2 N and v = (16.2 N) (2.20 N ⋅ s m) = 7.36 m s.
(d) When the net force is equal to zero, and hence the acceleration is zero, kvt = 18.0 N and vt = (18.0 N) (2.20 N ⋅ s m) = 8.18 m s.
(e) From Eq.(5.12),
⎡
⎤
3.00 kg
y = (8.18 m s) ⎢(2.00 s) −
(1 − e−((2.20 N⋅s m) (3.00 kg))(2.00 s) )⎥ = +7.78 m.
2.20 N ⋅ s m
⎣
⎦
From Eq. (5.10), v = (8.18 m s)[1 − e − ( (2.20 N ⋅s m) (3.00 kg))(2.00 s) ] = 6.29 m s.
From Eq.(5.11), but with a0 instead of g, a = (6.00 m s 2 )e − ((2.20 N⋅s m) (3.00 kg))(2.00 s) = 1.38 m s 2 .
(f) 1 − v
m
= 0.1 = e − ( k m ) t and t = ln (10) = 3.14 s.
k
vt EVALUATE: 5.102. The acceleration decreases with time until it becomes zero when v = vt . The speed increases with time and approaches vt as t → ∞ .
!
!
dv
dx
IDENTIFY: Apply ∑ F = ma to the rock. a =
and v =
yield differential equations that can be integrated to
dt
dt
give v(t ) and x (t ) .
SET UP: The retarding force of the surface is the only horizontal force acting.
v dv
F
F
−kv1 2 dv
dv
k
kt
EXECUTE: (a) Thus a = net = R =
=
and 1 2 = − dt . Integrating gives ∫ 1 2 = − ∫ dt and
v0 v
m
m
m
dt
v
m
m0
12
22
v kt k t
kt
+
2v1 2 v0 = − . This gives v = v0 − 0
.
v
m
4m 2
m
dx
v1 2 kt k 2t 2
v1 2 ktdt k 2t 2 dt
= v0 − 0
+
and dx = v0 dt − 0
+
For the rock’s position:
.
2
dt
m
4m
m
4m 2
v1 2 kt 2 k 2t 3
+
.
Integrating gives x = v0t − 0
2m
12m 2
v1 2 kt k 2t 2
(b) v = 0 = v0 − 0
+
. This is a quadratic equation in t; from the quadratic formula we can find the single
m
2m 2
2mv1 2
0
.
solution t =
k
(c) Substituting the expression for t into the equation for x:
x = v0 ⋅
EVALUATE: 1
3
3
2mv1 2 v0 2 k 4m 2v0
k 2 8m3v0 2 2mv0 2
0
−
⋅
+
⋅
=
2
2
3
k
2m
k
12m
k
3k The magnitude of the average acceleration is aav = Fav = maav = 1 kv1 / 2 , which is
0
2 1
2 times the initial value of the force. Δv
v0
1 kv1 / 2
0
−
=
. The average force is
1/ 2
Δt (2mv0 / k ) 2 m 5-46 Chapter 5 5.103. IDENTIFY:
SET UP:
EXECUTE: Apply ! ! ∑ F = ma to the object, with and without including the buoyancy force. At the terminal speed vt , a = 0 .
Without buoyancy, kvt = mg , so k = ⎛ 0.24 m s ⎞
B = mg − kvt = mg ⎜1 −
⎟ = mg 3 .
⎝ 0.36 m s ⎠
EVALUATE: At the terminal speed, B and f = kv together equal mg. The presence of B reduces the value of f
required, so the presence of B reduces the terminal speed.
IDENTIFY: The block has acceleration arad = v 2 / r , directed to the left in the figure in the problem. Apply
!
!
∑ F = ma to the block.
upward buoyancy force B, so B + kvt = mg 5.104. mg
mg
=
. With buoyancy included there is the additional
vt
0.36 s SET UP: . The block moves in a horizontal circle of radius r = (1.25 m) 2 − (1.00 m) 2 = 0.75 m . Each string 1.00 m
, so θ = 36.9° . The free-body diagram for the block is given in
1.25 m
Figure 5.104. Let + x be to the left and let + y be upward. makes an angle θ with the vertical. cosθ = EXECUTE: (a) ∑F y = ma y gives Tu cosθ − Tl cosθ − mg = 0 . mg
(4.00 kg)(9.80 m/s 2 )
= 80.0 N −
= 31.0 N .
cosθ
cos36.9°
v2
(b) ∑ Fx = max gives (Tu + Tl )sin θ = m .
r
Tl = Tu − v= r (Tu + Tl )sin θ
(0.75 m)(80.0 N + 31.0 N)sin 36.9°
=
= 3.53 m/s . The number of revolutions per second is
m
4.00 kg v
3.53 m/s
=
= 0.749 rev/s = 44.9 rev/min .
2π r 2π (0.75 m)
(c) If Tl → 0 , Tu cosθ = mg and Tu = v= v2
mg
(4.00 kg)(9.80 m/s 2 )
=
= 49.0 N . Tu sin θ = m .
r
cosθ
cos36.9° rTu sin θ
(0.75 m)(49.0 N)sin 36.9°
=
= 2.35 m/s . The number of revolutions per minute is
m
4.00 kg ⎛ 2.35 m/s ⎞
(44.9 rev/min) ⎜
⎟ = 29.9 rev/min
⎝ 3.53 m/s ⎠
EVALUATE: The tension in the upper string must be greater than the tension in the lower string so that together
they produce an upward component of force that balances the weight of the block. 5.105. Figure 5.104
!
!
IDENTIFY: Apply ∑ F = ma to the falling object.
SET UP: Follow the steps that lead to Eq.(5.10), except now v0 y = v0 and is not zero. Applying Newton’s Laws (a) Newton’s 2nd law gives m EXECUTE: dv y
dt 5-47 v = mg − kv y , where y
t
dv y
k
mg
= vt . ∫
= − ∫ dt . This is the same
m0
k
v0 v y − vt expression used in the derivation of Eq. (5.10), except the lower limit in the velocity integral is the initial speed v0
instead of zero. Evaluating the integrals and rearranging gives v = v0e − kt m + vt (1 − e − kt m ) . Note that at t = 0 this
expression says v y = v0 and at t → α it says v y → vt .
(b) The downward gravity force is larger than the upward fluid resistance force so the acceleration is downward,
until the fluid resistance force equals gravity when the terminal speed is reached. The object speeds up
until v y = vt . Take + y to be downward. The graph is sketched in Figure 5.105a.
(c) The upward resistance force is larger than the downward gravity force so the acceleration is upward and the
object slows down, until the fluid resistance force equals gravity when the terminal speed is reached. Take + y to
be downward. The graph is sketched in Figure 5.105b.
(d) When v0 = vt the acceleration at t = 0 is zero and remains zero; the velocity is constant and equal to the
terminal velocity.
EVALUATE: In all cases the speed becomes vt as t → ∞ . Figure 5.105a, b
5.106. !
!
IDENTIFY: Apply ∑ F = ma to the rock.
At the maximum height, v y = 0 . Let + y be upward. Suppress the y subscripts on v and a. SET UP: 2
(a) To find the maximum height and time to the top without fluid resistance: v 2 = v0 + 2a ( y − y0 ) and EXECUTE: y − y0 = 2
v 2 − v0 0 − (6.0 m s) 2
v − v0 0 − 6.0 m s
=
= 1.84 m . t =
=
= 0.61 s .
2a
2( − 9.8 m s 2 )
a
−9.8 m s 2 dv
= mg − kv . We rearrange and integrate, taking
dt
downward as positive as in the text and noting that the velocity at the top of the rock’s flight is zero:
0 dv
k
−vt
−2.0 m s
0
∫ v v − vt = − m t . ln(v − vt ) v = ln v − vt = ln −6.0 m s − 2.0 m s = ln(0.25) = −1.386
From Eq.(5.9), m k = vt g = (2.0 m s 2 ) (9.8 m s 2 ) = 0.204 s, and t = − m (−1.386) = (0.204 s) (1.386) = 0.283 s
k
dx
to the top. Equation 5.10 in the text gives us
= vt (1 − e− ( k m )t ) = vt − vt e − ( k m )t .
dt
(b) Starting from Newton’s Second Law for this situation m x t t 0 0 0 x = ∫ dx = ∫ vt dt − ∫ vt e− ( k m ) t dt = vtt + 5.107. vt m − ( k m )t
(e
− 1) .
k x = (2.0 m s) (0.283 s) + (2.0 m s) (0.204 s)(e −1.387 − 1) = 0.26 m .
EVALUATE: With fluid resistance present the maximum height is much less and the time to reach it is less.
!
!
IDENTIFY: Apply ∑ F = ma to the car.
SET UP: The forces on the car are the air drag force f D = Dv 2 and the rolling friction force μ r mg . Take the
velocity to be in the + x -direction. The forces are opposite in direction to the velocity.
EXECUTE: (a) ∑ Fx = max gives − Dv 2 − μ r mg = ma . We can write this equation twice, once with v = 32 m s and a = − 0.42 m s 2 and once with v = 24 m s and a = −0.30 m/s 2 . Solving these two simultaneous equations in
the unknowns D and μ r gives μ r = 0.015 and D = 0.36 N ⋅ s 2 m 2 .
(b) n = mg cos β and the component of gravity parallel to the incline is mg sin β , where β = 2.2°. For constant
speed, mg sin 2.2° − μ r mg cos 2.2° − Dv 2 = 0. Solving for v gives v = 29 m s. 5-48 Chapter 5 (c) For angle β , mg sin β − μ r mg cos β − Dv 2 = 0 and v = mg (sin β − μ r cosβ )
. The terminal speed for a falling
D object is derived from Dvt2 − mg = 0, so vt = mg D. v vt = sin β − μ r cos β . And since μ r = 0.015, v vt = sin β − (0.015) cosβ .
EVALUATE:
5.108. SET UP: In part (c), v → vt as β → 90° , since in that limit the incline becomes vertical.
!
!
∑ F = ma to the person and to the cart. Apply IDENTIFY: The apparent weight, wapp , which is the same as the upward force on the person exerted by the car seat. EXECUTE: (a) The apparent weight is the actual weight of the person minus the centripetal force needed to keep
him moving in its circular path: wapp = mg − 5.109. ⎡
mv 2
(12 m s) 2 ⎤
= (70 kg) ⎢(9.8 m s 2 ) −
⎥ = 434 N .
R
40 m ⎦
⎣ (b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to
mv 2
= 0 . v = Rg = (40 m) (9.8 m/s 2 ) = 19.8 m s . The answer doesn’t
exert any upward force on it: mg −
R
depend on the cart’s mass, because the centripetal force needed to hold it on the road is proportional to its mass and
so to its weight, which provides the centripetal force in this situation.
EVALUATE: At the speed calculated in part (b), the downward force needed for circular motion is provided by
gravity. For speeds greater than this more, downward force is needed and there is no source for it and the cart
leaves the circular path. For speeds less than this, less downward force than gravity is needed, so the roadway must
exert an upward vertical force.
(a) IDENTIFY: Use the information given about Jena to find the time t for one revolution of the merry-go-round.
!
Her acceleration is arad , directed in toward the axis. Let F1 be the horizontal force that keeps her from sliding off.
!
!
Let her speed be v1 and let R1 be her distance from the axis. Apply ∑ F = ma to Jena, who moves in uniform circular motion.
SET UP: The free-body diagram for Jena is sketched in Figure 5.109a
EXECUTE:
∑ Fx = max
F1 = marad
F1 = m v12
, v1 =
R1 R1F1
= 1.90 m/s
m Figure 5.109a The time for one revolution is t = 2π R1
m
. Jackie goes around once in the same time but her speed
= 2π R1
v1
R1F1 (v2 ) and the radius of her circular path ( R2 ) are different.
⎛ 1 ⎞ R1F1 R2 R1F1
2π R2
= 2π R2 ⎜
=
.
⎟
t
R1
m
⎝ 2π R1 ⎠ m
!
!
IDENTIFY: Now apply ∑ F = ma to Jackie. She also moves in uniform circular motion.
v2 = SET UP: The free-body diagram for Jackie is sketched in Figure 5.109b.
EXECUTE:
∑ Fx = max F2 = marad
Figure 5.109b F2 = m 2
2
v2 ⎛ m ⎞⎛ R2 ⎞ ⎛ R1F1 ⎞ ⎛ R2 ⎞
⎛ 3.60 m ⎞
= ⎜ ⎟⎜ 2 ⎟ ⎜
⎟ = ⎜ ⎟ F1 = ⎜
⎟ (60.0 N) = 120.0 N
R2 ⎝ R2 ⎠⎝ R1 ⎠ ⎝ m ⎠ ⎝ R1 ⎠
⎝ 1.80 m ⎠ (b) F2 = m 2
v2
, so v2 =
R2 F2 R2
(120.0 N)(3.60 m)
=
= 3.79 m/s
m
30.0 kg Applying Newton’s Laws EVALUATE:
5.110. 5-49 Both girls rotate together so have the same period T. By Eq.(5.16), arad is larger for Jackie so the force on her is larger. Eq.(5.15) says R1 / v1 = R2 / v2 so v2 = v1 ( R2 / R1 ); this agrees with our result in (a).
!
!
IDENTIFY: Apply ∑ F = ma to the passenger. The passenger has acceleration arad , directed inward toward the
center of the circular path.
SET UP: The passenger’s velocity is v = 2π R t = 8.80 m s. The vertical component of the seat’s force must
balance the passenger’s weight and the horizontal component must provide the centripetal force.
mv 2
EXECUTE: (a) Fseat sin θ = mg = 833 N and Fseat cosθ =
= 188 N . Therefore
R
tan θ = (833 N) (188 N) = 4.43; θ = 77.3° above the horizontal. The magnitude of the net force exerted by the
seat (note that this is not the net force on the passenger) is
Fseat = (833 N) 2 + (188 N) 2 = 854 N 5.111. (b) The magnitude of the force is the same, but the horizontal component is reversed.
v2
EVALUATE: At the highest point in the motion, Fseat = mg − m = 645 N . At the lowest point in the motion,
R
v2
Fseat = mg + m = 1021 N . The result in parts (a) and (b) lies between these extreme values.
R
!
!
IDENTIFY: Apply ∑ F = ma to the person. The person moves in a horizontal circle so his acceleration is arad = v 2 / R, directed toward the center of the circle. The target variable is the coefficient of static friction between
⎛ 2π R ⎞
⎛ 2π (2.5 m) ⎞
the person and the surface of the cylinder. v = (0.60 rev/s) ⎜
⎟ = (0.60 rev/s) ⎜
⎟ = 9.425 m/s
⎝ 1 rev ⎠
⎝ 1 rev ⎠
(a) SET UP: The problem situation is sketched in Figure 5.111a. Figure 5.111a The free-body diagram for the person is
sketched in Figure 5.111b.
The person is held up against gravity by
the static friction force exerted on him
by the wall. The acceleration of the person
is arad , directed in towards the axis of rotation.
Figure 5.111b
(b) EXECUTE: ∑F y To calculate the minimum μs required, take f s to have its maximum value, f s = μs n. = ma y f s − mg = 0 μs n = mg ∑F x = max n = mv 2 / R
Combine these two equations to eliminate n:
μs mv 2 / R = mg μs = Rg (2.5 m)(9.80 m/s 2 )
=
= 0.28
v2
(9.425 m/s) 2 5-50 Chapter 5 5.112. v must be to keep the person from sliding down. For smaller μs the cylinder must rotate faster to make n larger
enough.
!
!
IDENTIFY: Apply ∑ F = ma to the combined object of motorcycle plus rider. (c) EVALUATE: No, the mass of the person divided out of the equation for μs . Also, the smaller μs is, the larger SET UP: The object has acceleration arad = v 2 / r , directed toward the center of the circular path.
EXECUTE: (a) For the tires not to lose contact, there must be a downward force on the tires. Thus, the
v2
(downward) acceleration at the top of the sphere must exceed mg, so m > mg , and
R 5.113. v > gR = (9.80 m s 2 ) (13.0 m) = 11.3 m s.
(b) The (upward) acceleration will then be 4g, so the upward normal force must be
5mg = 5(110 kg) (9.80 m s 2 ) = 5390 N.
EVALUATE: At any nonzero speed the normal force at the bottom of the path exceeds the weight of the object.
!
!
IDENTIFY: Apply ∑ F = ma to your friend. Your friend moves in the arc of a circle as the car turns.
(a) Turn to the right. The situation is sketched in Figure 5.113a. As viewed in an inertial frame,
in the absence of sufficient friction
your friend doesn’t make the turn
completely and you move to the right
toward your friend.
Figure 5.113a
(b) The maximum radius of the turn is the one that makes arad just equal to the maximum acceleration that static friction can give to your friend, and for this situation f s has its maximum value f s = μs n.
SET UP: The free-body diagram for your friend, as viewed by someone standing behind the car, is sketched in
Figure 5.113b.
EXECUTE:
∑ Fy = ma y n − mg = 0
n = mg
Figure 5.113b ∑F x = max f s = marad μs n = mv 2 / R
μs mg = mv 2 / R
R= 5.114. v2
(20 m/s) 2
=
= 120 m
μs g (0.35)(9.80 m/s 2 ) EVALUATE: The larger μs is, the smaller the radius R must be.
IDENTIFY: The tension F in the string must be the same as the weight of the hanging block, and must also
provide the resultant force necessary to keep the block on the table in uniform circular motion.
SET UP: The acceleration of the block is arad = v 2 / r , directed toward the hole. v2
, so v = gr M m.
r
EVALUATE: The larger M is the greater must be the speed v, if r remains the same.
EXECUTE: Mg = F = m Applying Newton’s Laws 5.115. 5-51 !
!
IDENTIFY: Apply ∑ F = ma to the circular motion of the bead. Also use Eq.(5.16) to relate arad to the period of
rotation T.
SET UP: The bead and hoop are sketched in Figure 5.115a. The bead moves in a circle of radius
R = r sin β .
The normal force exerted on the bead by
the hoop is radially inward. Figure 5.115a The free-body diagram for the bead is sketched in Figure 5.115b.
EXECUTE:
∑ Fy = ma y n cos β − mg = 0
n = mg / cos β ∑F x = max n sin β = marad
Figure 5.115b Combine these two equations to eliminate n:
⎛ mg ⎞
⎜
⎟ sin β = marad
⎝ cos β ⎠
sin β arad
=
cos β
g
arad = v 2 / R and v = 2π R / T , so arad = 4π 2 R / T 2 , where T is the time for one revolution.
R = r sin β , so arad = 4π 2 r sin β
T2 sin β 4π 2 r sin β
=
cos β
T 2g
This equation is satisfied by sin β = 0, so β = 0, or by
Use this in the above equation: 1
4π 2 r
T 2g
= 2 , which gives cos β = 2
cos β T g
4π r
(a) 4.00 rev/s implies T = (1/ 4.00) s = 0.250 s
(0.250 s) 2 (9.80 m/s 2 )
and β = 81.1°.
4π 2 (0.100 m)
(b) This would mean β = 90°. But cos90° = 0, so this requires T → 0. So β approaches 90° as the hoop rotates
very fast, but β = 90° is not possible.
(c) 1.00 rev/s implies T = 1.00 s
(1.00 s) 2 (9.80 m/s 2 )
T 2g
The cos β = 2 equation then says cos β =
= 2.48, which is not possible. The only way to
4π 2 (0.100 m)
4π r
!
!
have the ∑ F = ma equations satisfied is for sin β = 0. This means β = 0; the bead sits at the bottom of the hoop. Then cos β = 5-52 Chapter 5 β → 90° as T → 0 (hoop moves faster). The largest value T can have is given by T 2 g/(4π 2 r ) = 1 so EVALUATE: 5.116. T = 2π r / g = 0.635 s. This corresponds to a rotation rate of (1/ 0.635) rev/s = 1.58 rev/s. For a rotation rate less
than 1.58 rev/s, β = 0 is the only solution and the bead sits at the bottom of the hoop. Part (c) is an example of this.
!
d 2x
d2y
!
IDENTIFY: ax = 2 and a y = 2 . Then apply ∑ F = ma to calculate the components of the net force.
dt
dt
!
SET UP: The components of F determine its magnitude and direction.
EXECUTE: (a) Differentiating twice, ax = −6 βt and a y = −2δ , so
Fx = max = ( 2.20 kg) ( − 0.72 N s)t = −(1.58 N/s)t and Fy = ma y = (2.20 kg) ( − 2.00 m s 2 ) = −4.40 N .
(b) The graph is given in Figure 5.116.
(c) At t = 3.00 s, Fx = −4.75 N and Fy = −4.40 N, so F = (−4.75 N) 2 + (−4.40 N) 2 = 6.48 N at an angle of ( ) arctan −4.40 = 223°.
−4.75
EVALUATE: Fy is constant and negative. Fx is zero at t = 0 and becomes increasingly more negative as t
increases. Figure 5.116
5.117. 5.118. IDENTIFY: The velocity is tangent to the path. The acceleration has a tangential component when the speed is
changing and a radial component when the path is curving.
!
!
!
SET UP: arad is toward the center of curvature of the path. atan is parallel to v when the speed is increasing and
!
!
!
antiparallel to v when the speed is decreasing. The net force F is proportional to a .
EXECUTE: The diagram is sketched in Figure 5.117.
!
!!
EVALUATE: v , a , and F all change during the motion. Figure 5.117
!
!
!
IDENTIFY: Apply ∑ F = ma to the car. It has acceleration arad , directed toward the center of the circular path. The analysis is the same as in Example 5.24.
⎛
⎛
v2 ⎞
(12.0 m/s) 2 ⎞
EXECUTE: (a) FA = m ⎜ g + ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 +
⎟ = 61.8 N.
R⎠
5.00 m ⎠
⎝
⎝
SET UP: ⎛
⎛
v2 ⎞
(12.0 m/s) 2 ⎞
(b) FB = m ⎜ g − ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 −
⎟ = −30.4 N. , where the minus sign indicates that the track
R⎠
5.00 m ⎠
⎝
⎝
pushes down on the car. The magnitude of this force is 30.4 N.
EVALUATE: FA > FB . FA − 2mg . Applying Newton’s Laws 5.119. 5-53 IDENTIFY: The analysis is the same as for Problem 5.96.
SET UP: The speed is related to the period by v = 2π R T = 2π h(tan β ) / T , or T = 2π h(tan β ) / v .
EXECUTE: The maximum and minimum speeds are the same as those found in Problem 5.96, vmax = gh tan β cos β + μs sin β
cos β − μs sin β
and vmin = gh tan β
.
sin β − μs cos β
sin β + μs cos β The minimum and maximum values of the period T are then
Tmin = 2π R
. The result for v then
tan β
agrees with the result in Example 5.23, if we take into account that in this problem β is measured from the vertical
whereas in Example 5.23 it is measured relative to the horizontal.
!
!
IDENTIFY: Apply ∑ F = ma to the block and to the wedge.
EVALUATE: 5.120. h tan β sin β − μs cos β
h tan β sin β + μs cos β
and Tmax = 2π
.
g cos β + μs sin β
g cos β − μs sin β For μs = 0 the results for the speeds reduce to vmin = vmax = gh . h = SET UP: For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be
vertical and positive upward. The normal force between the block and the wedge is n; the normal force between the
wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The
horizontal acceleration of the wedge is A, and the components of acceleration of the block are ax and a y .
EXECUTE: (a) The equations of motion are then MA = − n sin α , max = n sin α and ma y = n cos α − mg . Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are
three equations in four unknowns, A, ax , a y and n. Solution is possible with the imposition of the relation between
A, ax and a y . An observer on the wedge is not in an inertial frame, and should not apply Newton’s laws, but the
kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical
acceleration of the block is a y , but the horizontal acceleration of the block is ax − A. To this observer, the block
descends at an angle α , so the relation needed is ay
ax − A = − tan α. At this point, algebra is unavoidable. A possible approach is to eliminate ax by noting that ax = − M
A , using this in the kinematic constraint to eliminate
m a y and then eliminating n. The results are:
A= − gm
( M + m) tanα + ( M tan α ) ax = gM
( M + m) tanα + ( M tan α ) ay = − g ( M + m) tan α
( M + m) tanα + ( M tan α ) (b) When M >> m, A → 0, as expected (the large block won’t move). Also,
g
tan α
ax →
=g
= g sin α cos α which is the acceleration of the block ( gsinα in this case),
tan α + (1 tan α )
tan 2α + 1 with the factor of cos α giving the horizontal component. Similarly, a y → − g sin 2 α .
(c) The trajectory is a spiral.
EVALUATE: If m >> M , our general results give ax = 0 and a y = − g . The massive block accelerates straight
5.121. downward, as if it were in free-fall.
!
!
IDENTIFY: Apply ∑ F = ma to the block and to the wedge.
SET UP: From Problem 5.120, max = n sin α and ma y = n cos α − mg for the block. a y = 0 gives ax = g tan α . EXECUTE: If the block is not to move vertically, both the block and the wedge have this horizontal acceleration
and the applied force must be F = ( M + m)a = ( M + m) gtanα .
EVALUATE: F → 0 as α → 0 and F → ∞ as α → ∞ . 5-54 Chapter 5 5.122. IDENTIFY: Apply ! ! ∑ F = ma . SET UP: Let + x be directed up the ramp.
EXECUTE: The normal force that the ramp exerts on the box will be n = wcosα − Tsinα . The rope provides a force of
Tcosθ up the ramp, and the component of the weight down the ramp is wsinα . Thus, the net force up the ramp is F = Tcosθ − w sin α − μ k ( w cos α − Tsinθ ) = T (cosθ + μ k sinθ ) − w(sin α + μ k cosα ) The acceleration will be the greatest when the first term in parentheses is greatest and this occurs when tan θ = μ k . 5.123. EVALUATE: Small θ means F is more nearly in the direction of the motion. But θ → 90° means F is directed to
reduce the normal force and thereby reduce friction. The optimum value of θ is somewhere in between and
depends on μ k . When μ k = 0 , the optimum value of θ is θ = 0° .
IDENTIFY: Use the results of Problem 5.44.
d2 f
df
>0.
SET UP: f ( x) is a minimum when
= 0 and
dx 2
dx
EXECUTE: (a) F = μ k w /(cosθ + μ k sinθ )
(b) The graph of F versus θ is given in Figure 5.123.
(c) F is minimized at tan θ = μ k . For μ k = 0.25 , θ = 14.0° .
EVALUATE: Small θ means F is more nearly in the direction of the motion. But θ → 90° means F is directed to
reduce the normal force and thereby reduce friction. The optimum value of θ is somewhere in between and
depends on μ k . 5.124. Figure 5.123
!
!
IDENTIFY: Apply ∑ F = ma to the ball. At the terminal speed, a = 0 .
SET UP: For convenience, take the positive direction to be down, so that for the baseball released from rest, the
acceleration and velocity will be positive, and the speed of the baseball is the same as its positive component of
velocity. Then the resisting force, directed against the velocity, is upward and hence negative.
EXECUTE: (a) The free-body diagram for the falling ball is sketched in Figure 5.124.
(b) Newton’s Second Law is then ma = mg − Dv 2 . Initially, when v = 0, the acceleration is g, and the speed
increases. As the speed increases, the resistive force increases and hence the acceleration decreases. This continues
as the speed approaches the terminal speed.
mg
(c) At terminal velocity, a = 0, so vt =
in agreement with Eq. (5.13).
D
dv g 2 2
= (vt − v ) . This is a separable equation and may be
(d) The equation of motion may be rewritten as
dt vt2 expressed as
EVALUATE: ⎛ v ⎞ gt
dv
g
1
= 2 ∫ dt or
arctanh ⎜ ⎟ = 2 . v = vt tanh ( gt vt ) .
2
−v
vt
vt
⎝ vt ⎠ vt
x
−x
e −e
tanh x = x
. At t → 0 , tanh( gt / vt ) → 0 and v → 0 . At t → ∞ , tanh( gt / vt ) → ∞ and v → vt .
e + e− x ∫v 2
t Figure 5.124 Applying Newton’s Laws 5.125. 5-55 !
!
IDENTIFY: Apply ∑ F = ma to each of the three masses and to the pulley B.
SET UP: Take all accelerations to be positive downward. The equations of motion are straightforward, but the
kinematic relations between the accelerations, and the resultant algebra, are not immediately obvious. If the
acceleration of pulley B is aB , then aB = −a3 , and aB is the average of the accelerations of masses 1 and 2, or a1 + a2 = 2aB = −2a3 .
EXECUTE: (a) There can be no net force on the massless pulley B, so TC = 2TA . The five equations to be solved are then m1 g − TA = m1a1 , m2 g − TA = m2 a2 , m3 g − TC = m3a3 , a1 + a2 + 2a3 = 0 and 2TA − TC = 0 . These are five
equations in five unknowns, and may be solved by standard means.
The accelerations a1 and a2 may be eliminated by using 2a3 = −(a1 + a2 ) = −(2 g − TA ((1 m1 ) + (1 m2 ))).
The tension TA may be eliminated by using TA = (1 2)TC = (1 2)m3 ( g − a3 ).
Combining and solving for a3 gives a3 = g −4m1m2 + m2 m3 + m1m3
.
4m1m2 + m2 m3 + m1m3 (b) The acceleration of the pulley B has the same magnitude as a3 and is in the opposite direction.
(c) a1 = g − a1 = g TA
T
m
= g − C = g − 3 ( g − a3 ). Substituting the above expression for a3 gives
m1
2m1
2m1 4m1m2 − 3m2 m3 + m1m3
.
4m1m2 + m2 m3 + m1m3 4m1m2 − 3m1m3 + m2 m3
.
4m1m2 + m2 m3 + m1m3
(e), (f) Once the accelerations are known, the tensions may be found by substitution into the appropriate equation
4m1m2 m3
8m1m2 m3
of motion, giving TA = g
, TC = g
.
4m1m2 + m2 m3 + m1m3
4m1m2 + m2 m3 + m1m3
(d) A similar analysis (or, interchanging the labels 1 and 2) gives a2 = g (g) If m1 = m2 = m and m3 = 2m, all of the accelerations are zero, TC = 2mg and TA = mg . All masses and pulleys
are in equilibrium, and the tensions are equal to the weights they support, which is what is expected.
EVALUATE: It is useful to consider special cases. For example, when m1 = m2 >> m3 our general result gives
5.126. a1 = a2 = + g and a3 = − g .
!
!
IDENTIFY: Apply ∑ F = ma to each block. The tension in the string is the same at both ends. If T < w for a block, that block remains at rest.
SET UP: In all cases, the tension in the string will be half of F.
EXECUTE: (a) F 2 = 62 N, which is insufficient to raise either block; a1 = a2 = 0.
(b) F 2 = 62 N. The larger block (of weight 196 N) will not move, so a1 = 0, but the smaller block, of weight 98 N, has a net upward force of 49 N applied to it, and so will accelerate upwards with a2 = 49 N
= 4.9 m s 2 .
10.0 kg (c) F 2 = 212 N, so the net upward force on block A is 16 N and that on block B is 114 N, so 16 N
114 N
= 0.8 m s 2 and a2 =
= 11.4 m s 2 .
20.0 kg
10.0 kg
EVALUATE: The two blocks need not have accelerations with the same magnitudes.
!
!
IDENTIFY: Apply ∑ F = ma to the ball at each position.
a1 = 5.127. SET UP: When the ball is at rest, a = 0 . When the ball is swinging in an arc it has acceleration component
v2
arad = , directed inward.
R
EXECUTE: Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension
force must balance the weight, so TA cos β = w or TA = w cos β . At point B, the ball is not in equilibrium; its speed
is instantaneously 0, so there is no radial acceleration, and the tension force must balance the radial component of
the weight, so TB = w cos β and the ratio (TB TA ) = cos 2 β .
EVALUATE: At point B the net force on the ball is not zero; the ball has a tangential acceleration. WORK AND KINETIC ENERGY 6.1. 6.2. 6 IDENTIFY: Apply Eq.(6.2).
SET UP: The bucket rises slowly, so the tension in the rope may be taken to be the bucket’s weight.
EXECUTE: (a) W = Fs = mgs = (6.75 kg) (9.80 m / s 2 )(4.00 m) = 265 J.
(b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq.(6.2) gives the negative of the result of
part (a), or −265 J .
(c) The total work done on the bucket is zero.
EVALUATE: When the force is in the direction of the displacement, the force does positive work. When the force is
directed opposite to the displacement, the force does negative work.
IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so W = Fs cos φ .
SET UP: In part (a), when the cable pulls horizontally φ = 0° and when it pulls at 35.0° above the horizontal
φ = 35.0° . In part (b), if the cable pulls horizontally φ = 180° . If the cable pulls on the car at 35.0° above the
horizontal it pulls on the truck at 35.0° below the horizontal and φ = 145.0° . For the gravity force φ = 90° , since the
force is vertical and the displacement is horizontal.
EXECUTE: (a) When the cable is horizontal, W = (850 N)(5.00 × 103 m)cos 0° = 4.25 × 106 J . When the cable is 35.0° above the horizontal, W = (850 N)(5.00 × 103 m)cos35.0° = 3.48 × 106 J . 6.3. (b) cos180° = − cos0° and cos145.0° = − cos35.0° , so the answers are −4.26 × 106 J and −3.48 × 106 J .
(c) Since cos φ = cos90° = 0 , W = 0 in both cases.
EVALUATE: If the car and truck are taken together as the system, the tension in the cable does no net work.
IDENTIFY: Each force can be used in the relation W = F! s = ( F cosφ ) s for parts (b) through (d). For part (e), apply the net work relation as Wnet = Wworker + Wgrav + Wn + W f .
SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of
friction, Fworker = f k = μ k n.
EXECUTE: (a) The magnitude of the force the worker must apply is: Fworker = f k = μ k n = μ k mg = ( 0.25 ) ( 30.0 kg ) ( 9.80 m/s 2 ) = 74 N
(b) Since the force applied by the worker is horizontal and in the direction of the displacement, φ = 0° and the
work is: Wworker = ( Fworker cos φ ) s = [( 74 N ) ( cos0° )] ( 4.5 m ) = +333 J
(c) Friction acts in the direction opposite of motion, thus φ = 180° and the work of friction is: W f = ( f k cos φ ) s = [( 74 N ) ( cos180° )] ( 4.5 m ) = −333 J
(d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any
work on the crate and Wgrav = Wn = 0.0 J.
(e) Substituting into the net work relation, the net work done on the crate is: Wnet = Wworker + Wgrav + Wn + W f = +333 J + 0.0 J + 0.0 J − 333 J = 0.0 J
EVALUATE:
6.4. The net work done on the crate is zero because the two contributing forces, Fworker and Ff , are equal in magnitude and opposite in direction.
IDENTIFY: The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies a = 0. We
"
"
must use ∑ F = ma applied to the crate to find the forces acting on it. 6-1 6-2 Chapter 6 (a) SET UP: The free-body diagram for the crate is given in Figure 6.4. EXECUTE: ∑F y = ma y n − mg − F sin 30° = 0
n = mg + F sin 30°
f k = μ k n = μ k mg + F μ k sin 30°
Figure 6.4 ∑F x = max F cos30° − f k = 0
F cos30° − μ k mg − μ k sin 30° F = 0 μk mg
0.25(30.0 kg)(9.80 m/s 2 )
=
= 99.2 N
cos30° − μ k sin 30°
cos30° − (0.25)sin 30°
(b) WF = ( F cos φ ) s = (99.2 N)(cos30°)(4.5 m) = 387 J
F= "
"
"
( F cos30° is the horizontal component of F ; the work done by F is the displacement times the component of F
in the direction of the displacement.)
(c) We have an expression for f k from part (a):
f k = μ k ( mg + F sin 30°) = (0.250)[(30.0 kg)(9.80 m/s 2 ) + (99.2 N)(sin 30°)] = 85.9 N φ = 180° since f k is opposite to the displacement. Thus W f = ( f k cos φ ) s = (85.9 N)(cos180°)(4.5 m) = −387 J
(d) The normal force is perpendicular to the displacement so φ = 90° and Wn = 0. The gravity force (the weight) is perpendicular to the displacement so φ = 90° and Ww = 0
(e) Wtot = WF + W f + Wn + Ww = +387 J + (−387 J) = 0
EVALUATE: Forces with a component in the direction of the displacement do positive work, forces opposite to the
displacement do negative work and forces perpendicular to the displacement do zero work. The total work, obtained
as the sum of the work done by each force, equals the work done by the net force. In this problem, Fnet = 0 since
6.5. 6.6. a = 0 and Wtot = 0, which agrees with the sum calculated in part (e).
IDENTIFY: The gravity force is constant and the displacement is along a straight line, so W = Fs cosφ .
SET UP: The displacement is upward along the ladder and the gravity force is downward, so
φ = 180.0° − 30.0° = 150.0° . w = mg = 735 N .
EXECUTE: (a) W = (735 N)(2.75 m)cos150.0° = −1750 J .
(b) No, the gravity force is independent of the motion of the painter.
EVALUATE: Gravity is downward and the vertical component of the displacement is upward, so the gravity force
does negative work.
IDENTIFY and SET UP: WF = ( F cos φ ) s, since the forces are constant. We can calculate the total work by summing
the work done by each force. The forces are sketched in Figure 6.6.
EXECUTE: W1 = F1s cos φ1 W1 = (1.80 × 106 N)(0.75 × 103 m)cos14°
W1 = 1.31 × 109 J
W2 = F2 s cos φ2 = W1
Figure 6.6 Wtot = W1 + W2 = 2(1.31× 109 J) = 2.62 × 109 J
EVALUATE: Only the component F cos φ of force in the direction of the displacement does work. These
"
components are in the direction of s so the forces do positive work. Work and Kinetic Energy 6.7. 6-3 IDENTIFY: All forces are constant and each block moves in a straight line. so W = Fs cosφ . The only direction the
system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right.
SET UP: Since the 12.0 N block moves at constant speed, a = 0 for it and the tension T in the string is T = 12.0 N .
Since the 20.0 N block moves to the right at constant speed the friction force f k on it is to the left and
f k = T = 12.0 N .
EXECUTE: (a) (i) φ = 0° and W = (12.0 N)(0.750 m)cos0° = 9.00 J . (ii) φ = 180° and W = (12.0 N)(0.750 m)cos180° = −9.00 J .
(b) (i) φ = 90° and W = 0 . (ii) φ = 0° and W = (12.0 N)(0.750 m)cos0° = 9.00 J . (iii) φ = 180° and
W = (12.0 N)(0.750 m)cos180° = −9.00 J . (iv) φ = 90° and W = 0 . 6.8. 6.9. (c) Wtot = 0 for each block.
EVALUATE: For each block there are two forces that do work, and for each block the two forces do work of equal
magnitude and opposite sign. When the force and displacement are in opposite directions, the work done is
negative.
IDENTIFY: Apply Eq.(6.5).
ˆˆ jj
ˆj jˆ
SET UP: i ⋅ i = ˆ ⋅ ˆ = 1 and i ⋅ ˆ = ˆ ⋅ i = 0
""
ˆ
ˆ
EXECUTE: The work you do is F ⋅ s = ((30 N)i − (40 N) ˆ) ⋅ ((−9.0 m)i − (3.0 m) ˆ)
j
j
""
F ⋅ s = (30 N)(−9.0 m) + (−40 N)( −3.0 m) = −270 N ⋅ m + 120 N ⋅ m = −150 J .
"
"
EVALUATE: The x-component of F does negative work and the y-component of F does positive work. The total
"
work done by F is the sum of the work done by each of its components.
IDENTIFY: Apply Eq.(6.2) or (6.3).
""
SET UP: The gravity force is in the − y -direction , so Fmg ⋅ s = − mg ( y2 − y1 )
EXECUTE: (a) (i) Tension force is always perpendicular to the displacement and does no work.
(ii) Work done by gravity is − mg ( y2 − y1 ). When y1 = y2 , Wmg = 0 .
(b) (i) Tension does no work. (ii) Let l be the length of the string. Wmg = − mg ( y2 − y1 ) = −mg (2l ) = −25.1 J 6.10. EVALUATE: In part (b) the displacement is upward and the gravity force is downward, so the gravity force does
negative work.
IDENTIFY: K = 1 mv 2
2
SET UP: 65 mi/h = 29.1 m/s
EXECUTE: (a) K = 1 (750 kg)(29.1 m/s) 2 = 3.18 × 105 J
2
2
(b) K1 = 1 mv12 . K 2 = 1 mv2 , with v2 = v1 / 2 , so K 2 = 1 m(v1 / 2) 2 = 1 ( 1 mv12 ) = K1 / 4 . The change in kinetic energy is a
2
2
2
42 decrease of 3
4 K1 . (c) K 2 = 1 K1 .
2 6.11. K
K
Km
= = constant , so 21 = 22 . v2 = v1 K 2 / K1 = (65 mi/h)
v1
v2
v2 2 1
2 K1 / K1 = 46 mi/h . EVALUATE: Since K ∼ v 2 , to have half the kinetic energy the speed must be less than half of the original speed.
IDENTIFY: K = 1 mv 2 . Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic
2
energy.
SET UP: v = 12 km/s = 1.2 × 104 m/s . A 1.0 megaton bomb releases 4.184 × 1015 J of energy.
EXECUTE: (a) K = 1 (1.4 × 108 kg)(1.2 × 104 m/s) 2 = 1.0 × 1016 J .
2 1.0 × 1016 J
= 2.4 . The energy is equivalent to 2.4 one-megaton bombs.
4.184 × 1015 J
EVALUATE: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large
crater.
IDENTIFY: K = 1 mv 2 . Use the equations for free-fall to find the speed of the weight when it reaches the ground.
2
SET UP: Estimate that a person has speed 2 m/s when walking and 6 m/s when running. The mass of an electron is
9.11 × 10−31 kg . In part (c) take + y downward, so a y = +9.80 m/s 2 . Estimate a shoulder height of 1.6 m.
(b) 6.12. EXECUTE: (a) Walking: K = 1 (75 kg)(2 m/s) 2 = 150 J . Running: K = 1 (75 kg)(6 m/s) 2 = 1400 J .
2
2 (b) K = 1 (9.11 × 10−31 kg)(2.19 × 106 m/s) 2 = 2.2 × 10−18 J .
2
2
2
(c) v y = v0 y + 2a y ( y − y0 ) gives v y = 2(9.80 m/s 2 )(1.6 m) = 5.6 m/s . K = 1 (1.0 kg)(5.6 m/s) 2 = 16 J .
2 6-4 Chapter 6 (d) v = 6.13. 2K
2(100 J)
=
= 2.6 m/s . Yes, this is reasonable.
30 kg
m EVALUATE: A walking speed of 2 m/s corresponds to walking a mile in about 13 min. A running speed of 6 m/s
corresponds to running a 100 m dash in about 17 s.
IDENTIFY: K = 1 mv 2 . Set up a ratio that relates K, m and v.
2
SET UP:
EXECUTE: mp = 1836me
2
(a) K p = K e gives meve2 = mpvp . ve = vp mp / me = V 1836 = 42.85V . (b) vp = ve gives 6.14. Kp
mp = Ke
. K p = K e ( mp / me ) = 1836 K .
me EVALUATE: The electron has less mass so must travel faster to have the same kinetic energy. And with equal speeds
the proton has more kinetic energy.
IDENTIFY: Only gravity does work on the watermelon, so Wtot = Wgrav . Wtot = ΔK and K = 1 mv 2 .
2
SET UP:
EXECUTE: Since the watermelon is dropped from rest, K1 = 0 .
(a) Wgrav = mgs = (4.80 kg)(9.80 m/s 2 )(25.0 m) = 1180 J (b) Wtot = K 2 − K1 so K 2 = 1180 J . v = 2K2
2(1180 J)
=
= 22.2 m/s .
4.80 kg
m (c) The work done by gravity would be the same. Air resistance would do negative work and Wtot would be less than Wgrav . The answer in (a) would be unchanged and both answers in (b) would decrease.
6.15. EVALUATE: The gravity force is downward and the displacement is downward, so gravity does positive work.
IDENTIFY: Wtot = K 2 − K1 . In each case calculate Wtot from what we know about the force and the displacement.
SET UP: The gravity force is mg, downward. The friction force is f k = μ k n = μ k mg and is directed opposite to the
displacement. The mass of the object isn't given, so we expect that it will divide out in the calculation.
EXECUTE: 2
(a) K1 = 0 . Wtot = Wgrav = mgs . mgs = 1 mv2 and v2 = 2 gs = 2(9.80 m/s 2 )(95.0 m) = 43.2 m/s .
2 (b) K 2 = 0 (at the maximum height). Wtot = Wgrav = − mgs . − mgs = − 1 mv12 and
2 v1 = 2 gs = 2(9.80 m/s 2 )(525 m) = 101 m/s .
(c) K1 = 1 mv12 . K 2 = 0 . Wtot = W f = − μ k mgs . − μ k mgs = − 1 mv12 . s =
2
2
2
(d) K1 = 1 mv12 . K 2 = 1 mv2 . Wtot = W f = − μ k mgs . K 2 = Wtot + K1 .
2
2 1
2 v12 2μk g = (5.00 m/s) 2
= 5.80 m .
2(0.220)(9.80 m/s 2 ) 2
mv2 = − μ k mgs + 1 mv12
2 v2 = v12 − 2 μk gs = (5.00 m/s) 2 − 2(0.220)(9.80 m/s 2 )(2.90 m) = 3.53 m/s .
(e) K1 = 1 mv12 . K 2 = 0 . Wgrav = −mgy2 , where y2 is the vertical height. − mgy2 = − 1 mv12 and
2
2 (12.0 m/s) 2
v12
=
= 7.35 m .
2 g 2(9.80 m/s 2 )
EVALUATE: In parts (c) and (d), friction does negative work and the kinetic energy is reduced. In part (a), gravity
does positive work and the speed increases. In parts (b) and (e), gravity does negative work and the speed decreases.
The vertical height in part (e) is independent of the slope angle of the hill.
IDENTIFY: From the work-energy relation, W = Wgrav = ΔK rock .
y2 = 6.16. SET UP: As the rock rises, the gravitational force, F = mg , does work on the rock. Since this force acts in the
direction opposite to the motion and displacement, s, the work is negative. Let h be the vertical distance the rock
travels.
2
EXECUTE: (a) Applying Wgrav = K 2 − K1 we obtain − mgh = 1 mv2 − 1 mv12 . Dividing by m and solving for v1 ,
2
2
2
v1 = v2 + 2 gh . Substituting h = 15.0 m and v2 = 25.0 m/s,
2
v1 = ( 25.0 m/s ) + 2 ( 9.80 m/s 2 ) (15.0 m ) = 30.3 m/s (b) Solve the same work-energy relation for h. At the maximum height v2 = 0 .
2
− mgh = 1 mv2 − 1 mv12 and h =
2
2 2
2
2
v12 − v2 ( 30.3 m/s ) − ( 0.0 m/s )
=
= 46.8 m .
2
2g
2 ( 9.80 m/s ) Work and Kinetic Energy 6.17. 6-5 EVALUATE: Note that the weight of 20 N was never used in the calculations because both gravitational potential
and kinetic energy are proportional to mass, m. Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have
an initial velocity of 30.3 m/s. As the rock moves upward gravity does negative work and this reduces the kinetic
energy of the rock.
IDENTIFY and SET UP: Apply Eq.(6.6) to the box. Let point 1 be at the bottom of the incline and let point 2 be at
the skier. Work is done by gravity and by friction. Solve for K1 and from that obtain the required initial speed.
EXECUTE: Wtot = K 2 − K1 K1 = mv , K 2 = 0
1
2 2
0 Work is done by gravity and friction, so Wtot = Wmg + W f .
Wmg = − mg ( y2 − y1 ) = −mgh
W f = − fs. The normal force is n = mg cos α and s = h / sin α , where s is the distance the box travels along the
incline.
W f = −( μk mg cos α )(h / sin α ) = − μk mgh / tan α
Substituting these expressions into the work-energy theorem gives
2
− mgh − μk mgh / tan α = − 1 mv0 .
2
Solving for v0 then gives v0 = 2 gh(1 + μk / tan α ). 6.18. EVALUATE: The result is independent of the mass of the box. As α → 90°, h = s and v0 = 2 gh , the same as
throwing the box straight up into the air. For α = 90° the normal force is zero so there is no friction.
IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK . Parallel to incline: force component W! = mg sin α , down incline; displacement s = h/ sin α , down incline. SET UP: Perpendicular to the incline: s = 0 .
EXECUTE: (a) W|| = (mg sin α )(h / sin α ) = mgh . W⊥ = 0 , since there is no displacement in this direction.
Wmg = W|| + W⊥ = mgh , same as falling height h. 6.19. (b) Wtot = K 2 − K1 gives mgh = 1 mv 2 and v = 2 gh , same as if had been dropped from height h. The work done by
2
gravity depends only on the vertical displacement of the object. When the slope angle is small, there is a small force
component in the direction of the displacement but a large displacement in this direction. When the slope angle is
large, the force component in the direction of the displacement along the incline is larger but the displacement in this
direction is smaller.
(c) h = 15.0 m , so v = 2 gh = 17.1 s .
EVALUATE: The acceleration and time of travel are different for an object sliding down an incline and an object in
free-fall, but the final velocity is the same in these two cases.
IDENTIFY: Wtot = K 2 − K1 with Wtot = W f . The car stops, so K 2 = 0 . In each case identify what is constant and set up a ratio.
2
SET UP: W f = − fs , so − fs = − 1 mv0 .
2
EXECUTE: 6.20. (a) v0b = 3v0 a . sa = D . f is constant. 2 2
⎛v ⎞
v2
v2
v0 2 f
=
= constant , so 0 a = 0b . sb = sa ⎜ 0b ⎟ = D(3) 2 = 9 D .
sa
sb
s
m
⎝ v0 a ⎠ ⎛f ⎞
2
(b) f b = 3 f a . v0 is constant. fs = 1 mv0 = constant , so f a sa = fb sb . sb = sa ⎜ a ⎟ = D / 3 .
2
⎝ fb ⎠
EVALUATE: The stopping distance is proportional to the square of the initial speed. When the friction force
increases, the stopping distance decreases.
IDENTIFY and SET UP: Apply Eq.(6.6). The relation between the speeds v1 and v2 tells us the relation between K1
and K 2 .
EXECUTE: (a) W = K 2 − K1 2
K1 = 1 mv , K 2 = 1 mv2
2
2
2
1 1
1
v2 = 1 v1 gives that K 2 = 1 m ( 1 v1 ) = 16 ( 1 mv12 ) = 16 K1
2
4
2
4
2 15
1
W = K 2 − K1 = 16 K1 − K1 = − 16 K1
"
(b) EVALUATE: K depends only on the magnitude of v not on its direction, so the answer for W in part (a) does not
depend on the final direction of the electron’s motion. The electron slows down, so its kinetic energy decreases and
the total work done on it is negative. 6-6 6.21. Chapter 6 IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK . SET UP: φ = 0°
EXECUTE: From Equations (6.1), (6.5) and (6.6), and solving for F, F= 6.22. 2
ΔK 1 m(v2 − v12 ) 1 (8.00 kg)((6.00 m / s) 2 − (4.00 m / s) 2 )
=2
=2
= 32.0 N.
(2.50 m)
s
s EVALUATE: The force is in the direction of the displacement, so the force does positive work and the kinetic energy
of the object increases.
IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work done by the foot on the ball. Then use Eq.(6.2) to find the
distance over which this force acts.
EXECUTE: Wtot = K 2 − K1 K1 = 1 mv12 = 1 (0.420 kg)(2.00 m/s) 2 = 0.84 J
2
2
2
K 2 = 1 mv2 = 1 (0.420 kg)(6.00 m/s) 2 = 7.56 J
2
2 Wtot = K 2 − K1 = 7.56 J − 0.84 J = 6.72 J The 40.0 N force is the only force doing work on the ball, so it must do 6.72 J of work. WF = ( F cosφ ) s gives that
s= 6.23. W
6.72 J
=
= 0.168 m
F cos φ (40.0 N)(cos0) EVALUATE: The force is in the direction of the motion so positive work is done and this is consistent with an
increase in kinetic energy.
IDENTIFY: Apply Wtot = ΔK .
SET UP: v1 = 0 , v2 = v . f k = μk mg and f k does negative work. The force F = 36.0 N is in the direction of the
motion and does positive work.
EXECUTE: (a) If there is no work done by friction, the final kinetic energy is the work done by the applied force,
and solving for the speed, v= 2W
2 Fs
2(36.0 N)(1.20 m)
=
=
= 4.48 m / s.
(4.30 kg)
m
m (b) The net work is Fs − f k s = ( F − μk mg ) s , so v= 6.24. EVALUATE: The total work done is larger in the absence of friction and the final speed is larger in that case.
IDENTIFY: Apply W = Fs cos φ and Wtot = ΔK
SET UP: The gravity force has magnitude mg and is directed downward.
EXECUTE: (a) On the way up, gravity is opposed to the direction of motion, and so
W = − mgs = −(0.145 kg)(9.80 m / s 2 )(20.0 m) = −28.4 J .
(b) v2 = v12 + 2 6.25. 2( F − μ k mg ) s
2(36.0 N − (0.30)(4.30 kg)(9.80 m / s 2 ))(1.20 m)
=
= 3.61 m/s
m
(4.30 kg) W
2(−28.4 J)
= (25.0 m / s) 2 +
= 15.3 m / s .
m
(0.145 kg) (c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the
way down, gravity will have done both negative and positive work on the ball, but the net work at this height will be
the same.
EVALUATE: As the baseball moves upward, gravity does negative work and the speed of the baseball decreases.
(a) IDENTIFY and SET UP: Use Eq.(6.2) to find the work done by the positive force. Then use Eq.(6.6) to find the
2
final kinetic energy, and then K 2 = 1 mv2 gives the final speed.
2
EXECUTE: Wtot = K 2 − K1 , so K 2 = Wtot + K1 K1 = mv = 1 (7.00 kg)(4.00 m/s) 2 = 56.0 J
2
The only force that does work on the wagon is the 10.0 N force. This force is in the direction of the displacement so
φ = 0° and the force does positive work:
1
2 2
1 WF = ( F cos φ ) s = (10.0 N)(cos0)(3.0 m) = 30.0 J
Then K 2 = Wtot + K1 = 30.0 J + 56.0 J = 86.0 J.
2
K 2 = 1 mv2 ; v2 =
2 2K2
2(86.0 J)
=
= 4.96 m/s
m
7.00 kg Work and Kinetic Energy (b) IDENTIFY: 6-7 "
"
Apply ∑ F = ma to the wagon to calculate a. Then use a constant acceleration equation to calculate the final speed. The free-body diagram is given in Figure 6.25.
SET UP:
EXECUTE: ∑F x = max F = max
ax = F 10.0 N
=
= 1.43 m/s 2
m 7.00 kg Figure 6.25
2
v2 x = v12x + 2a2 ( x − x0 ) v2 x = v12x + 2ax ( x − x0 ) = (4.00 m/s) 2 + 2(1.43 m/s 2 )(3.0 m) = 4.96 m/s 6.26. EVALUATE: This agrees with the result calculated in part (a). The force in the direction of the motion does positive
work and the kinetic energy and speed increase. In part (b), the equivalent statement is that the force produces an
acceleration in the direction of the velocity and this causes the magnitude of the velocity to increase.
IDENTIFY: Apply Wtot = K 2 − K1 .
SET UP: K1 = 0 . The normal force does no work. The work W done by gravity is W = mgh , where h = L sin θ is
the vertical distance the block has dropped when it has traveled a distance L down the incline and θ is the angle the
plane makes with the horizontal.
2K
2W
EXECUTE: The work-energy theorem gives v =
=
= 2 gh = 2 gL sin θ . Using the given numbers,
m
m 6.27. v = 2(9.80 m / s 2 )(0.75 m)sin 36.9° = 2.97 m / s.
EVALUATE: The final speed of the block is the same as if it had been dropped from a height h.
IDENTIFY: Wtot = K 2 − K1 . Only friction does work.
SET UP:
EXECUTE: 2
Wtot = W fk = − μ k mgs . K 2 = 0 (car stops). K1 = 1 mv0 .
2
2
(a) Wtot = K 2 − K1 gives − μ k mgs = − 1 mv0 . s =
2 (b) (i) μ kb = 2 μ ka . sμ k = 2
v0 2μk g . 2
⎛μ ⎞
v0
= constant so sa μ ka = sb μ kb . sb = ⎜ ka ⎟ sa = sa / 2 . The minimum stopping distance
2g
⎝ μ kb ⎠
2 would be halved. (ii) v0b = 2v0 a . ⎛v ⎞
s
1
s
sb
=
= constant , so 2a = 2 . sb = sa ⎜ 0b ⎟ = 4 sa . The stopping distance
2
v0 2 μ k g
v0 a v0b
⎝ v0 a ⎠ would become 4 times as great. (iii) v0b = 2v0 a , μ kb = 2 μ ka . sμk
1
sμ
sμ
=
= constant , so a 2 ka = b 2 kb .
2
v0
2g
v0 a
v0b 2 6.28. ⎛ μ ⎞⎛ v ⎞
⎛1⎞
sb = sa ⎜ ka ⎟⎜ 0b ⎟ = sa ⎜ ⎟ (2) 2 = 2sa . The stopping distance would double.
⎝2⎠
⎝ μkb ⎠⎝ v0 a ⎠
EVALUATE: The stopping distance is directly proportional to the square of the initial speed and indirectly
proportional to the coefficient of kinetic friction.
2
IDENTIFY: The work that must be done to move the end of a spring from x1 to x2 is W = 1 kx2 − 1 kx12 . The force
2
2
required to hold the end of the spring at displacement x is Fx = kx .
SET UP: When the spring is at its unstretched length, x = 0 . When the spring is stretched, x > 0 , and when the
spring is compressed, x < 0 .
2W
2(12.0 J)
2
= 2.67 × 104 N/m .
EXECUTE: (a) x1 = 0 and W = 1 kx2 . k = 2 =
2
x2
(0.0300 m) 2
(b) Fx = kx = (2.67 × 104 N/m)(0.0300 m) = 801 N .
(c) x1 = 0 , x2 = −0.0400 m . W = 1 (2.67 × 104 N/m)( −0.0400 m) 2 = 21.4 J .
2 Fx = kx = (2.67 × 104 N/m)(0.0400 m) = 1070 N .
EVALUATE: When a spring, initially unstretched, is either compressed or stretched, positive work is done by the
force that moves the end of the spring. 6-8 6.29. Chapter 6 IDENTIFY and SET UP: Use Eq.(6.8) to calculate k for the spring. Then Eq.(6.10), with x1 = 0, can be used to calculate the work done to stretch or compress the spring an amount x2 .
EXECUTE: Use the information given to calculate the force constant of the spring.
F
160 N
Fx = kx gives k = x =
= 3200 N/m
x 0.050 m
(a) Fx = kx = (3200 N/m)(0.015 m) = 48 N
Fx = kx = (3200 N/m)( − 0.020 m) = −64 N (magnitude 64 N)
(b) W = 1 kx 2 = 1 (3200 N/m)(0.015 m) 2 = 0.36 J
2
2 6.30. 6.31. W = 1 kx 2 = 1 (3200 N/m)( − 0.020 m) 2 = 0.64 J
2
2
Note that in each case the work done is positive.
EVALUATE: The force is not constant during the displacement so Eq.(6.2) cannot be used. A force in the + x
direction is required to stretch the spring and a force in the opposite direction to compress it. The force Fx is in the
same direction as the displacement, so positive work is done in both cases.
IDENTIFY: The magnitude of the work can be found by finding the area under the graph.
SET UP: The area under each triangle is 1/2 base × height . Fx > 0 , so the work done is positive when x increases
during the displacement.
EXECUTE: (a) 1 / 2 (8 m)(10 N) = 40 J .
(b) 1 / 2 (4 m)(10 N) = 20 J .
(c) 1 / 2 (12 m)(10 N) = 60 J .
EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c).
IDENTIFY: Use the work-energy theorem and the results of Problem 6.30.
SET UP: For x = 0 to x = 8.0 m , Wtot = 40 J . For x = 0 to x = 12.0 m , Wtot = 60 J .
EXECUTE: (a) v = (2)(40 J)
= 2.83 m / s
10 kg (2)(60 J)
= 3.46 m / s .
10 kg
"
"
EVALUATE: F is always in the + x -direction. For this motion F does positive work and the speed continually
increases during the motion.
(b) v = 6.32. IDENTIFY: x2 The force has only an x-component and the motion is along the x-direction, so W = ∫ Fx dx .
x1 SET UP: x1 = 0 and x2 = 6.9 m .
EXECUTE: The work you do with your changing force is
x2 x2 x2 x1 x1 x1 W = ∫ F ( x)dx = ∫ (−20.0 N) dx − ∫ (3.0 N/m)xdx = (−20.0 N) x |x12 −(3.0 N/m)( x 2 /2) |x12
x
x 6.33. W = −138 N ⋅ m − 71.4 N ⋅ m = −209 J .
EVALUATE: The work is negative because the cow continues to move forward (in the + x -direction ) as you vainly
attempt to push her backward.
IDENTIFY: Apply Eq.(6.6) to the box.
SET UP: Let point 1 be just before the box reaches the end of the spring and let point 2 be where the spring has
maximum compression and the box has momentarily come to rest.
EXECUTE: Wtot = K 2 − K1
2
K1 = 1 mv0 , K 2 = 0
2
2
Work is done by the spring force. Wtot = − 1 kx2 , where x2 is the amount the spring is compressed.
2
2
2
− 1 kx2 = − 1 mv0 and x2 = v0 m / k = (3.0 m/s) (6.0 kg)/(7500 N/m) = 8.5 cm
2
2 6.34. EVALUATE: The compression of the spring increases when either v0 or m increases and decreases when k increases
(stiffer spring).
IDENTIFY: The force applied to the springs is Fx = kx . The work done on a spring to move its end from x1 to x2 is
2
W = 1 kx2 − 1 kx12 . Use the information that is given to calculate k.
2
2 SET UP: When the springs are compressed 0.200 m from their uncompressed length, x1 = 0 and x2 = −0.200 m . When the platform is moved 0.200 m farther, x2 becomes −0.400 m . Work and Kinetic Energy 6-9 2W
2(80.0 J)
=
= 4000 N/m . Fx = kx = (4000 N/m)(−0.200 m) = −800 N . The
2
x − x1 (0.200 m) 2 − 0
magnitude of force that is required is 800 N.
(b) To compress the springs from x1 = 0 to x2 = −0.400 m , the work required is
(a) k = EVALUATE: 2
2 2
W = 1 kx2 − 1 kx12 = 1 (4000 N/m)(−0.400 m) 2 = 320 J . The additional work required is 320 J − 80 J = 240 J . For
2
2
2 6.35. x = −0.400 m , Fx = kx = −1600 N . The magnitude of force required is 1600 N.
EVALUATE: More work is required to move the end of the spring from x = −0.200 m to x = −0.400 m than to move
it from x = 0 to x = −0.200 m , even though the displacement of the platform is the same in each case. The magnitude
of the force increases as the compression of the spring increases.
"
"
IDENTIFY: Apply ∑ F = ma to calculate the μs required for the static friction force to equal the spring force.
SET UP: (a) The free-body diagram for the glider is given in Figure 6.35.
EXECUTE: ∑F y = ma y n − mg = 0
n = mg
f s = μs mg
Figure 6.35 ∑F x = max f s − Fspring = 0 μs mg − kd = 0
μs = kd
(20.0 N/m)(0.086 m)
=
= 1.76
mg (0.100 kg)(9.80 m/s 2 ) (b) IDENTIFY and SET UP: Apply " " ∑ F = ma to find the maximum amount the spring can be compressed and still have the spring force balanced by friction. Then use Wtot = K 2 − K1 to find the initial speed that results in this
compression of the spring when the glider stops.
EXECUTE: μs mg = kd μs mg (0.60)(0.100 kg)(9.80 m/s 2 )
= 0.0294 m
20.0 N/m
k
Now apply the work-energy theorem to the motion of the glider:
Wtot = K 2 − K1
d= = K1 = 1 mv12 , K 2 = 0 (instantaneously stops)
2
Wtot = Wspring + Wfric = − 1 kd 2 − μ k mgd (as in Example 6.8)
2
Wtot = − 1 (20.0 N/m)(0.0294 m) 2 − 0.47(0.100 kg)(9.80 m/s 2 )(0.0294 m) = −0.02218 J
2
Then Wtot = K 2 − K1 gives −0.02218 J = − 1 mv12 .
2
v1 = 6.36. 2(0.02218 J)
= 0.67 m/s
0.100 kg EVALUATE: In Example 6.8 an initial speed of 1.50 m/s compresses the spring 0.086 m and in part (a) of this
problem we found that the glider doesn’t stay at rest. In part (b) we found that a smaller displacement of 0.0294 m
when the glider stops is required if it is to stay at rest. And we calculate a smaller initial speed (0.67 m/s) to produce
this smaller displacement.
2
IDENTIFY: For the spring, W = 1 kx12 − 1 kx2 . Apply Wtot = K 2 − K1 .
2
2
SET UP:
EXECUTE: x1 = −0.025 m and x2 = 0 .
(a) W = 1 kx12 = 1 (200 N / m)(−0.025 m) 2 = 0.060 J .
2
2 2W
2(0.060 J)
=
= 0.18 m / s.
m
(4.0 kg)
EVALUATE: The block moves in the direction of the spring force, the spring does positive work and the kinetic
energy of the block increases.
(b) The work-energy theorem gives v2 = 6-10 6.37. Chapter 6 IDENTIFY and SET UP: The magnitude of the work done by Fx equals the area under the Fx versus x curve. The work is positive when Fx and the displacement are in the same direction; it is negative when they are in opposite
directions.
EXECUTE: (a) Fx is positive and the displacement Δx is positive, so W > 0.
W = 1 (2.0 N)(2.0 m) + (2.0 N)(1.0 m) = +4.0 J
2
(b) During this displacement Fx = 0, so W = 0. 6.38. (c) Fx is negative, Δx is positive, so W < 0. W = − 1 (1.0 N)(2.0 m) = −1.0 J
2
(d) The work is the sum of the answers to parts (a), (b), and (c), so W = 4.0 J + 0 − 1.0 J = +3.0 J
(e) The work done for x = 7.0 m to x = 3.0 m is +1.0 J. This work is positive since the displacement and the force
are both in the − x-direction. The magnitude of the work done for x = 3.0 m to x = 2.0 m is 2.0 J, the area under Fx
versus x. This work is negative since the displacement is in the − x -direction and the force is in the + x -direction.
Thus W = +1.0 J − 2.0 J = −1.0 J
EVALUATE: The work done when the car moves from x = 2.0 m to x = 0 is − 1 (2.0 N)(2.0 m) = −2.0 J. Adding
2
this to the work for x = 7.0 m to x = 2.0 m gives a total of W = −3.0 J for x = 7.0 m to x = 0. The work for
x = 7.0 m to x = 0 is the negative of the work for x = 0 to x = 7.0 m.
IDENTIFY: Apply Wtot = K 2 − K1 .
SET UP: K1 = 0 . From Exercise 6.37, the work for x = 0 to x = 3.0 m is 4.0 J. W for x = 0 to x = 4.0 m is also
4.0 J. For x = 0 to x = 7.0 m , W = 3.0 J .
EXECUTE: (a) K = 4.0 J , so v = 2 K m = 2(4.0 J) (2.0 kg) = 2.00 m / s .
(b) No work is done between x = 3.0 m and x = 4.0 m , so the speed is the same, 2.00 m/s. 6.39. (c) K = 3.0 J , so v = 2 K / m = 2(3.0 J) /(2.0 kg) = 1.73 m / s .
EVALUATE: In each case the work done by F is positive and the car gains kinetic energy.
IDENTIFY and SET UP: Apply Eq.(6.6). Let point 1 be where the sled is released and point 2 be at x = 0 for part (a)
2
and at x = −0.200 m for part (b). Use Eq.(6.10) for the work done by the spring and calculate K 2 . Then K 2 = 1 mv2
2 gives v2 .
EXECUTE: (a) Wtot = K 2 − K1 so K 2 = K1 + Wtot 2
K1 = 0 (released with no initial velocity), K 2 = 1 mv2
2 The only force doing work is the spring force. Eq.(6.10) gives the work done on the spring to move its end from x1 to
x2 . The force the spring exerts on an object attached to it is F = − kx , so the work the spring does is 2
2
Wspr = − ( 1 kx2 − 1 kx12 ) = 1 kx12 − 1 kx2 . Here x1 = −0.375 m and x2 = 0. Thus Wspr = 1 (4000 N/m)( − 0.375 m) 2 − 0 = 281 J.
2
2
2
2
2 K 2 = K1 + Wtot = 0 + 281 J = 281 J
2
Then K 2 = 1 mv2 implies v2 =
2 2K2
2(281 J)
=
= 2.83 m/s.
m
70.0 kg (b) K 2 = K1 + Wtot K1 = 0
2
Wtot = Wspr = 1 kx12 − 1 kx2 . Now x2 = 0.200 m, so
2
2 Wspr = 1 (4000 N/m)( − 0.375 m) 2 − 1 (4000 N/m)( − 0.200 m) 2 = 281 J − 80 J = 201 J
2
2
2
Thus K 2 = 0 + 201 J = 201 J and K 2 = 1 mv2 gives v2 =
2 6.40. 2K2
2(201 J)
=
= 2.40 m/s.
m
70.0 kg EVALUATE: The spring does positive work and the sled gains speed as it returns to x = 0. More work is done
during the larger displacement in part (a), so the speed there is larger than in part (b).
IDENTIFY: Fx = kx
SET UP: When the spring is in equilibrium, the same force is applied to both ends of any segment of the spring.
EXECUTE: (a) When a force F is applied to each end of the original spring, the end of the spring is displaced a
distance x. Each half of the spring elongates a distance xh , where xh = x / 2 . Since F is also the force applied to each ⎛x
half of the spring, F = kx and F = kh xh . kx = kh xh and kh = k ⎜
⎝ xh ⎞
⎟ = 2k .
⎠ Work and Kinetic Energy 6-11 (b) The same reasoning as in part (a) gives kseg = 3k , where kseg is the force constant of each segment. 6.41. EVALUATE: For half of the spring the same force produces less displacement than for the original spring. Since
k = F / x , smaller x for the same F means larger k.
IDENTIFY and SET UP: Apply Eq.(6.6) to the glider. Work is done by the spring and by gravity. Take point 1 to be
where the glider is released. In part (a) point 2 is where the glider has traveled 1.80 m and K 2 = 0. There two points
are shown in Figure 6.41a. In part (b) point 2 is where the glider has traveled 0.80 m.
EXECUTE: (a) Wtot = K 2 − K1 = 0. Solve for x1 , the amount the spring is initially compressed. Wtot = Wspr + Ww = 0
So Wspr = −Ww
(The spring does positive work on
the glider since the spring force is
directed up the incline, the same as
the direction of the displacement.)
Figure 6.41a The directions of the displacement and of the gravity force are shown in Figure 6.41b.
Ww = ( w cos φ ) s = ( mg cos130.0°) s
Ww = (0.0900 kg)(9.80 m/s 2 )(cos130.0°)(1.80 m) = −1.020 J
(The component of w parallel to the incline is
directed down the incline, opposite to the
displacement, so gravity does negative work.)
Figure 6.41b Wspr = −Ww = +1.020 J
2Wspr 2(1.020 J)
= 0.0565 m
k
640 N/m
(b) The spring was compressed only 0.0565 m so at this point in the motion the glider is no longer in contact with the
spring. Points 1 and 2 are shown in Figure 6.41c.
Wspr = 1 kx12 so x1 =
2 = Wtot = K 2 − K1
K 2 = K1 + Wtot
K1 = 0
Figure 6.41c Wtot = Wspr + Ww
From part (a), Wspr = 1.020 J and
Ww = (mg cos130.0°) s = (0.0900 kg)(9.80 m/s 2 )(cos130.0°)(0.80 m) = −0.454 J
Then K 2 = Wspr + Ww = +1.020 J − 0.454 J = +0.57 J. 6.42. EVALUATE: The kinetic energy in part (b) is positive, as it must be. In part (a), x2 = 0 since the spring force is no
longer applied past this point. In computing the work done by gravity we use the full 0.80 m the glider moves.
IDENTIFY: Apply Wtot = K 2 − K1 to the brick. Work is done by the spring force and by gravity.
SET UP: At the maximum height. v = 0 . Gravity does negative work, Wgrav = − mgh . The work done by the spring 2 is kd , where d is the distance the spring is compressed initially.
EXECUTE: The initial and final kinetic energies of the brick are both zero, so the net work done on the brick by the
spring and gravity is zero, so (1 2)kd 2 − mgh = 0 , or
1
2 d = 2mgh / k = 2(1.80 kg)(9.80 m / s 2 )(3.6 m) /(450 N / m) = 0.53 m. The spring will provide an upward force
while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length.
But when the spring reaches its uncompressed length the brick has an upward velocity and leaves the spring.
EVALUATE: Gravity does negative work because the gravity force is downward and the brick moves upward. The
spring force does positive work on the brick because the spring force is upward and the brick moves upward. 6-12 6.43. Chapter 6 IDENTIFY:
SET UP: Apply the relation between energy and power.
W
to solve for W, the energy the bulb uses. Then set this value equal to
Use P =
Δt speed.
EXECUTE: mv 2 and solve for the W = PΔt = (100 W)(3600 s) = 3.6 × 105 J K = 3.6 × 105 J so v =
6.44. 1
2 2K
2(3.6 × 105 J)
=
= 100 m/ s
70 kg
m EVALUATE: Olympic runners achieve speeds up to approximately 36 m/s, or roughly one third the result calculated.
IDENTIFY: Energy is power times time.
SET UP: 1 W = 1 J/s . 1 yr = 3.16 × 107 s .
EXECUTE:
(b) (a) (1.0 × 1019 J / yr)
= 3.2 × 1011 W.
(3.16 × 107 s / yr) 3.2 × 1011 W
= 1.1 kW/person.
3.0 × 108 folks 3.2 × 1011 W
= 8.0 × 108 m 2 = 800 km 2 .
(0.40)1.0 × 103 W / m 2
EVALUATE: The area in part (c) corresponds to a square about 28 km on a side, which is about 18 miles. The space
required is not an impediment.
ΔW
. ΔW is the energy released.
IDENTIFY: Pav =
Δt
SET UP: ΔW is to be the same. 1 y = 3.156 × 107 s .
(c) A = 6.45. Pav Δt = ΔW = constant , so Pav-sun Δtsun = Pav-m Δtm . EXECUTE: 6.46. ⎛ Δt ⎞ ⎛ [2.5 × 105 y][3.156 × 107 s/y] ⎞
13
Pav-m = Pav-sun ⎜ sun ⎟ = ⎜
⎟ = 3.9 × 10 P .
0.20 s
Δt m ⎠
⎠
⎝
EVALUATE: Since the power output of the magnetar is so much larger than that of our sun, the mechanism by which
it radiates energy must be quite different.
IDENTIFY: The thermal energy is produced as a result of the force of friction, F = μ k mg . The average thermal
power is thus the average rate of work done by friction or P = F!vav .
SET UP: v2 + v1 ⎛ 8.00 m/s + 0 ⎞
=⎜
⎟ = 4.00 m/s
2
2
⎝
⎠
P = Fvav = ⎡( 0.200 ) ( 20.0 kg ) ( 9.80 m/s 2 ) ⎤ ( 4.00 m/s ) = 157 W
⎣
⎦ vav = EXECUTE:
EVALUATE: The power could also be determined as the rate of change of kinetic energy, ΔK t , where the time is calculated from vf = vi + at and a is calculated from a force balance,
6.47. 1 hp = 746 W EXECUTE: 6.48. k Use the relation P = F!v to relate the given force and velocity to the total power developed. IDENTIFY:
SET UP: ∑ F = ma = μ mg. The total power is P = F!v = (165 N ) ( 9.00 m/s ) = 1.49 × 103 W. Each rider therefore contributes Peach rider = (1.49 × 103 W ) / 2 = 745 W ≈ 1 hp.
EVALUATE: The result of one horsepower is very large; a rider could not sustain this output for long periods of
time.
IDENTIFY and SET UP: Calculate the power used to make the plane climb against gravity. Consider the vertical
motion since gravity is vertical.
EXECUTE: The rate at which work is being done against gravity is
P = Fv = mgv = (700 kg)(9.80 m/s 2 )(2.5 m/s) = 17.15 kW.
This is the part of the engine power that is being used to make the airplane climb. The fraction this is of the total is
17.15 kW/75 kW = 0.23.
EVALUATE:
engine. The power we calculate for making the airplane climb is considerably less than the power output of the Work and Kinetic Energy 6.49. 6-13 ΔW
. The work you do in lifting mass m a height h is mgh.
Δt
SET UP: 1 hp = 746 W
EXECUTE: (a) The number per minute would be the average power divided by the work (mgh) required to lift one
(0.50 hp) (746 W hp)
box,
= 1.41 s, or 84.6 min.
(30 kg) (9.80 m s 2 ) (0.90 m)
Pav = IDENTIFY: (100 W)
= 0.378 s, or 22.7 min.
(30 kg) (9.80 m s 2 ) (0.90 m)
EVALUATE: A 30-kg crate weighs about 66 lbs. It is not possible for a person to perform work at this rate.
IDENTIFY and SET UP: Use Eq.(6.15) to relate the power provided and the amount of work done against gravity in
16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers.
EXECUTE: Find the total mass that can be lifted:
Pt
ΔW mgh
Pav =
=
, so m = av
Δt
t
gh
(b) Similarly, 6.50. ⎛ 746 W ⎞
4
Pav = (40 hp) ⎜
⎟ = 2.984 × 10 W
1 hp ⎠
⎝
m= Pavt (2.984 × 104 W)(16.0 s)
=
= 2.436 × 103 kg
gh
(9.80 m/s 2 )(20.0 m) This is the total mass of elevator plus passengers. The mass of the passengers is 2.436 ×103 kg − 600 kg = 1.836 ×103 kg.
1.836 × 103 kg
= 28.2. 28 passengers can ride.
65.0 kg
EVALUATE: Typical elevator capacities are about half this, in order to have a margin of safety.
IDENTIFY: Calculate the gallons of gasoline consumed and from that the energy consumed. Find the time Δt for the
ΔW
, where ΔW is the energy consumed.
trip and use Pav =
Δt
SET UP: 200 km = 124 mi
124 mi
EXECUTE: (a) The gallons of gasoline consumed is
= 4.13 gal . The energy consumed is
30 mi/gal The number of passengers is
6.51. (4.13 gal)(1.3 × 109 J/gal) = 5.4 × 109 J .
ΔW 5.4 × 109 J
124 mi
= 2.07 h = 7450 s . Pav =
=
= 7.2 × 105 W = 720 kW .
60 mi/h
7450 s
Δt
720 × 103 W
The rate of energy consumption is
= 970 hp .
746 W/hp (b) The time for the trip is
EVALUATE:
6.52. Apply P = F!v . F! is the force F of water resistance. IDENTIFY:
SET UP: 1 hp = 746 W . 1 km/h = 0.228 m/s
F= EXECUTE:
EVALUATE:
6.53. (0.70) P
(0.70) (280,000 hp)(746 W hp)
=
= 8.1 × 106 N.
v
(65 km h) ((0.228 m/s) (1 km/h)) The power required depends on speed, because of the factor of v in P = F!v and also because the resistive force increases with speed.
IDENTIFY: To lift the skiers, the rope must do positive work to counteract the negative work developed by the
component of the gravitational force acting on the total number of skiers,
Frope = Nmg sin α .
SET UP: P = F!v = Fropev EXECUTE: Prope = Fropev = ⎡ + Nmg ( cos φ ) ⎤ v .
⎣
⎦ ⎡
⎛ 1 m/s ⎞ ⎤
Prope = ⎡( 50 riders ) ( 70.0 kg ) ( 9.80 m/s 2 ) ( cos75.0º ) ⎤ ⎢(12.0 km/h ) ⎜
⎟.
⎣
⎦
⎝ 3.60 km/h ⎠ ⎥
⎣
⎦
Prope = 2.96 × 104 W = 29.6 kW .
EVALUATE:
rest. Some additional power would be needed to give the riders kinetic energy as they are accelerated from 6-14 6.54. 6.55. Chapter 6 IDENTIFY: Relate power, work and time.
SET UP: Work done in each stroke is W = Fs and Pav = W t .
EXECUTE: 100 strokes per second means Pav = 100 Fs t with t = 1.00 s, F = 2mg and s = 0.010 m. Pav = 0.20 W.
EVALUATE: For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per
second, the average power output would be 7.0 × 105 W . This power output is very far beyond the capability of a
person.
IDENTIFY: For mass dm located a distance x from the axis and moving with speed v, the kinetic energy is
K = 1 (dm)v 2 . Follow the procedure specified in the hint.
2
SET UP: The bar and an infinitesimal mass element along the bar are sketched in Figure 6.55. Let M = total mass
2πx
.
and T = time for one revolution . v =
T
1
M
dx , so
EXECUTE: K = ∫ (dm)v 2 . dm =
2
L
2 L
1 ⎛ M ⎞ ⎛ 2πx ⎞ 1 ⎛ M ⎞ ⎛ 4π 2 ⎞
K = ∫ ⎜ dx ⎟ ⎜
⎟
⎟ = ⎜ ⎟⎜
2⎝ L ⎠ ⎝ T ⎠ 2⎝ L ⎠ ⎝ T2 ⎠
0 1 ⎛ M ⎞ ⎛ 4π 2 ⎞ ⎛ L3 ⎞ 2
x 2 dx = ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎟ = π 2 ML2 T 2
∫
2⎝ L ⎠ ⎝ T ⎠ ⎝ 3 ⎠ 3
0
L There are 5 revolutions in 3 seconds, so T = 3 5 s = 0.60 s
2
K = π 2 (12.0 kg) (2.00 m)2 (0.60 s) 2 = 877 J.
3
EVALUATE: If a point mass 12.0 kg is 2.00 m from the axis and rotates at the same rate as the bar,
2π (2.00 m)
v=
= 20.9 m/s and K = 1 mv 2 = 1 (12 kg)(20.9 m/s) 2 = 2.62 × 103 J . K for the bar is smaller by a factor of
2
2
0.60 s
0.33. The speed of a segment of the bar decreases toward the axis. Figure 6.55
6.56. IDENTIFY: Density is mass per unit volume, ρ = m / V , so we can calculate the mass of the asteroid. K = 1 mv 2 .
2
Since the asteroid comes to rest, the kinetic energy it delivers equals its initial kinetic energy.
1
SET UP: The volume of a sphere is related to its diameter by V = π d 3 .
6 π (320 m)3 = 1.72 × 107 m3 . m = ρV = (2600 kg/m3 )(1.72 × 107 m 3 ) = 4.47 × 1010 kg .
6
K = 1 mv 2 = 1 (4.47 × 1010 kg)(12.6 × 103 m/s) 2 = 3.55 × 1018 J .
2
2 EXECUTE: (a) V = 3.55 × 1018 J
= 56.5 devices .
6.28 × 1016 J
EVALUATE: If such an asteroid were to hit the earth the effect would be catastrophic.
IDENTIFY and SET UP: Since the forces are constant, Eq.(6.2) can be used to calculate the work done by each force.
The forces on the suitcase are shown in Figure 6.57a.
(b) The yield of a Castle/Bravo device is (1 s)(4.184 × 1015 J) = 6.28 × 1016 J . 6.57. Figure 6.57a In part (f ) , Eq.(6.6) is used to relate the total work to the initial and final kinetic energy.
EXECUTE: (a) WF = ( F cos φ ) s
"
"
Both F and s are parallel to the incline and in the same direction, so φ = 90° and WF = Fs = (140 N)(3.80 m) = 532 J Work and Kinetic Energy 6-15 (b) The directions of the displacement and of the gravity force are shown in Figure 6.57b.
Ww = ( w cos φ ) s
φ = 115°, so
Ww = (196 N)(cos115°)(3.80 m) Ww = −315 J
Figure 6.57b
Alternatively, the component of w parallel to the incline is w sin 25°. This component is down the incline so its angle
"
with s is φ = 180°. Ww sin 25° = (196 Nsin 25°)(cos180°)(3.80 m) = −315 J. The other component of w, w cos 25°, is
"
perpendicular to s and hence does no work. Thus Ww = Ww sin 25° = −315 J, which agrees with the above.
(c) The normal force is perpendicular to the displacement (φ = 90°), so Wn = 0.
(d) n = w cos 25° so f k = μ k n = μ k w cos 25° = (0.30)(196 N)cos 25° = 53.3 N W f = ( f k cos φ ) x = (53.3 N)(cos180°)(3.80 m) = −202 J
(e) Wtot = WF + Ww + Wn + W f = +532 J − 315 J + 0 − 202 J = 15 J
(f ) Wtot = K 2 − K1 , K1 = 0, so K 2 = Wtot
1
2 6.58. 2
mv2 = Wtot so v2 = 2Wtot
2(15 J)
=
= 1.2 m/s
20.0 kg
m EVALUATE: The total work done is positive and the kinetic energy of the suitcase increases as it moves up the incline.
IDENTIFY: The work he does to lift his body a distance h is W = mgh . The work per unit mass is (W m) = gh.
SET UP: The quantity gh has units of N/kg.
EXECUTE: (a) The man does work, (9.8 N kg ) (0.4 m) = 3.92 J kg.
(b) (3.92 J kg) (70 J kg ) × 100 = 5.6%.
(c) The child does work (9.8 N kg)(0.2 m) = 1.96 J kg. (1.96 J kg) (70 J kg ) × 100 = 2.8%.
(d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg 6.59. instead of 3.92 J kg, the child should be able to do more chin-ups.
EVALUATE: Since the child has arms half the length of his father’s arms, the child must lift his body only 0.20 m to
do a chin-up.
IDENTIFY: Apply the definitions of IMA and AMA given in the problem.
SET UP: When the object moves a distance L along the ramp, it rises a vertical distance L sin α .
1
.
EXECUTE: (a) sin = L, sout = L sin α, so IMA =
sin α
(b) If AMA = IMA, ( Fout Fin ) = ( sin sout ) and so ( Fout ) (sout ) = ( Fin ) (sin ) , or Wout = Win .
(c) The pulley is sketched in Figure 6.59.
W
( F )( s ) F F
AMA
(d) e = out = out out = out in =
.
Win
( Fin )( sin )
sin sout
IMA
EVALUATE: Fin = w sin α and Fout = w . ( Fin )( sin ) = ( w sin α ) L . ( Fout )( sout ) = w(sin α L) . Therefore, ( Fin )( sin ) = ( Fout )( sout ) . A smaller force acting over a larger distance does the same amount of work as a larger force
acting over a smaller distance. 6.60. IDENTIFY:
SET UP: mB = Apply Figure 6.59
"
F = ma to each block to find the tension in the string. Each force is constant and W = Fs cosφ .
∑ The free-body diagram for each block is given in Figure 6.60. m A = 12.0 N
= 1.22 kg .
g 20.0 N
= 2.04 kg and
g 6-16 Chapter 6 EXECUTE: T − f k = m Aa . wB − T = mB a . wB − f k = ( mA + mB )a . ⎛ wB ⎞
⎛ mA ⎞
⎛ wA ⎞
fk = 0 . a = ⎜
⎟ and T = wB ⎜
⎟ = wB ⎜
⎟ = 7.50 N .
⎝ mA + mB ⎠
⎝ mA + mB ⎠
⎝ wA + wB ⎠
20.0 N block: Wtot = Ts = (7.50 N)(0.750 m) = 5.62 J .
12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 7.50 N)(0.750 m) = 3.38 J
(b) f k = μ k wA = 6.50 N . a = ⎛ mA ⎞
⎛ wA ⎞
wB − μ k wA
. T = f k + ( wB − μ k wA ) ⎜
⎟ = μ k wA + ( wB − μ k wA ) ⎜
⎟.
m A + mB
⎝ mA + mB ⎠
⎝ wA + wB ⎠ T = 6.50 N + (5.50 N)(0.625) = 9.94 N .
20.0 N block: Wtot = (T − f k ) s = (9.94 N − 6.50 N)(0.750 m) = 2.58 J .
12.0 N block: Wtot = ( wB − T ) s = (12.0 N − 9.94 N)(0.750 m) = 1.54 J .
EVALUATE: Since the two blocks move with equal speeds, for each block Wtot = K 2 − K1 is proportional to the mass
(or weight) of that block. With friction the gain in kinetic energy is less, so the total work on each block is less. Figure 6.60
6.61. IDENTIFY: K = 1 mv 2 . Find the speed of the shuttle relative to the earth and relative to the satellite.
2
SET UP: Velocity is distance divided by time. For one orbit the shuttle travels a distance 2π R .
2 EXECUTE: 6.62. (a) 2 ⎛ 2π (6.66 × 106 m)
⎞
1 2 1 ⎛ 2πR ⎞ 1
12
mv = m ⎜
⎟ = 2.59 × 10 J.
⎟ = (86,400 kg) ⎜
2
2 ⎝T⎠ 2
⎝ (90.1 min) (60 s min) ⎠ (b) (1 2) mv 2 = (1 2) (86,400 kg) ((1.00 m) (3.00 s))2 = 4.80 × 103 J.
EVALUATE: The kinetic energy of an object depends on the reference frame in which it is measured.
IDENTIFY: W = Fs cos φ . Wtot = K 2 − K1 .
SET UP: f k = μ k n . The normal force is n = mg cosθ , with θ = 12.0° . The component of the weight parallel to the
incline is mg sin θ .
EXECUTE: (a) φ = 180° and W f = − f k s = −( μk mg cos θ ) s = −(0.31)(5.00 kg)(9.80 m s 2 )(cos 12.0°)(1.50 m) = −22.3 J (b) (5.00 kg)(9.80 m s 2 )(sin12.0°)(1.50 m) = 15.3 J.
(c) The normal force does no work.
(d) Wtot = 15.3 J − 22.3 J = −7.0 J. 6.63. (e) K 2 = K1 + Wtot = (1 2)(5.00 kg)(2.2 m s)2 − 7.0 J = 5.1 J , and so v2 = 2(5.1 J) /(5.00 kg) = 1.4 m / s .
EVALUATE: Friction does negative work and gravity does positive work. The net work is negative and the kinetic
energy of the object decreases.
IDENTIFY: The effective force constant is defined by keff = F / x , where F is the force applied to each end of the
spring combination and x is the amount the spring combination is stretched.
SET UP: Consider a force F applied to each end of the combination. Then F1 and F2 are the forces applied to each spring and F = F1 + F2 . Each spring stretches the same amount x.
EXECUTE: (a) F = keff x . F = F1 + F2 = k1 x + k2 x . Equating the two expressions for F gives keff = k1 + k2 . (b) The same procedure as in part (a) gives keff = k1 + k2 + $ + k N .
EVALUATE: The effective force constant of the configuration is greater than any of the force constants of the
individual springs. More force is required to stretch the parallel combination that is required to stretch each separate
spring the same amount. Work and Kinetic Energy 6.64. 6-17 IDENTIFY: The effective force constant is defined by keff = F/x , where F is the force applied to each end of the
spring combination and x is the amount the spring combination is stretched.
SET UP: Consider a force F applied to each end of the combination. The same force F is applied to each spring.
Spring 1 stretches a distance x1 and spring 2 stretches a distance x2 , where x1 = F / k1 and x2 = F / k2 . The total distance the combination stretches is x = x1 + x2 .
EXECUTE: (a) x = x1 + x2 gives F
FF
1
11
=+.
= = and
keff k1 k2
keff k1 k2 (b) The same procedure as in part (a) gives k1k2
. The effective force constant for
k1 + k2
the two springs in series is less than the force constant for each individual spring. It takes less force to stretch the
combination an amount x than to stretch either separate spring an amount x.
IDENTIFY: Apply Eq.(6.7).
dx
1
SET UP: ∫ 2 = − .
x
x
EVALUATE: 6.65. 1
11
1
.
= + +$+
keff k1 k2
kN For two springs the result in part (a) can be written as keff = x 2
⎛1 1⎞
dx
⎡ 1⎤
= − k ⎢ − ⎥ = k ⎜ − ⎟ . The force is given to be attractive, so Fx < 0 ,
2
x1
x1 x
x ⎦ x1
⎣
⎝ x2 x1 ⎠
11
< , and W < 0 .
and k must be positive. If x2 > x1 ,
x2 x1
(b) Taking “slowly” to be constant speed, the net force on the object is zero. The force applied by the hand is
⎛1 1⎞
opposite Fx , and the work done is negative of that found in part (a), or k ⎜ − ⎟ , which is positive if x2 > x1 .
⎝ x1 x2 ⎠ EXECUTE: 6.66. x2 (a) W = ∫ Fx dx = −k ∫ x2 (c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero.
EVALUATE: Your force is directed away from the origin, so when the object moves away from the origin your force
does positive work.
IDENTIFY: Apply Eq.(6.6) to the motion of the asteroid.
SET UP: Let point 1 be at a great distance and let point 2 be at the surface of the earth. Assume K1 = 0. From the
information given about the gravitational force its magnitude as a function of distance r from the center of the earth
ˆ
must be F = mg ( RE / r ) 2 . This force is directed in the − r direction since it is a “pull”. F is not constant so Eq.(6.7)
must be used to calculate the work it does.
2
RE ⎛ mgR ⎞
2
R
2
EXECUTE: W = − ∫ F ds = − ∫ ⎜ 2 E ⎟ dr = − mgRE −(1/ r ) ∞E = mgRE
1
∞
⎝r ⎠ ( ) Wtot = K 2 − K1 , K1 = 0
This gives K 2 = mgRE = 1.25 × 1012 J
2
K 2 = 1 mv2 so v2 = 2 K 2 / m = 11,000 m/s
2 EVALUATE:
6.67. Note that v2 = 2 gRE , the impact speed is independent of the mass of the asteroid. IDENTIFY: Calculate the work done by friction and apply Wtot = K 2 − K1 . Since the friction force is not constant,
use Eq.(6.7) to calculate the work.
SET UP: Let x be the distance past P. Since μ k increases linearly with x, μ k = 0.100 + Ax . When x = 12.5 m , μ k = 0.600 , so A = 0.500 /(12.5 m) = 0.0400 /m
1
(a) Wtot = ΔK = K 2 − K1 gives − ∫ μk mgdx = 0 − mv12 . Using the above expression for μ k ,
2
2
x2
⎡
⎡
x⎤ 1
x2 ⎤ 1
1
g ∫ (0.100 + Ax)dx = v12 and g ⎢(0.100) x2 + A 2 ⎥ = v12 . (9.80 m/s 2 ) ⎢ (0.100) xf + (0.0400/m) f ⎥ = (4.50 m/s)2 .
0
2
2⎦ 2
2⎦ 2
⎣
EXECUTE: Solving for x2 gives x2 = 5.11 m .
(b) μ k = 0.100 + (0.0400/m)(5.11 m) = 0.304 6-18 6.68. Chapter 6 v2
(4.50 m/s) 2
1
(c) Wtot = K 2 − K1 gives − μk mgx2 = 0 − mv12 . x2 = 1 =
= 10.3 m .
2
2 μk g 2(0.100)(9.80 m/s 2 )
EVALUATE: The box goes farther when the friction coefficient doesn’t increase.
IDENTIFY: Use Eq.(6.7) to calculate W.
SET UP: x1 = 0 . In part (a), x2 = 0.050 m . In part (b), x2 = −0.050 m .
k2 b3 c4
x2 − x2 + x2 .
2
3
4
2
2
3
3
4
W = (50.0 N / m) x2 − (233 N / m ) x2 + (3000 N / m ) x2 . When x2 = 0.050 m , W = 0.12 J .
x2 x2 (a) W = ∫ Fdx = ∫ ( kx − bx 2 + cx 3 )dx = EXECUTE: 0 0 (b) When x2 = −0.050 m, W = 0.17 J .
(c) It’s easier to stretch the spring; the quadratic −bx 2 term is always in the − x -direction, and so the needed force,
and hence the needed work, will be less when x2 > 0 . When x = 0.050 m , Fx = 4.75 N . When x = −0.050 m , Fx = 8.25 N .
"
"
IDENTIFY and SET UP: Use ∑ F = ma to find the tension force T. The block moves in uniform circular motion and
""
a = arad .
(a) The free-body diagram for the block is given in Figure 6.69.
EVALUATE: 6.69. EXECUTE: T =m ∑F x = max v2
R T = (0.120 kg) (0.70 m/s) 2
= 0.15 N
0.40 m Figure 6.69
(b) T = m v2
(2.80 m/s) 2
= (0.120 kg)
= 9.4 N
0.10 m
R (c) SET UP: x2 The tension changes as the distance of the block from the hole changes. We could use W = ∫ Fx dx to
x1 calculate the work. But a much simpler approach is to use Wtot = K 2 − K1.
EXECUTE: The only force doing work on the block is the tension in the cord, so Wtot = WT .
K1 = 1 mv12 = 1 (0.120 kg)(0.70 m/s) 2 = 0.0294 J
2
2
2
K 2 = 1 mv2 = 1 (0.120 kg)(2.80 m/s) 2 = 0.470 J
2
2 Wtot = K 2 − K1 = 0.470 J − 0.029 J = 0.44 J 6.70. This is the amount of work done by the person who pulled the cord.
EVALUATE: The block moves inward, in the direction of the tension, so T does positive work and the kinetic energy
increases.
IDENTIFY: Use Eq.(6.7) to find the work done by F. Then apply Wtot = K 2 − K1 .
SET UP: dx ∫x 2 1
=− .
x ⎛1 1⎞
dx = α ⎜ − ⎟ . W = (2.12 × 10−26 N ⋅ m 2 )((0.200 m −1 ) − (1.25 × 109 m −1 )) = −2.65 × 10−17 J .
⎝ x1 x2 ⎠
Note that x1 is so large compared to x2 that the term 1/ x1 is negligible. Then, using Eq. (6.13)) and solving for v2 , EXECUTE: W =∫ x2 α x1 x2 v2 = v12 + 2W
2(−2.65 × 10−17 J)
= (3.00 × 105 m/s) 2 +
= 2.41 × 105 m/s.
(1.67 × 10−27 kg)
m (b) With K 2 = 0, W = − K1 . Using W = − x2 = α
K1 = α
x2 , 2α
2(2.12 × 10−26 N ⋅ m 2 )
=
= 2.82 × 10−10 m.
2
mv1 (1.67 × 10−27 kg)(3.00 × 105 m/s) 2 Work and Kinetic Energy 6.71. 6-19 (c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their
original values, and the speed is 3.00 × 105 m/s .
EVALUATE: As the proton moves toward the uranium nucleus the repulsive force does negative work and the
kinetic energy of the proton decreases. As the proton moves away from the uranium nucleus the repulsive force does
positive work and the kinetic energy of the proton increases.
"
"
"
"
IDENTIFY and SET UP: Use vx = dx/dt and ax = dvx /dt. Use ∑ F = ma to calculate F from a. dx
= 2α t + 3β t 2
dt
t = 4.00 s: vx = 2(0.200 m/s 2 )(4.00 s) + 3(0.0200 m/s3 )(4.00 s) 2 = 2.56 m/s. EXECUTE: (a) x(t ) = α t 2 + β t 3 , vx (t ) = dvx
= 2α + 6 β t
dt
Fx = max = m(2α + 6 β t ) (b) ax (t ) = t = 4.00 s: Fx = 6.00 kg(2(0.200 m/s 2 ) + 6(0.0200 m/s3 )(4.00 s)) = 5.28 N
(c) IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work.
EXECUTE: Wtot = K 2 − K1
At t1 = 0, v1 = 0 so K1 = 0.
Wtot = WF
2
K 2 = 1 mv2 = 1 (6.00 kg)(2.56 m/s) 2 = 19.7 J
2
2 6.72. Then Wtot = K 2 − K1 gives that WF = 19.7 J
EVALUATE: v increases with t so the kinetic energy increases and the work done is positive. We can also calculate
WF directly from Eq.(6.7), by writing dx as vx dt and performing the integral.
IDENTIFY: Since the capsule comes to rest, the amount of work the capsule does on the ground equals its original
kinetic energy. Use constant acceleration kinematic equations to calculate the stopping time t; Δt = t .
SET UP: 311 km/h = 86.4 m/s . Let + y be the direction the capsule is traveling before the crash.
EXECUTE: 6.73. ΔW = K1 = 1 mv12 = 1 (210 kg)(86.4 m/s) 2 = 7.84 × 105 J . y − y0 = 0.810 m , v0 y = 86.4 m/s and v y = 0 .
2
2 ⎛ v + vy ⎞
2( y − y0 ) 2(0.810 m)
ΔW 7.84 × 105 J
y − y0 = ⎜ 0 y
=
= 0.01875 s .
=
= 4.18 × 107 W
⎟ t gives t =
v0 y
86.4 m/s
0.01875 s
Δt
⎝2⎠
EVALUATE: A large amount of work is done in a very small amount of time.
IDENTIFY and SET UP: Use Eq.(6.6). You do positive work and gravity does negative work. Let point 1 be at the
base of the bridge and point 2 be at the top of the bridge.
EXECUTE: (a) Wtot = K 2 − K1
K1 = 1 mv12 = 1 (80.0 kg)(5.00 m/s) 2 = 1000 J
2
2
2
K 2 = 1 mv2 = 1 (80.0 kg)(1.50 m/s)2 = 90 J
2
2 Wtot = 90 J − 1000 J = −910 J
(b) Neglecting friction, work is done by you (with the force you apply to the pedals) and by gravity:
Wtot = Wyou + Wgravity . The gravity force is w = mg = (80.0 kg)(9.80 m/s 2 ) = 784 N, downward. The displacement is
5.20 m, upward. Thus φ = 180° and
Wgravity = ( F cos φ ) s = (784 N)(5.20 m)cos180° = −4077 J
Then Wtot = Wyou + Wgravity gives
Wyou = Wtot − Wgravity = −910 J − (−4077 J) = +3170 J
6.74. EVALUATE: The total work done is negative and you lose kinetic energy.
IDENTIFY: Use Eq.(6.7) to calculate W.
1 − ( n −1)
SET UP: ∫ x − n dx = −
x
n −1
EXECUTE: (a) W = ∫ ∞
x0 b
b
dx =
( n − 1)) x n −1
xn ∞ =
x0 b
. Note that for this part, for n > 1, x1− n → 0 as x → ∞ .
n
(n − 1) x0 −1 6-20 Chapter 6 ⎡b
⎤
n
n
(b) When 0 < n < 1 , the improper integral must be used, W = lim ⎢
( x2 −1 − x0 −1 ) ⎥ , and because the exponent on
x2 →∞ ( n − 1)
⎣
⎦
n
the x2 −1 is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an 6.75. infinite amount of work, even though F → 0 as x2 → ∞.
EVALUATE: The work-energy theorem says that an object gains an infinite amount of kinetic energy when an
infinite amount of work is done on it.
IDENTIFY: The negative work done by the spring equals the change in kinetic energy of the car.
SET UP: The work done by a spring when it is compressed a distance x from equilibrium is − 1 kx 2 . K 2 = 0 .
2
EXECUTE: 6.76. − 1 kx 2 = K 2 − K1 gives
2 1
2 kx 2 = 1 mv12 and
2 2
2
k = ( mv12 ) x 2 = ⎡(1200 kg ) ( 0.65 m/s ) ⎤ ( 0.070 m ) = 1.0 × 105 N/m .
⎣
⎦
EVALUATE: When the spring is compressed, the spring force is directed opposite to the displacement of the object
and the work done by the spring is negative.
IDENTIFY: Apply Wtot = K 2 − K1 .
2
SET UP: Let x0 be the initial distance the spring is compressed. The work done by the spring is 1 kx0 − 1 kx 2 , where
2
2
x is the final distance the spring is compressed.
2
EXECUTE: (a) Equating the work done by the spring to the gain in kinetic energy, 1 kx0 = 1 mv 2 , so
2
2 v= 400 N / m
k
(0.060 m) = 6.93 m/s.
x0 =
0.0300 kg
m (b) Wtot must now include friction, so v= 1
2 2
mv 2 = Wtot = 1 kx0 − fx0 , where f is the magnitude of the friction force. Then,
2 400 N/m
2(6.00 N)
k 2 2f
(0.060 m) 2 −
(0.060 m) = 4.90 m/s.
x0 −
x0 =
0.0300 kg
(0.0300 kg)
m
m (c) The greatest speed occurs when the acceleration (and the net force) are zero. Let x be the amount the spring is still
f
6.00 N
compressed, so the distance the ball has moved is x0 − x . kx = f , x = =
= 0.0150 m . To find the speed,
k 400 N/m
2
the net work is Wtot = 1 k ( x0 − x 2 ) − f ( x0 − x) , so the maximum speed is vmax =
2 vmax = 6.77. k2
2f
( x0 − x 2 ) −
( x0 − x) .
m
m 400 N / m
2(6.00 N)
((0.060 m) 2 − (0.0150 m) 2 ) −
(0.060 m − 0.0150 m) = 5.20 m/s
(0.0300 kg)
(0.0300 kg) EVALUATE: The maximum speed with friction present (part (c)) is larger than the result of part (b) but smaller than
the result of part (a).
IDENTIFY and SET UP: Use Eq.(6.6). Work is done by the spring and by gravity. Let point 1 be where the textbook
is released and point 2 be where it stops sliding. x2 = 0 since at point 2 the spring is neither stretched nor
compressed. The situation is sketched in Figure 6.77.
EXECUTE: Wtot = K 2 − K1
K1 = 0, K 2 = 0
Wtot = Wfric + Wspr
Figure 6.77 Wspr = 1 kx12 , where x1 = 0.250 m (Spring force is in direction of motion of block so it does positive work.)
2
Wfric = − μ k mgd
Then Wtot = K 2 − K1 gives 1
2 kx12 − μ k mgd = 0 (250 N/m)(0.250 m) 2
kx12
=
= 1.1 m, measured from the point where the block was released.
2μ k mg 2(0.30)(2.50 kg)(9.80 m/s 2 )
EVALUATE: The positive work done by the spring equals the magnitude of the negative work done by friction. The
total work done during the motion between points 1 and 2 is zero and the textbook starts and ends with zero kinetic
energy.
d= Work and Kinetic Energy 6.78. 6-21 IDENTIFY: Apply Wtot = K 2 − K1 to the cat.
SET UP: Let point 1 be at the bottom of the ramp and point 2 be at the top of the ramp.
EXECUTE: The work done by gravity is Wg = −mgL sin θ (negative since the cat is moving up), and the work done by the applied force is FL, where F is the magnitude of the applied force. The total work is
Wtot = (100 N)(2.00 m) − (7.00 kg)(9.80 m/s 2 )(2.00 m)sin 30° = 131.4 J .
The cat’s initial kinetic energy is 1
2 mv12 = 1 (7.00 kg)(2.40 m/s) 2 = 20.2 J , and
2 v2 = 2( K1 + W )
2(20.2 J + 131.4 J)
=
= 6.58 m/s.
(7.00 kg)
m EVALUATE: The net work done on the cat is positive and the cat gains speed. Without your push,
Wtot = Wgrav = −68.6 J and the cat wouldn’t have enough initial kinetic energy to reach the top of the ramp.
6.79. Apply Wtot = K 2 − K1 to the vehicle. IDENTIFY:
SET UP: Call the bumper compression x and the initial speed v0 . The work done by the spring is − 1 kx 2 and
2 K2 = 0 .
1212
kx = mv0 , kx < 5 mg . Combining to eliminate k and then x, the two
2
2
(20.0 m/s) 2
v2
mg 2
inequalities are x >
and k < 25 2 . Using the given numerical values, x >
= 8.16 m and
5g
v
5(9.80 m/s 2 )
(a) The necessary relations are EXECUTE: (1700 kg)(9.80 m/s 2 ) 2
= 1.02 × 104 N/m.
(20.0 m/s) 2
(b) A distance of 8 m is not commonly available as space in which to stop a car. Also, the car stops only momentarily
and then returns to its original speed when the spring returns to its original length.
EVALUATE: If k were doubled, to 2.04 × 104 N/m , then x = 5.77 m . The stopping distance is reduced by a factor of
k < 25 1/ 2 , but the maximum acceleration would then be kx/m = 69.2 m/s 2 , which is 7.07 g .
6.80. IDENTIFY: Apply Wtot = K 2 − K1 . W = Fs cos φ .
SET UP: The students do positive work, and the force that they exert makes an angle of 30.0° with the direction of
motion. Gravity does negative work, and is at an angle of 120.0° with the chair’s motion,
EXECUTE: The total work done is Wtot = ((600 N)cos30.0° + (85.0 kg)(9.80 m/s 2 )cos120.0°)(2.50 m) = 257.8 J , and so the speed at the top of the ramp is v2 = v12 +
EVALUATE: 6.81. 2Wtot
2(257.8 J)
= (2.00 m/s) 2 +
= 3.17 m/s.
(85.0 kg)
m The component of gravity down the incline is mg sin 30° = 417 N and the component of the push up the incline is (600 N)cos30° = 520 N . The force component up the incline is greater than the force component down
the incline, the net work done is positive and the speed increases.
IDENTIFY: Apply Wtot = K 2 − K1 to the blocks.
SET UP: If X is the distance the spring is compressed, the work done by the spring is − 1 kX 2 . At maximum
2 compression, the spring (and hence the block) is not moving, so the block has no kinetic energy and x2 = 0 .
EXECUTE: (a) The work done by the block is equal to its initial kinetic energy, and the maximum compression is
5.00 kg
m
2
found from 1 kX 2 = 1 mv0 and X =
(6.00 m/s) = 0.600 m.
v=
2
2
500 N/m
k
(b) Solving for v0 in terms of a known X, v0 =
6.82. 500 N/m
k
(0.150 m) = 1.50 m/s.
X=
5.00 kg
m EVALUATE: The negative work done by the spring removes the kinetic energy of the block.
IDENTIFY: Apply Wtot = K 2 − K1 to the system of the two blocks. The total work done is the sum of that done by
gravity (on the hanging block) and that done by friction (on the block on the table).
SET UP: Let h be the distance the 6.00 kg block descends. The work done by gravity is (6.00 kg)gh and the work
done by friction is − μk (8.00 kg)gh . 6-22 Chapter 6 EXECUTE: 6.83. Wtot = (6.00 kg − (0.25)(8.00 kg)) (9.80 m/s 2 ) (1.50 m) = 58.8 J. This work increases the kinetic energy 2(58.8 J)
1
of both blocks: Wtot = ( m1 + m2 )v 2 , so v =
= 2.90 m/s.
2
(14.00 kg)
EVALUATE: Since the two blocks are connected by the rope, they move the same distance h and have the same
speed v.
IDENTIFY and SET UP: Apply Wtot = K 2 − K1 to the system consisting of both blocks. Since they are connected by
the cord, both blocks have the same speed at every point in the motion. Also, when the 6.00-kg block has moved
downward 1.50 m, the 8.00-kg block has moved 1.50 m to the right. The target variable, μ k , will be a factor in the
work done by friction. The forces on each block are shown in Figure 6.83. EXECUTE: K1 = 1 mAv12 + 1 mB v12 = 1 (mA + mB )v12
2
2
2 K2 = 0 Figure 6.83 The tension T in the rope does positive work on block B and the same magnitude of negative work on block A, so T
does no net work on the system. Gravity does work Wmg = mA gd on block A, where d = 2.00 m. (Block B moves
horizontally, so no work is done on it by gravity.) Friction does work Wfric = − μ k mB gd on block B. Thus
Wtot = Wmg + Wfric = mA gd − μ k mB gd . Then Wtot = K 2 − K1 gives mA gd − μ k mB gd = − 1 (mA + mB )v12 and
2 μk = mA 1 (mA + mB )v12 6.00 kg (6.00 kg + 8.00 kg)(0.900 m/s) 2
+2
=
+
= 0.786
mB
mB gd
8.00 kg 2(8.00 kg)(9.80 m/s 2 )(2.00 m) EVALUATE: The weight of block A does positive work and the friction force on block B does negative work, so the
net work is positive and the kinetic energy of the blocks increases as block A descends. Note that K1 includes the 6.84. kinetic energy of both blocks. We could have applied the work-energy theorem to block A alone, but then Wtot
includes the work done on block A by the tension force.
IDENTIFY: Apply Wtot = K 2 − K1 . The work done by the force from the bow is the area under the graph of Fx versus
the draw length.
SET UP: One possible way of estimating the work is to approximate the F versus x curve as a parabola which goes
2
to zero at x = 0 and x = x0 , and has a maximum of F0 at x = x0 / 2 , so that F ( x) = (4 F0 / x0 ) x( x0 − x). This may
seem like a crude approximation to the figure, but it has the advantage of being easy to integrate.
x0
4 F x0
4 F ⎛ x 2 x3 ⎞ 2
EXECUTE: ∫ Fdx = 20 ∫ ( x0 x − x 2 ) dx = 20 ⎜ x0 0 − 0 ⎟ = F0 x0 . With F0 = 200 N and x0 = 0.75 m,
0
x0 0
x0 ⎝ 2 3 ⎠ 3
W = 100 J. The speed of the arrow is then 6.85. 2W
2(100 J)
=
= 89 m/s .
m
(0.025 kg) EVALUATE: We could alternatively represent the area as that of a rectangle 180 N by 0.55 m. This gives W = 99 J ,
in close agreement with our more elaborate estimate.
IDENTIFY: Apply Eq.(6.6) to the skater.
SET UP: Let point 1 be just before she reaches the rough patch and let point 2 be where she exits from the patch.
Work is done by friction. We don’t know the skater’s mass so can’t calculate either friction or the initial kinetic
energy. Leave her mass m as a variable and expect that it will divide out of the final equation.
EXECUTE: f k = 0.25mg so W f = Wtot = −(0.25mg ) s, where s is the length of the rough patch. Wtot = K 2 − K1 2
2
2
K1 = 1 mv0 , K 2 = 1 mv2 = 1 m(0.45v0 ) 2 = 0.2025 ( 1 mv0 )
2
2
2
2 2
The work-energy relation gives −(0.25mg ) s = ( 0.2025 − 1) 1 mv0
2 The mass divides out, and solving gives s = 1.5 m.
EVALUATE: Friction does negative work and this reduces her kinetic energy. Work and Kinetic Energy 6.86. Pav = F!vav . Use F = ma to calculate the force. IDENTIFY:
SET UP: 6-23 vav = 0 + 6.00 m/s
= 3.00 m/s
2 Your friend’s average acceleration is a = EXECUTE: v − v0 6.00 m/s
=
= 2.00 m/s 2 . Since there are no other
3.00 s
t horizontal forces acting, the force you exert on her is given by Fnet = ma = (65.0 kg)(2.00 m/s 2 ) = 130 N .
Pav = (130 N)(3.00 m/s) = 390 W .
EVALUATE: We could also use the work-energy theorem: W = K 2 − K1 = 1 (65.0 kg)(6.00 m/s)2 = 1170 J .
2 W 1170 J
=
= 390 W , the same as obtained by our other approach.
t
3.00 s
IDENTIFY: To lift a mass m a height h requires work W = mgh . To accelerate mass m from rest to speed v requires
Pav = 6.87. W = K 2 − K1 = 1 mv 2 . Pav =
2 ΔW
.
Δt SET UP: t = 60 s
EXECUTE: (a) (800 kg)(9.80 m/s 2 )(14.0 m) = 1.10 × 105 J
(b) (1/ 2)(800 kg)(18.0 m/s 2 ) = 1.30 × 105 J. 1.10 × 105 J + 1.30 × 105 J
= 3.99 kW.
60 s
EVALUATE: Approximately the same amount of work is required to lift the water against gravity as to accelerate it
to its final speed.
IDENTIFY: P = F!v and F! = ma .
(c) 6.88. SET UP:
EXECUTE: 6.89. From Problem 6.71, v = 2α t + 3β t 2 and a = 2α + 6β t .
P = F!v = mav = m(2α + 6 βt )(2α t + 3 βt 2 ) = m(4α 2t + 18α βt 2 + 18 β 2t 3 ) . P = (0.96 N/s)t + (0.43 N/s 2 )t 2 + (0.043 N/s3 )t 3 . At t = 4.00 s, the power output is 13.5 W.
EVALUATE: P increases in time because v increase and because a increases.
IDENTIFY and SET UP: Energy is Pavt. The total energy expended in one day is the sum of the energy expended in
each type of activity.
EXECUTE: 1 day = 8.64 × 104 s
Let twalk be the time she spends walking and tother be the time she spends in other activities; tother = 8.64 × 104 s − t walk .
The energy expended in each activity is the power output times the time, so
E = Pt = (280 W)t walk + (100 W)tother = 1.1 × 107 J
(280 W)twalk + (100 W)(8.64 × 104 s − twalk ) = 1.1 × 107 J
(180 W)twalk = 2.36 × 106 J
t walk = 1.31 × 104 s = 218 min = 3.6 h. 6.90. 6.91. EVALUATE: Her average power for one day is (1.1 × 107 J)/([24][3600 s]) = 127 W. This is much closer to her
100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate.
IDENTIFY and SET UP: W = Pt
EXECUTE: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must
expend between 0.07 J/beat and 0.175 J/beat.
(b) The steady output of the athlete is (500 W)/(70 kg) = 7 W/kg, which is below the 10 W/kg necessary to stay aloft.
Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could
stay aloft for very long.
EVALUATE: Movies of early attempts at human-powered flight bear out our results.
IDENTIFY and SET UP: Use Eq.(6.15). The work done on the water by gravity is mgh, where h = 170 m. Solve for
the mass m of water for 1.00 s and then calculate the volume of water that has this mass. 6-24 Chapter 6 ΔW
and 92% of the work done on the water
Δt
by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is
P Δt (2.00 × 109 W)(1.00 s)
W = av =
= 2.174 × 109 J
0.92
0.92
W = mgh, so the mass of water flowing over the dam in 1.00 s must be
EXECUTE: m= 2.174 × 109 J
W
=
= 1.30 × 106 kg
gh (9.80 m/s 2 )(170 m) density = 6.92. The power output is Pav = 2000 MW = 2.00 × 109 W. Pav = 1.30 × 106 kg
m
m
so V =
=
= 1.30 × 103 m3 .
V
density 1.00 × 103 kg/m 3 EVALUATE: The dam is 1270 m long, so this volume corresponds to about a m3 flowing over each 1 m length of
the dam, a reasonable amount.
W
dv
and W = 1 mv 2 , if the object starts from rest. a =
and x − x0 = ∫ vdt .
IDENTIFY: P =
2
t
dt
d 1 / 2 1 −1/ 2
2
t = 2t
SET UP:
. ∫ t1 / 2 dt = 3 t 3 / 2 .
dt
EXECUTE:
(b) a = (a) The power P is related to the speed by Pt = K = 1 mv 2 , so v =
2 2 Pt
.
m 2P d
2P 1
dv d 2 Pt
P
.
=
=
t=
=
2mt
dt dt m
m dt
m2t (c) x − x0 = ∫ v dt =
EVALUATE: 2P 1
2P 2 3
8P 3
∫ t 2 dt = m 3 t 2 = 9m t 2 .
m v, a, and x − x0 at a particular time are all proportional to P1 / 2 . The result in part (b) could also be P
.
vm
IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by the heart in one day. For
part (b) use W calculated in part (a) in Eq.(6.15).
EXECUTE: (a) W = mgh, as in Example 6.11. We need the mass of blood lifted; we are given the volume obtained from P = Fv and a = F / m , so a =
6.93. ⎛ 1 × 10−3 m3 ⎞
3
V = (7500 L) ⎜
⎟ = 7.50 m .
1L
⎝
⎠
m = density × volume = (1.05 × 103 kg/m3 )(7.50 m3 ) = 7.875 × 103 kg
Then W = mgh = (7.875 × 103 kg)(9.80 m/s 2 )(1.63 m) = 1.26 × 105 J.
1.26 × 105 J
ΔW
=
= 1.46 W.
(24 h)(3600 s/h)
Δt
EVALUATE: Compared to light bulbs or common electrical devices, the power output of the heart is rather small.
IDENTIFY: P = F!v = Mav . To overcome gravity on a slope that is at an angle α above the horizontal, P = (Mg sin α )v.
(b) Pav =
6.94. SET UP: 1 MW = 106 W . 1 kN = 103 N . When α is small, tan α ≈ sin α .
EXECUTE: (a) The number of cars is the total power available divided by the power needed per car,
13.4 × 106 W
= 177, rounding down to the nearest integer.
(2.8 × 103 N)(27 m/s)
(b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav. To climb a hill of
angle α , the extra power needed is (Mg sin α )v. This will be nearly the same if a ~ g sin α ; if g sin α ~ g tan α ~ 0.10 m/s 2 , the power is about the same as that needed to accelerate at 0.10 m/s 2 .
(c) P = ( Mg sin α )v , where M is the total mass of the diesel units.
P = (1.10 × 106 kg)(9.80 m/s 2 )(0.010)(27 m/s) = 2.9 MW.
(d) The power available to the cars is 13.4 MW, minus the 2.9 MW needed to maintain the speed of the diesel units
13.4 × 106 W − 2.9 × 106 W
on the incline. The total number of cars is then
= 36,
(2.8 × 103 N + (8.2 × 104 kg)(9.80 m/s 2 )(0.010))(27 m/s)
rounding to the nearest integer. Work and Kinetic Energy 6.95. 6-25 EVALUATE: For a single car, Mg sin α = (8.2 × 104 kg)(9.80 m/s 2 )(0.010) = 8.0 × 103 N , which is over twice the
2.8 kN required to pull the car at 27 m/s on level tracks. Even a slope as gradual as 1.0% greatly increases the power
requirements, or for constant power greatly decreases the number of cars that can be pulled.
IDENTIFY: P = F!v . The force required to give mass m an acceleration a is F = ma . For an incline at an angle α above the horizontal, the component of mg down the incline is mg sin α .
SET UP: For small α , sin α ≈ tan α .
EXECUTE: (a) P0 = Fv = (53 × 103 N)(45 m/s) = 2.4 MW.
(b) P = mav = (9.1 × 105 kg)(1.5 m/s 2 )(45 m/s) = 61 MW.
1
(c) Approximating sinα , by tanα , and using the component of gravity down the incline as mgsinα , P2 = (mgsinα )v = (9.1 × 105 kg)(9.80 m/s 2 )(0.015)(45 m/s) = 6.0 MW.
EVALUATE: From Problem 6.94, we would expect that a 0.15 m/s 2 acceleration and a 1.5% slope would require the
same power. We found that a 1.5 m/s 2 acceleration requires ten times more power than a 1.5% slope, which is
consistent.
6.96. IDENTIFY:
SET UP:
EXECUTE: x2 W = ∫ Fx dx , and Fx depends on both x and y.
x1 In each case, use the value of y that applies to the specified path.
(a) Along this path, y is constant, with the value y = 3.00 m . ∫ xdx = 1
2 x2 . ∫ x dx =
2 1
3 x3 x2
⎛ 2.00 m ⎞
W = αy ∫ xdx = (2.50 N/m 2 )(3.00 m) ⎜
⎟ = 15.0 J , since x1 = 0 and x2 = 2.00 m .
x1
⎝2⎠
(b) Since the force has no y-component, no work is done moving in the y-direction.
(c) Along this path, y varies with position along the path, given by y = 1.5 x, so Fx = α (1.5 x) x = 1.5α x 2 , and (2.00 m)3
= 10.0 J.
x1
x1
3
EVALUATE: The force depends on the position of the object along its path.
IDENTIFY and SET UP: Use Eq.(6.18) to relate the forces to the power required. The air resistance force is
Fair = 1 CAρ v 2 , where C is the drag coefficient.
2
x2 x2 W = ∫ Fdx = 1.5α ∫ x 2 dx = 1.5(2.50 N/m 2 ) 6.97. EXECUTE: (a) P = Ftot v, with Ftot = Froll + Fair Fair = CAρ v 2 = 1 (1.0)(0.463 m3 )(1.2 kg/m3 )(12.0 m/s) 2 = 40.0 N
2
1
2 Froll = μr n = μr w = (0.0045)(490 N + 118 N) = 2.74 N
P = ( Froll + Fair )v = (2.74 N + 40.0 N)(12.0 s) = 513 W
(b) Fair = 1 CAρ v 2 = 1 (0.88)(0.366 m3 )(1.2 kg/m3 )(12.0 m/s) 2 = 27.8 N
2
2 Froll = μr n = μr w = (0.0030)(490 N + 88 N) = 1.73 N
P = ( Froll + Fair )v = (1.73 N + 27.8 N)(12.0 s) = 354 W
(c) Fair = 1 CAρ v 2 = 1 (0.88)(0.366 m3 )(1.2 kg/m3 )(6.0 m/s) 2 = 6.96 N
2
2 Froll = μr n = 1.73 N (unchanged)
P = ( Froll + Fair )v = (1.73 N + 6.96 N)(6.0 s) = 52.1 W 6.98. EVALUATE: Since Fair is proportional to v 2 and P = Fv, reducing the speed greatly reduces the power
required.
IDENTIFY: P = F!v 1 m/s = 3.6 km/h
28.0 × 103 W
P
EXECUTE: (a) F = =
= 1.68 × 103 N.
v (60.0 km/h)((1 m/s)/(3.6 km/h))
(b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35/ 4),
and so the power at the lower speed is (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp.
(c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 + 0.35 × 4) = 114.8 kW = 154 hp.
EVALUATE: At low speeds rolling friction dominates the power requirement but at high speeds air resistance
dominates.
SET UP: 6-26 6.99. Chapter 6 IDENTIFY and SET UP: Use Eq.(6.18) to relate F and P. In part (a), F is the retarding force. In parts (b) and (c),
F includes gravity.
EXECUTE: (a) P = Fv, so F = P / v. ⎛ 746 W ⎞
P = (8.00 hp) ⎜
⎟ = 5968 W
⎝ 1 hp ⎠
⎛ 1000 m ⎞⎛ 1 h ⎞
v = (60.0 km/h) ⎜
⎟⎜
⎟ = 16.67 m/s
⎝ 1 km ⎠⎝ 3600 s ⎠
P 5968 W
F= =
= 358 N.
v 16.67 m/s
(b) The power required is the 8.00 hp of part (a) plus the power Pg required to lift the car against gravity. The
situation is sketched in Figure 6.99.
10 m
= 0.10
100 m
α = 5.71°
tan α = Figure 6.99
The vertical component of the velocity of the car is v sin α = (16.67 m/s)sin 5.71° = 1.658 m/s. Then Pg = F (v sin a) = mgv sin α = (1800 kg)(9.80 m/s 2 )(1.658 m/s) = 2.92 × 104 W
⎛ 1 hp ⎞
Pg = 2.92 × 104 W ⎜
⎟ = 39.1 hp
⎝ 746 W ⎠
The total power required is 8.00 hp + 39.1 hp = 47.1 hp.
(c) The power required from the engine is reduced by the rate at which gravity does positive work. The road incline
angle α is given by tan α = 0.0100, so α = 0.5729°.
Pg = mg (v sin α ) = (1800 kg)(9.80 m/s 2 )(16.67 m/s)sin 0.5729° = 2.94 × 103 W = 3.94 hp.
The power required from the engine is then 8.00 hp − 3.94 hp = 4.06 hp.
(d) No power is needed from the engine if gravity does work at the rate of Pg = 8.00 hp = 5968 W Pg
5968 W
=
= 0.02030
mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s)
α = 1.163° and tan α = 0.0203, a 2.03% grade.
EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle
the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force
from the engine is required. The retarding force depends on the speed so it is the same in parts (a), (b), and (c).
IDENTIFY: Apply Wtot = K 2 − K1 to relate the initial speed v0 to the distance x along the plank that the box moves
before coming to rest.
SET UP: The component of weight down the incline is mg sin α , the normal force is mg cos α and the friction force
is f = μ mg cos α .
Pg = mgv sin α , so sin α = 6.100. x EXECUTE: 12
ΔK = 0 − mv0 and W = ∫ ( −mg sin α − μmg cos α )dx. Then,
2
0
x
⎡
⎤
Ax 2
cos α ⎥ .
W = − mg ∫ (sin α + Ax cos α )dx, W = −mg ⎢sin α x +
2
⎣
⎦
0 ⎡
⎤
12
Ax 2
Set W = ΔK : − mv0 = −mg ⎢sin α x +
cos α ⎥ . To eliminate x, note that the box comes to a rest when the force of
2
2
⎣
⎦
static friction balances the component of the weight directed down the plane. So, mg sin α = Ax mg cos α . Solve this for
x and substitute into the previous equation: x = 2
⎡
⎤
sin α
12
sin α
1 ⎛ sin α ⎞
. Then, v0 = + g ⎢sin α
+ A⎜
⎟ cos α ⎥ , and
A cos α
2
A cos α 2 ⎝ A cos α ⎠
⎢
⎥
⎣
⎦ 2
upon canceling factors and collecting terms, v0 = 3g sin 2 α
3g sin 2 α
2
. The box will remain stationary whenever v0 ≥
.
A cos α
A cos α Work and Kinetic Energy EVALUATE:
6.101. 6-27 If v0 is too small the box stops at a point where the friction force is too small to hold the box in place. sin α increases and cos α decreases as α increases, so the v0 required increases as α increases.
IDENTIFY: In part (a) follow the steps outlined in the problem. For parts (b), (c) and (d) apply the work-energy theorem.
SET UP: ∫ x 2 dx = 1 x3
3 EXECUTE: (a) Denote the position of a piece of the spring by l; l = 0 is the fixed point and l = L is the moving end of
the spring. Then the velocity of the point corresponding to l, denoted u, is u (l ) = v(I/L) (when the spring is moving, l
will be a function of time, and so u is an implicit function of time). The mass of a piece of length dl is dm = ( M/L)dl , 1
1 Mv 2 2
Mv 2 L
Mv 2
.
and so dK = (dm)u 2 =
l dl , and K = ∫ dK = 3 ∫ l 2 dl =
3
0
2L
6
2
2L
(b) 1 kx 2 = 1 mv 2 , so v = (k/m) x = (3200 N/m)/(0.053 kg)(2.50 × 10−2 m) = 6.1 m/s.
2
2
(c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball
and the spring, so 1 kx 2 = 1 mv 2 + 1 Mv 2 . Solving for v,
2
2
6 v=
(d) Algebraically, 12
(1/2)kx 2
1
(1/2) kx 2
mv =
= 0.40 J and Mv 2 =
= 0.60 J.
2
(1 + M/3m)
6
(1 + 3m/M ) ⎛ 0.053 kg ⎞
K ball
3m
=
= 3⎜
⎟ = 0.65 . The percentage of the final kinetic energy
K spring M
⎝ 0.243 kg ⎠
that ends up with each object depends on the ratio of the masses of the two objects. As expected, when the mass of the
spring is a small fraction of the mass of the ball, the fraction of the kinetic energy that ends up in the spring is small.
IDENTIFY: In both cases, a given amount of fuel represents a given amount of work W0 that the engine does in
EVALUATE: 6.102. k
(3200 N/m)
(2.50 × 10−2 m) = 3.9 m/s.
x=
(0.053 kg) + (0.243 kg)/3
m + M/3 For this ball and spring, moving the plane forward against the resisting force. Write W0 in terms of the range R and speed v and in terms of the
time of flight T and v.
SET UP: In both cases assume v is constant, so W0 = RF and R = vT .
EXECUTE: β⎞
⎛
In terms of the range R and the constant speed v, W0 = RF = R ⎜ α v 2 + 2 ⎟ .
v⎠
⎝ β⎞
⎛
In terms of the time of flight T , R = vt , so W0 = vTF = T ⎜ α v 3 + ⎟ .
v⎠
⎝
(a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, setting
dW0
dR
dF
dR
dF
= 0 to obtain
F+R
= 0. For the maximum range,
= 0, so
= 0. Performing the differentiation,
dv
dv
dv
dv
dv
dF
= 2α v − 2 β/v3 = 0, which is solved for
dv
14 ⎛β ⎞
v=⎜ ⎟
⎝α ⎠ 14 ⎛ 3.5 × 105 N ⋅ m 2 /s 2 ⎞
=⎜
⎟ = 32.9 m/s = 118 km/h.
2
2
⎝ 0.30 N ⋅ s /m ⎠
d
(b) Similarly, the maximum time is found by setting
( Fv) = 0; performing the differentiation, 3α v 2 − β /v 2 = 0 .
dv
1/ 4 ⎛β⎞
v=⎜
⎟
⎝ 3α ⎠ EVALUATE: 1/ 4 ⎛ 3.5 × 105 N ⋅ m 2 /s 2 ⎞
=⎜
2
2⎟
⎝ 3(0.30 N ⋅ s /m ) ⎠ = 25 m/s = 90 km/h. When v = ( β /α )1/ 4 , Fair has its minimum value Fair = 2 αβ . For this v, R1 = (0.50) T1 = (0.50)α −1/ 4 β −3/ 4 . When v = ( β /3α )1/ 4 , Fair = 2.3 αβ . For this v, R2 = (0.43) 6.103. W0 αβ W0 αβ and and T2 = (0.57)α −1/ 4 β −3/ 4 . R1 > R2 and T2 > T1 , as they should be.
IDENTIFY: For each speed, calculate the time. Then use the graph to find the oxygen consumption and from that the
energy consumption.
SET UP: t = d/v 6-28 Chapter 6 EXECUTE: (a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate
appears to be about 12 cm3 /kg ⋅ min, and so the total energy is (12 cm3/kg ⋅ min) (70 kg) (12 min) (20 J/cm3 ) = 2.0 × 105 J.
(b) The run will take 6 min. Using an estimation of the rate from the graph of about 33 cm3 /kg ⋅ min gives an energy 6.104. consumption of about 2.8 × 105 J.
(c) The run takes 4 min, and with an estimated rate of about 50 cm3 /kg ⋅ min, the energy used is about 2.8 × 105 J.
(d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the
point on the graph is the smallest is the most efficient speed; about 5 km/h.
EVALUATE: In an exercise program, for a fixed distance, running burns more energy than walking.
IDENTIFY: Write equations similar to (6.11) for each component. Eq.(6.12) will now involve the sum of three
integrals, one for each component.
2
2
SET UP: v 2 = vx + v y + vz2
"
"
EXECUTE: From F = ma , Fx = max , Fy = ma y and Fz = maz . The generalization of Eq. (6.11) is then
a x = vx dv
dvx
dv
, a y = v y y , az = vz z . The total work is then
dx
dy
dz y2
z2
dv
⎛ x2 dv
⎞
dv
Fx dx + Fy dy + Fz dz = m ⎜ ∫ vx x dx + ∫ v y y dy + ∫ vz z dz ⎟ .
x1
y1
z1
dx
dy
dz ⎠
⎝
vx 2
vy 2
vz 2
1
1 21
2
2
2
2
Wtot = m ⎛ ∫ vx dvx + ∫ v y dv y + ∫ vz dvz ⎞ = m(vx 2 − vx1 + v y 2 − v y1 + vz22 − vz21 = mv2 − mv12 .
⎜v
⎟
vy1
vz 1
x1
⎝
⎠2
2
2
"
"
EVALUATE: F and dl are vectors and have components. W and K are scalars and we never speak of their
components. Wtot = ∫ ( x2 , y2 , z2 ) ( x1 , y1 , z1 ) POTENTIAL ENERGY AND ENERGY CONSERVATION 7.1. U grav = mgy so ΔU grav = mg ( y2 − y1 ) IDENTIFY:
SET UP: 7 + y is upward. EXECUTE: (a) ΔU = (75 kg)(9.80 m/s 2 )(2400 m − 1500 m) = +6.6 × 105 J (b) ΔU = (75 kg)(9.80 m/s 2 )(1350 m − 2400 m) = −7.7 × 105 J
EVALUATE:
7.2. 7.3. IDENTIFY: U grav increases when the altitude of the object increases.
!
!
Apply ∑ F = ma to the sack to find the force. W = Fs cos φ . SET UP: The lifting force acts in the same direction as the sack’s motion, so φ = 0°
EXECUTE: (a) For constant speed, the net force is zero, so the required force is the sack’s weight,
(5.00 kg)(9.80 m/s 2 ) = 49.0 N.
(b) W = (49.0 N) (15.0 m) = 735 J . This work becomes potential energy.
EVALUATE: The results are independent of the speed.
IDENTIFY: Use the free-body diagram for the bag and Newton's first law to find the force the worker applies.
Since the bag starts and ends at rest, K 2 − K1 = 0 and Wtot = 0 . 2.0 m
A sketch showing the initial and final positions of the bag is given in Figure 7.3a. sin φ =
and
3.5 m
!
φ = 34.85° . The free-body diagram is given in Figure 7.3b. F is the horizontal force applied by the worker. In the
calculation of U grav take + y upward and y = 0 at the initial position of the bag.
SET UP: EXECUTE: (a) ∑F y = 0 gives T cos φ = mg and ∑F x = 0 gives F = T sin φ . Combining these equations to eliminate T gives F = mg tan φ = (120 kg)(9.80 m/s ) tan 34.85° = 820 N .
(b) (i) The tension in the rope is radial and the displacement is tangential so there is no component of T in the
direction of the displacement during the motion and the tension in the rope does no work. (ii) Wtot = 0 so
2 Wworker = −Wgrav = U grav,2 − U grav,1 = mg ( y2 − y1 ) = (120 kg)(9.80 m/s 2 )(0.6277 m) = 740 J .
EVALUATE: The force applied by the worker varies during the motion of the bag and it would be difficult to
calculate Wworker directly. Figure 7.3
7.4. IDENTIFY: Only gravity does work on him from the point where he has just left the board until just before he
enters the water, so Eq.(7.4) applies.
SET UP: Let point 1 be just after he leaves the board and point 2 be just before he enters the water. + y is upward
and y = 0 at the water.
7-1 7-2 Chapter 7 EXECUTE: 2
(a) K1 = 0 . y2 = 0 . y1 = 3.25 m . K1 + U grav,1 = K 2 + U grav,2 gives U grav,1 = K 2 and mgy1 = 1 mv2 .
2 v2 = 2 gy1 = 2(9.80 m/s 2 )(3.25 m) = 7.98 m/s .
(b) v1 = 2.50 m/s , y2 = 0 , y1 = 3.25 m . K1 + U grav,1 = K 2 and 1
2 2
mv12 + mgy1 = 1 mv2 .
2 v2 = v12 + 2 gy1 = (2.50 m/s) 2 + 2(9.80 m/s 2 )(3.25 m) = 8.36 m/s . 7.5. (c) v1 = 2.5 m/s and v2 = 8.36 m/s , the same as in part (b).
EVALUATE: Kinetic energy depends only on the speed, not on the direction of the velocity.
IDENTIFY and SET UP: Use energy methods.
2
(a) K1 + U1 + Wother = K 2 + U 2 . Solve for K 2 and then use K 2 = 1 mv2 to obtain v2 .
2 Wother = 0 (The only force on the
ball while it is in the air is gravity.)
2
K1 = 1 mv12 ; K 2 = 1 mv2
2
2
U1 = mgy1 , y1 = 22.0 m
U 2 = mgy2 = 0, since y2 = 0
for our choice of coordinates.
Figure 7.5
EXECUTE: 1
2 2
mv + mgy1 = 1 mv2
2
2
1 v2 = v12 + 2 gy1 = (12.0 m/s) 2 + 2(9.80 m/s 2 )(22.0 m) = 24.0 m/s
EVALUATE: The projection angle of 53.1° doesn’t enter into the calculation. The kinetic energy depends only on
the magnitude of the velocity; it is independent of the direction of the velocity.
(b) Nothing changes in the calculation. The expression derived in part (a) for v2 is independent of the angle, so 7.6. v2 = 24.0 m/s, the same as in part (a).
(c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect.
IDENTIFY: The normal force does no work, so only gravity does work and Eq.(7.4) applies.
SET UP: K1 = 0 . The crate’s initial point is at a vertical height of d sin α above the bottom of the ramp.
EXECUTE: 2
(a) y2 = 0, y1 = d sin α . K1 + U grav,1 = K 2 + U grav,2 gives U grav,1 = K 2 . mgd sin α = 1 mv2 and
2 v2 = 2 gd sin α .
2
(b) y1 = 0 , y2 = − d sin α . K1 + U grav,1 = K 2 + U grav,2 gives 0 = K 2 + U grav,2 . 0 = 1 mv2 + (− mgd sin α ) and
2 v2 = 2 gd sin α , the same as in part (a).
(c) The normal force is perpendicular to the displacement and does no work.
EVALUATE: When we use U grav = mgy we can take any point as y = 0 but we must take + y to be upward.
7.7. IDENTIFY:
SET UP: As in Example 7.6, K 2 = 0, U 2 = 94 J, and U 3 = 0. EXECUTE: 7.8. Apply Eq.(7.7) to points 2 and 3. Take results from Example 7.6. Wother = − fs, the work done by friction. The work done by friction is −(35 N) (1.6 m) = −56 J . K 3 = 38 J, and v3 = 2(38 J)
= 2.5 m/s.
12 kg EVALUATE: The value of v3 we obtained is the same as calculated in Example 7.6. For the motion from point 2 to
point 3, gravity does positive work, friction does negative work and the net work is positive.
IDENTIFY and SET UP: Apply Eq.(7.7) and consider how each term depends on the mass.
EXECUTE: The speed is v and the kinetic energy is 4K. The work done by friction is proportional to the normal
force, and hence to the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the
speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same
speed but four times the mass, the kinetic energy is quadrupled.
!
!
EVALUATE: The same result is obtained if we apply ∑ F = ma to the motion. Each force is proportional to m and m divides out, so a is independent of m. Potential Energy and Energy Conservation 7.9. IDENTIFY:
SET UP: 7-3 Wtot = K B − K A . The forces on the rock are gravity, the normal force and friction. Let y = 0 at point B and let + y be upward. y A = R = 0.50 m . The work done by friction is negative; W f = −0.22 J . K A = 0 . The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of
the rock at this point is arad = v 2 / R , upward.
EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work.
(ii) Wgrav = U grav, A − U grav, B = mgy A = (0.20 kg)(9.80 m/s 2 )(0.50 m) = 0.98 J .
(b) Wtot = Wn + W f + Wgrav = 0 + (−0.22 J) + 0.98 J = 0.76 J . Wtot = K B − K A gives vB = 1
2 2
mvB = Wtot . 2Wtot
2(0.76 J)
=
= 2.8 m/s .
m
0.20 kg (c) Gravity is constant and equal to mg. n is not constant; it is zero at A and not zero at B. Therefore, f k = μ k n is
also not constant.
(d) ∑ Fy = ma y applied to Figure 7.9 gives n − mg = marad . ⎛
⎛
[2.8 m/s]2 ⎞
v2 ⎞
n = m ⎜ g + ⎟ = (0.20 kg) ⎜ 9.80 m/s 2 +
⎟ = 5.1 N .
0.50 m ⎠
R⎠
⎝
⎝
EVALUATE: In the absence of friction, the speed of the rock at point B would be 2 gR = 3.1 m/s . As the rock
slides through point B, the normal force is greater than the weight mg = 2.0 N of the rock. Figure 7.9
7.10. IDENTIFY: Only gravity does work, so Eq.(7.4) applies.
SET UP: Let point 1 be just after the rock leaves the thrower and point 2 be at the maximum height. Let
y1 = 0 and + y be upward. v1 = v0 . At the highest point, v2 = v0 cosθ . sin 2 θ + cos 2 θ = 1 .
EXECUTE: 7.11. K1 + U grav,1 = K 2 + U grav,2 gives 1
2 2
mv0 = 1 m(v0 cosθ ) 2 + mgy2 . y2 =
2 2
v0
v 2 sin 2 θ
(1 − cos 2 θ ) = 0
, was to
2g
2g be shown.
EVALUATE: The initial kinetic energy is independent of the angle θ but the kinetic energy at the maximum
height depends on θ , so the maximum height depends on θ .
IDENTIFY: Apply Eq.(7.7) to the motion of the car.
SET UP: Take y = 0 at point A. Let point 1 be A and point 2 be B.
K1 + U1 + Wother = K 2 + U 2
EXECUTE: U1 = 0, U 2 = mg (2 R ) = 28,224 J, Wother = W f 2
K1 = 1 mv = 37,500 J, K 2 = 1 mv2 = 3840 J
2
2
2
1 The work-energy relation then gives W f = K 2 + U 2 − K1 = −5400 J.
EVALUATE:
7.12. Friction does negative work. The final mechanical energy ( K 2 + U 2 = 32,064 J) is less than the initial mechanical energy ( K1 + U1 = 37,500 J) because of the energy removed by friction work.
IDENTIFY: Only gravity does work, so apply Eq.(7.5).
2
SET UP: v1 = 0 , so 1 mv2 = mg ( y1 − y2 ) .
2
EXECUTE: Tarzan is lower than his original height by a distance y1 − y2 = l (cos30° − cos 45°) so his speed is v = 2 gl (cos30°− cos 45°) = 7.9 m/s, a bit quick for conversation.
EVALUATE: The result is independent of Tarzan’s mass. 7-4 Chapter 7 7.13. y1 = 0
y2 = (8.00 m)sin 36.9°
y2 = 4.80 m
Figure 7.13a
!
(a) IDENTIFY and SET UP: F is constant so Eq.(6.2) can be used. The situation is sketched in Figure 7.13a.
EXECUTE: WF = ( F cos φ ) s = (110 N)(cos0°)(8.00 m) = 880 J
!
EVALUATE: F is in the direction of the displacement and does positive work.
(b) IDENTIFY and SET UP: Calculate W using Eq.(6.2) but first must calculate the friction force. Use the freebody diagram for the oven sketched in Figure 7.13b to calculate the normal force n; then the friction force can be
calculated from f k = μ k n. For this calculation use coordinates parallel and perpendicular to the incline. EXECUTE: ∑F y = ma y n − mg cos36.9° = 0
n = mg cos36.9°
f k = μ k n = μ k mg cos36.9°
f k = (0.25)(10.0 kg)(9.80 m/s 2 )cos36.9° = 19.6 N
Figure 7.13b W f = ( f k cos φ ) s = (19.6 N)(cos180°)(8.00 m) = −157 J
EVALUATE: Friction does negative work.
(c) IDENTIFY and SET UP: U = mgy; take y = 0 at the bottom of the ramp.
EXECUTE: ΔU = U 2 − U1 = mg ( y2 − y1 ) = (10.0 kg)(9.80 m/s 2 )(4.80 m − 0) = 470 J
EVALUATE: The object moves upward and U increases.
(d) IDENTIFY and SET UP: Use Eq.(7.7). Solve for ΔK .
EXECUTE: K1 + U1 + Wother = K 2 + U 2 ΔK = K 2 − K1 = U1 − U 2 + Wother
ΔK = Wother − ΔU
Wother = WF + W f = 880 J − 157 J = 723 J
ΔU = 470 J
Thus ΔK = 723 J − 470 J = 253 J.
EVALUATE: Wother is positive. Some of Wother goes to increasing U and the rest goes to increasing K.
!
!
!
(e) IDENTIFY: Apply ∑ F = ma to the oven. Solve for a and then use a constant acceleration equation to
calculate v2 .
SET UP: We can use the free-body diagram that is in part (b):
∑ Fx = max
F − f k − mg sin 36.9° = ma
EXECUTE:
SET UP: a= F − f k − mg sin 36.9° 110 N − 19.6 N − (10 kg)(9.80 m/s 2 )sin 36.9°
= 3.16 m/s 2
=
10.0 kg
m v1x = 0, ax = 3.16 m/s 2 , x − x0 = 8.00 m, v2 x = ? 2
v2 x = v12x + 2ax ( x − x0 ) EXECUTE: v2 x = 2ax ( x − x0 ) = 2(3.16 m/s 2 )(8.00 m) = 7.11 m/s 2 2
Then ΔK = K 2 − K1 = 1 mv2 = 1 (10.0 kg)(7.11 m/s) 2 = 253 J.
2
2
EVALUATE: This agrees with the result calculated in part (d) using energy methods. Potential Energy and Energy Conservation 7.14. IDENTIFY:
SET UP: 7-5 !
!
Only gravity does work, so apply Eq.(7.4). Use ∑ F = ma to calculate the tension. Let y = 0 at the bottom of the arc. Let point 1 be when the string makes a 45° angle with the vertical and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration arad = v 2 / r
EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the
bottom of the circular arc) is mgl (1 − cos θ ), where l is the length of the string and θ is the angle the string makes
with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so
mgl (1 − cos θ ) = 1 mv 2 , or v = 2 gl (1 − cos θ ) = 2(9.80 m/s 2 ) (0.80 m) (1 − cos 45°) = 2.1 m/s .
2
(b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial
component of the weight, or mg cosθ = (0.12 kg) (9.80 m/s 2 ) cos 45° = 0.83 N.
(c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration,
2
mg + mv2 l = mg (1 + 2(1 − cos 45°)) = 1.9 N 7.15. EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the
acceleration is upward.
IDENTIFY: Apply U el = 1 kx 2 .
2
SET UP: kx = F , so U = 1 Fx ,where F is the magnitude of force required to stretch or compress the spring a
2
distance x.
EXECUTE: (a) (1 2)(800 N)(0.200 m) = 80.0 J.
(b) The potential energy is proportional to the square of the compression or extension;
(80.0 J) (0.050 m 0.200 m ) 2 = 5.0 J. F
800 N
=
= 4000 N/m and then used U el = 1 kx 2 directly.
2
x 0.200 m
1
IDENTIFY: Use the information given in the problem with F = kx to find k. Then U el = 2 kx 2 .
SET UP: x is the amount the spring is stretched. When the weight is hung from the spring, F = mg .
EVALUATE: 7.16. EXECUTE: We could have calculated k = k= F mg (3.15 kg)(9.80 m/s 2 )
=
=
= 2205 N/m .
x
x
0.1340 m − 0.1200 m 2U el
2(10.0 J)
=±
= ±0.0952 m = ±9.52 cm . The spring could be either stretched 9.52 cm or
k
2205 N/m
compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm + 9.52 cm = 21.52 cm .
If it were compressed, the total length of the spring would be 12.00 cm − 9.52 cm = 2.48 cm .
EVALUATE: To stretch or compress the spring 9.52 cm requires a force F = kx = 210 N .
IDENTIFY: Apply U el = 1 kx 2 .
2
x=± 7.17. SET UP:
EXECUTE: 2
U 0 = 1 kx0 . x is the distance the spring is stretched or compressed.
2
2
(a) (i) x = 2 x0 gives U el = 1 k (2 x0 ) 2 = 4( 1 kx0 ) = 4U 0 . (ii) x = x0 / 2 gives
2
2 2
U el = 1 k ( x0 / 2) 2 = 1 ( 1 kx0 ) = U 0 / 4 .
2
42 (b) (i) U = 2U 0 gives
7.18. 1
2 2
kx 2 = 2( 1 kx0 ) and x = x0 2 . (ii) U = U 0 / 2 gives
2 1
2 2
kx 2 = 1 ( 1 kx0 ) and x = x0 / 2 .
22 EVALUATE: U is proportional to x 2 and x is proportional to U .
IDENTIFY: Apply Eq.(7.13).
SET UP: Initially and at the highest point, v = 0 , so K1 = K 2 = 0 . Wother = 0 .
EXECUTE: (a) In going from rest in the slingshot’s pocket to rest at the maximum height, the potential energy
stored in the rubber band is converted to gravitational potential energy;
U = mgy = (10 × 10−3 kg)(9.80 m/s 2 ) (22.0 m) = 2.16 J.
(b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m.
(c) The lack of air resistance and no deformation of the rubber band are two possible assumptions.
EVALUATE: The potential energy stored in the rubber band depends on k for the rubber band and the maximum
distance it is stretched. 7-6 7.19. Chapter 7 IDENTIFY and SET UP: Use energy methods. There are changes in both elastic and gravitational potential energy;
elastic; U = 1 kx 2 , gravitational: U = mgy.
2 2U
2(3.20 J)
=
= 0.0632 m = 6.32 cm
k
1600 N/m
(b) Points 1 and 2 in the motion are sketched in Figure 7.19.
K1 + U1 + Wother = K 2 + U 2
EXECUTE: (a) U = 1 kx 2 so x =
2 Wother = 0 (Only work is that done
by gravity and spring force)
K1 = 0, K 2 = 0
y = 0 at final position of book
U1 = mg ( h + d ) , U 2 = 1 kd 2
2
Figure 7.19 0 + mg (h + d ) + 0 = 1 kd 2
2
The original gravitational potential energy of the system is converted into potential energy of the compressed
spring.
1
kd 2 − mgd − mgh = 0
2 ⎞
1⎛
⎛1 ⎞
d = ⎜ mg ± (mg ) 2 + 4 ⎜ k ⎟ (mgh) ⎟
⎜
⎟
k⎝
⎝2 ⎠
⎠
1
d must be positive, so d = mg + (mg ) 2 + 2kmgh
k
1
d=
((1.20 kg)(9.80 m/s 2 ) +
1600 N/m ( ) ((1.20 kg)(9.80 m/s 2 )) 2 + 2(1600 N/m)(1.20 kg)(9.80 m/s 2 )(0.80 m)
d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm 7.20. EVALUATE: It was important to recognize that the total displacement was h + d ; gravity continues to do work as
the book moves against the spring. Also note that with the spring compressed 0.12 m it exerts an upward force
(192 N) greater than the weight of the book (11.8 N). The book will be accelerated upward from this position.
IDENTIFY: Use energy methods. There are changes in both elastic and gravitational potential energy.
SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.20. The spring force and gravity are the
only forces doing work on the cheese,
so Wother = 0 and U = U grav + U el . Figure 7.20
EXECUTE: Cheese released from rest implies K1 = 0. At the maximum height v2 = 0 so K 2 = 0.
U1 = U1,el + U1, grav
y1 = 0 implies U1,grav = 0
U1,el = 1 kx12 = 1 (1800 N/m)(0.15 m) 2 = 20.25 J
2
2
(Here x1 refers to the amount the spring is stretched or compressed when the cheese is at position 1; it is not the
x-coordinate of the cheese in the coordinate system shown in the sketch.)
U 2 = U 2,el + U 2,grav Potential Energy and Energy Conservation 7-7 U 2,grav = mgy2 , where y2 is the height we are solving for. U 2, el = 0 since now the spring is no longer compressed.
Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el = U 2,grav y2 = 7.21. 20.25 J
20.25 J
=
= 1.72 m
mg
(1.20 kg)(9.80 m/s 2 ) EVALUATE: The description in terms of energy is very simple; the elastic potential energy originally stored in the
spring is converted into gravitational potential energy of the system.
IDENTIFY: Apply Eq.(7.13).
SET UP: Wother = 0 . As in Example 7.7, K1 = 0 and U1 = 0.0250 J.
EXECUTE: For v2 = 0.20 m s, K 2 = 0.0040 J . U 2 = 0.0210 J = 1 kx 2 , and x = ±
2 2(0.0210 J)
= ±0.092 m. The
5.00 N m glider has this speed when the spring is stretched 0.092 m or compressed 0.092 m.
EVALUATE: Example 7.7 showed that vx = 0.30 m/s when x = 0.0800 m . As x increases, vx decreases, so our
result of vx = 0.20 m/s at x = 0.092 m is consistent with the result in the example.
7.22. IDENTIFY and SET UP: Use energy methods. The elastic potential energy changes. In part (a) solve for K 2 and from this obtain v2 . In part (b) solve for U1 and from this obtain x1.
(a) K1 + U1 + Wother = K 2 + U 2 point 1: the glider is at its initial position, where x1 = 0.100 m and v1 = 0
point 2: the glider is at x = 0
2
EXECUTE: K1 = 0 (released from rest), K 2 = 1 mv2
2
U1 = 1 kx12 , U 2 = 0, Wother = 0 (only the spring force does work)
2
2
Thus 1 kx12 = 1 mv2 . (The initial potential energy of the stretched spring is converted entirely into kinetic energy of
2
2
the glider.) v2 = x1 k
5.00 N/m
= (0.100 m)
= 0.500 m/s
m
0.200 kg (b) The maximum speed occurs at x = 0, so the same equation applies.
1
2 x1 = v2
EVALUATE:
7.23. IDENTIFY: 2
kx12 = 1 mv2
2 m
0.200 kg
= 2.50 m/s
= 0.500 m
k
5.00 N/m Elastic potential energy is converted into kinetic energy. A larger x1 gives a larger v2 .
Only the spring does work and Eq.(7.11) applies. a = F − kx
=
, where F is the force the spring exerts
m
m on the mass.
SET UP: Let point 1 be the initial position of the mass against the compressed spring, so K1 = 0 and U1 = 11.5 J .
Let point 2 be where the mass leaves the spring, so U el,2 = 0 .
EXECUTE: (a) K1 + U el,1 = K 2 + U el,2 gives U el,1 = K 2 . 1
2 2
mv2 = U el,1 and v2 = 2U el,1
m = 2(11.5 J)
= 3.03 m/s .
2.50 kg K is largest when U el is least and this is when the mass leaves the spring. The mass achieves its maximum speed of
3.03 m/s as it leaves the spring and then slides along the surface with constant speed.
(b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has its
2U el
2(11.5 J)
maximum compression. U el = 1 kx 2 so x = −
=−
= −0.0959 m . The minus sign indicates
2
k
2500 N/m
kx
(2500 N/m)(−0.0959 m)
compression. F = − kx = max and ax = − = −
= 95.9 m/s 2 .
m
2.50 kg
EVALUATE: If the end of the spring is displaced to the left when the spring is compressed, then ax in part (b) is to
the right, and vice versa. 7-8 7.24. Chapter 7 (a) IDENTIFY and SET UP: Use energy methods. Both elastic and gravitational potential energy changes. Work is
done by friction.
Choose point 1 as in Example 7.9 and let that be the origin, so y1 = 0. Let point 2 be 1.00 m below point 1, so y2 = −1.00 m.
K1 + U1 + Wother = K 2 + U 2 EXECUTE: K1 = 1 mv = 1 (2000 kg)(25 m/s) 2 = 625,000 J, U1 = 0
2
2
2
1 Wother = − f y2 = −(17,000 N)(1.00 m) = −17,000 J
2
K 2 = 1 mg 2
2
2
U 2 = U 2,grav + U 2,el = mgy2 + 1 ky2
2 U 2 = (2000 kg)(9.80 m/s 2 )(−1.00 m) + 1 (1.41 × 105 N/m)(1.00 m)2
2
U 2 = −19,600 J + 70,500 J = +50,900 J
2
Thus 625,000 J − 17,000 J = 1 mv2 + 50,900 J
2
1
2 2
mv2 = 557,100 J v2 = 2(557,100 J)
= 23.6 m/s
2000 kg EVALUATE: The elevator stops after descending 3.00 m. After descending 1.00 m it is still moving but has
slowed down.
!
!
!
(b) IDENTIFY: Apply ∑ F = ma to the elevator. We know the forces and can solve for a . The free-body diagram for the elevator is given in Figure 7.24. SET UP: EXECUTE: Fspr = kd , where d is the distance the spring is compressed
∑ Fy = ma y
f k + Fspr − mg = ma
f k + kd − mg = ma
Figure 7.24 a= 7.25. f k + kd − mg 17,000 N + (1.41 × 105 N/m)(1.00 m) − (2000 kg)(9.80 m/s 2 )
=
= 69.2 m/s 2
m
2000 kg We calculate that a is positive, so the acceleration is upward.
EVALUATE: The velocity is downward and the acceleration is upward, so the elevator is slowing down at this
point. Note that a = 7.1g ; this is unacceptably high for an elevator.
IDENTIFY: Apply Eq.(7.13) and F = ma .
SET UP: Wother = 0 . There is no change in U grav . K1 = 0 , U 2 = 0 .
EXECUTE: 1
2 2
2
kx 2 = 1 mvx . The relations for m, vx , k and x are kx 2 = mvx and kx = 5mg .
2 Dividing the first equation by the second gives x =
(a) k = 25
(b) x = 2
vx
mg 2
, and substituting this into the second gives k = 25 2 .
vx
5g (1160 kg)(9.80 m/s 2 ) 2
= 4.46 × 105 N/m
(2.50 m/s) 2 (2.50 m/s) 2
= 0.128 m
5(9.80 m/s 2 ) EVALUATE: Our results for k and x do give the required values for ax and vx :
ax = kx (4.46 × 105 N/m)(0.128 m)
k
=
= 49.2 m/s 2 = 5.0 g and vx = x
= 2.5 m/s .
m
1160 kg
m Potential Energy and Energy Conservation 7.26. IDENTIFY: 7-9 Wgrav = mg cos φ . SET UP: When he moves upward, φ = 180° and when he moves downward, φ = 0° . When he moves parallel to
the ground, φ = 90° .
EXECUTE: (a) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos180° = −5100 J . (b) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos0° = +5100 J .
(c) φ = 90° in each case and Wgrav = 0 in each case. 7.27. (d) The total work done on him by gravity during the round trip is −5100 J + 5100 J = 0 .
(e) Gravity is a conservative force since the total work done for a round trip is zero.
EVALUATE: The gravity force is independent of the position and motion of the object. When the object moves
upward gravity does negative work and when the object moves downward gravity does positive work.
IDENTIFY: Apply W fk = f k s cos φ . f k = μ k n .
SET UP: For a circular trip the distance traveled is d = 2π r . At each point in the motion the friction force and the
displacement are in opposite directions and φ = 180° . Therefore, W fk = − f k d = − f k (2π r ) . n = mg so f k = μ k mg .
EXECUTE: 7.28. (a) W fk = − μ k mg 2π r = −(0.250)(10.0 kg)(9.80 m/s 2 )(2π )(2.00 m) = −308 J . (b) The distance along the path doubles so the work done doubles and becomes −616 J .
(c) The work done for a round trip displacement is not zero and friction is a nonconservative force.
EVALUATE: The direction of the friction force depends on the direction of motion of the object and that is why
friction is a nonconservative force.
IDENTIFY and SET UP: The force is not constant so we must use Eq.(6.14) to calculate W. The properties of work
done by a conservative force are described in Section 7.3.
!!
2!
ˆ
W = ∫ F ⋅ dl , F = −α x 2 i
1 !
(a) dl = dyˆ (x is constant; the displacement is in the + y -direction )
j EXECUTE:
!!
ˆj
F ⋅ dl = 0 (since i ⋅ ˆ = 0) and thus W = 0.
!
ˆ
(b) dl = dxi
!!
ˆ
ˆ
F ⋅ dl = (−α x 2 i ) ⋅ ( dxi ) = −α x 2 dx
x2 W = ∫ (−α x 2 )dx = − 1 ax3
3
x1 x2
x1 3
= − 1 α ( x2 − x13 ) = −
3 12 N/m 2
((0.300 m)3 − (0.10 m)3 ) = −0.10 J
3 !
ˆ
(c) dl = dxi as in part (b), but now x1 = 0.30 m and x2 = 0.10 m 3
W = − 1 α ( x2 − x13 ) = +0.10 J
3 (d) EVALUATE: The total work for the displacement along the x-axis from 0.10 m to 0.30 m and then back to
0.10 m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when the starting and
ending points are the same, so the force is conservative.
3
3
EXECUTE: Wx 1 → x2 = − 1 α ( x2 − x13 ) = 1 α x13 − 1 α x2
3
3
3 The definition of the potential energy function is Wx1 → x2 = U1 − U 2 . Comparison of the two expressions for W gives 7.29. U = 1 α x3 . This does correspond to U = 0 when x = 0.
3
EVALUATE: In part (a) the work done is zero because the force and displacement are perpendicular. In part (b)
the force is directed opposite to the displacement and the work done is negative. In part (c) the force and
displacement are in the same direction and the work done is positive.
IDENTIFY: Since the force is constant, use W = Fs cos φ .
SET UP: For both displacements, the direction of the friction force is opposite to the displacement and φ = 180° .
EXECUTE: (a) When the book moves to the left, the friction force is to the right, and the work is
−(1.2 N)(3.0 m) = −3.6 J.
(b) The friction force is now to the left, and the work is again −3.6 J.
(c) −7.2 J.
(d) The net work done by friction for the round trip is not zero, and friction is not a conservative force.
EVALUATE: The direction of the friction force depends on the motion of the object. For the gravity force, which
is conservative, the force does not depend on the motion of the object. 7-10 7.30. Chapter 7 IDENTIFY and SET UP: The friction force is constant during each displacement and Eq.(6.2) can be used to
calculate work, but the direction of the friction force can be different for different displacements.
!
f = μ k mg = (0.25)(1.5 kg)(9.80 m/s 2 ) = 3.675 N; direction of f is opposite to the motion.
EXECUTE: (a) The path of the book is sketched in Figure 7.30a. Figure 7.30a !
For the motion from you to Beth the friction force is directed opposite to the displacement s and
W1 = − fs = −(3.675 N)(8.0 m) = −29.4 J.
For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and
W2 = −29.4 J.
Wtot = W1 + W2 = −29.4 J − 29.4 J = −59 J
(b) The path of the book is sketched in Figure 7.30b. s = 2(8.0 m) 2 = 11.3 m Figure 7.30b
!
!
f is opposite to s , so W = − fs = −(3.675 N)(11.3 m) = −42 J
(c) For the motion from you to Kim
W = − fs
W = −(3.675 N)(8.0 m) = −29.4 J
Figure 7.30c For the motion from Kim
to you
W = − fs = −29.4 J
Figure 7.30d 7.31. The total work for the round trip is −29.4 J − 29.4 J = −59 J.
(d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by
friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero.
Both these results show that the friction force is nonconservative.
IDENTIFY: The work done by a spring on an object attached to its end when the object moves from xi to xf is
W = 1 kxi2 − 1 kxf2 . This result holds for any xi and xf .
2
2
SET UP: Assume for simplicity that x1 , x2 and x3 are all positive, corresponding to the spring being stretched. EXECUTE: (a) 1
2 2
k ( x12 − x2 ) 2
(b) − 1 k ( x12 − x2 ). The total work is zero; the spring force is conservative.
2
2
2
2
2
(c) From x1 to x3 , W = − 1 k ( x3 − x12 ). From x3 to x2 , W = − 1 k ( x2 − x3 ). The net work is − 1 k ( x2 − x12 ). This is
2
2
2
the same as the result of part (a).
EVALUATE: The results of part (c) illustrate that the work done by a conservative force is path independent. Potential Energy and Energy Conservation 7.32. 7.33. 7-11 IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U ( x ). Use coordinates where the origin is at
one atom. The other atom then has coordinate x.
EXECUTE:
dU
d ⎛ C6 ⎞
d⎛1⎞
6C
Fx = −
= − ⎜ − 6 ⎟ = +C6 ⎜ 6 ⎟ = − 76
dx
dx ⎝ x ⎠
dx ⎝ x ⎠
x The minus sign mean that Fx is directed in the − x -direction, toward the origin. The force has magnitude 6C6 / x 7
and is attractive.
!
EVALUATE: U depends only on x so F is along the x-axis; it has no y or z components.
IDENTIFY: Apply Eq.(7.16).
SET UP: The sign of Fx indicates its direction.
Fx = EXECUTE: dU
4
4
= −4α x3 = −(4.8 J m ) x3 . Fx ( −0.800 m) = −(4.8 J m )(−0.80 m)3 = 2.46 N. The force is
dx in the + x -direction.
EVALUATE: Fx > 0 when x < 0 and Fx < 0 when x > 0 , so the force is always directed towards the origin.
7.34. Apply F ( x) = − IDENTIFY: dU ( x )
.
dx d (1/ x)
1
=− 2
dx
x
d (−Gm1m2 / x )
Gm1m2
⎡ d (1/ x) ⎤
= Gm1m2 ⎢
EXECUTE: Fx ( x) = −
⎥ = − x 2 . The force on m2 is in the − x-direction . This
dx
⎣ dx ⎦
SET UP: is toward m1 , so the force is attractive. 7.35. EVALUATE: By Newton's 3rd law the force on m1 due to m2 is Gm1m2 / x 2 , in the + x -direction (toward m2 ). The
gravitational potential energy belongs to the system of the two masses.
∂U
∂U
and Fy = −
IDENTIFY: Apply Fx = −
.
∂x
∂y
SET UP: r = ( x 2 + y 2 )1/ 2 . EXECUTE: Fy = − (a) U (r ) = − ∂ (1/ r )
x
∂ (1/ r )
y
=− 2
and
=− 2
.
2 3/ 2
∂x
(x + y )
∂y
( x + y 2 )3 / 2 ∂U
Gm1m2 x
Gm1m2
⎡ ∂ (1/ r ) ⎤
= +Gm1m2 ⎢
. Fx = −
⎥ = − ( x 2 + y 2 )3 / 2 and
r
∂x
⎣ ∂x ⎦ ⎡ ∂ (1/ r ) ⎤
∂U
Gm1m2 y
= +Gm1m2 ⎢
.
⎥=− 2
( x + y 2 )3 / 2
∂y
∂y ⎦
⎣ Gm1m2 x
Gm1m2 y
Gm1m2 2
Gm1m2
and Fy = −
. F = Fx2 + Fy2 =
x + y2 =
.
3
3
3
r
r
r
r2
!
!
(c) Fx and Fy are negative. Fx = α x and Fy = α y , where α is a constant, so F and the vector r from m1 to m2 are
!
!
in the same direction. Therefore, F is directed toward m1 at the origin and F is attractive.
(b) ( x 2 + y 2 )3 / 2 = r 3 so Fx = − !
x
y
If θ is the angle between the vector r that points from m1 to m2 , then = cosθ and = sin θ . This
r
r
gives Fx = − F cosθ and Fy = − F sin θ , our more usual way of writing the components of a vector.
EVALUATE: 7.36. Apply Eq.(7.18).
d⎛1⎞
2
d⎛1⎞
2
SET UP:
⎜ ⎟=− 3 .
⎜ ⎟ = − 3 and
dx ⎝ x 2 ⎠
x
dy ⎝ y 2 ⎠
y
!
∂U ˆ ∂U ˆ
α
α
EXECUTE: F = −
i−
j since U has no z-dependence. ∂U = −23 and ∂U = −23 , so
∂x
∂y
x
y
∂x
∂y
!
!
!
⎛i
⎛ −2 ˆ −2 ⎞
j⎞
F = −α ⎜ 3 i + 3 ˆ ⎟ = 2α ⎜ 3 + 3 ⎟ .
j
y⎠
y⎠
⎝x
⎝x
EVALUATE: Fx and x have the same sign and Fy and y have the same sign. When x > 0 , Fx is in the
IDENTIFY: + x-direction, and so forth. 7-12 7.37. Chapter 7 IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U. At equilibrium F = 0.
(a) EXECUTE: The graphs are sketched in Figure 7.37. a
b
−
r 12 r 6
dU
12a 6b
F =−
= + 13 − 7
dr
r
r U= Figure 7.37
(b) At equilibrium F = 0, so dU
=0
dr
F = 0 implies +12a 6b
− 7 =0
r 13
r 6br 6 = 12a; solution is the equilibrium distance r0 = (2a / b)1 / 6
U is a minimum at this r; the equilibrium is stable.
(c) At r = (2a / b)1 / 6 , U = a / r12 − b / r 6 = a (b / 2a ) 2 − b(b / 2a ) = −b 2 / 4a.
At r → ∞, U = 0. The energy that must be added is −ΔU = b 2 / 4a.
(d) r0 = (2a / b)1 / 6 = 1.13 × 10−10 m gives that 2a / b = 2.082 × 10−60 m 6 and b / 4a = 2.402 × 1059 m −6
b 2 / 4a = b(b / 4a ) = 1.54 × 10−18 J
b(2.402 × 1059 m −6 ) = 1.54 × 10−18 J and b = 6.41 × 10−78 J ⋅ m 6 .
Then 2a / b = 2.082 × 10−60 m 6 gives a = (b / 2)(2.082 × 10−60 m 6 ) =
(6.41 × 10−78 J ⋅ m 6 )(2.082 × 10−60 m 6 ) = 6.67 × 10−138 J ⋅ m12
EVALUATE: As the graphs in part (a) show, F (r ) is the slope of U (r ) at each r. U (r ) has a minimum where
F = 0.
IDENTIFY: Apply Eq.(7.16).
dU
SET UP:
is the slope of the U versus x graph.
dx
dU
EXECUTE: (a) Considering only forces in the x-direction, Fx = −
and so the force is zero when the slope of
dx
the U vs x graph is zero, at points b and d.
(b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is
stable.
(c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the
marble tends to move further away, and so d is an unstable point.
EVALUATE: At point b, Fx is negative when the marble is displaced slightly to the right and Fx is positive when
the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a
small displacement in either direction produces a force directed away from d and the equilibrium is unstable.
!
!
IDENTIFY: Apply ∑ F = ma to the bag and to the box. Apply Eq.(7.7) to the motion of the system of the box
1
2 7.38. 7.39. and bucket after the bag is removed.
SET UP: Let y = 0 at the final height of the bucket, so y1 = 2.00 m and y2 = 0 . K1 = 0 . The box and the bucket
move with the same speed v, so K 2 = 1 ( mbox + mbucket )v 2 . Wother = − f k d , with d = 2.00 m and f k = μk mbox g .
2
Before the bag is removed, the maximum possible friction force the roof can exert on the box is
(0.700)(80.0 kg + 50.0 kg)(9.80 m/s 2 ) = 892 N . This is larger than the weight of the bucket (637 N), so before the
bag is removed the system is at rest.
EXECUTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag
for friction to oppose. The static friction force on the box equals the weight of the bucket, 637 N. Potential Energy and Energy Conservation (b) Eq.(7.7) gives mbucket gy1 − f k d = 1 mtot v 2 , with mtot = 145.0 kg . v =
2 v= 2
( mbucket gy1 − μ k mbox gd ) .
mtot 2
⎡(65.0 kg)(9.80 m/s 2 )(2.00 m) − (0.400)(80.0 kg)(9.80 m/s 2 )(2.00 m) ⎤ .
⎦
145.0 kg ⎣ v = 2.99 m/s .
EVALUATE: If we apply 7.40. 7-13 ! ! ∑ F = ma to the box and to the bucket we can calculate their common acceleration a. Then a constant acceleration equation applied to either object gives v = 2.99 m/s , in agreement with our result
obtained using energy methods.
IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq.(7.5).
SET UP: Call the blocks A and B, where A is the more massive one. v A1 = vB1 = 0 . Let y = 0 for each block to be
at the initial height of that block, so y A1 = yB1 = 0 . y A 2 = −1.20 m and yB 2 = +1.20 m . v A 2 = vB 2 = v2 = 3.00 m/s .
2
Eq.(7.5) gives 0 = 1 (m A + mB )v2 + g (1.20 m)( mB − mA ) . m A + mB = 15.0 kg .
2 EXECUTE:
1
2 7.41. (15.0 kg)(3.00 m/s) 2 + (9.80 m/s 2 )(1.20 m)(15.0 kg − 2m A ) . Solving for mA gives mA = 10.4 kg . And then mB = 4.6 kg .
EVALUATE: The final kinetic energy of the two blocks is 68 J. The potential energy of block A decreases by 122 J.
The potential energy of block B increases by 54 J. The total decrease in potential energy is 122 J − 54 J = 68 J, and
this equals the increase in kinetic energy of the system.
IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2
SET UP: U1 = U 2 = K 2 = 0 . Wother = W f = − μ k mgs, with s = 280 ft = 85.3 m
(a) The work-energy expression gives EXECUTE: 1
2 mv12 − μ k mgs = 0 . v1 = 2 μk gs = 22.4 m/s = 50 mph; the driver was speeding. 7.42. (b) 15 mph over speed limit so $150 ticket.
EVALUATE: The negative work done by friction removes the kinetic energy of the object.
IDENTIFY: Apply Eq.(7.14).
SET UP: Only the spring force and gravity do work, so Wother = 0 . Let y = 0 at the horizontal surface.
(a) Equating the potential energy stored in the spring to the block's kinetic energy, EXECUTE: v= 7.43. 1
2 kx 2 = 1 mv 2 , or
2 k
400 N/m
x=
(0.220 m) = 3.11 m/s.
m
2.00 kg (b) Using energy methods directly, the initial potential energy of the spring equals the final gravitational potential
1
1
kx 2
(400 N/m)(0.220 m) 2
2
energy, 1 kx 2 = mgL sin θ , or L = 2
=
= 0.821 m.
2
mg sin θ (2.00 kg)(9.80 m/s 2 )sin 37.0°
EVALUATE: The total energy of the system is constant. Initially it is all elastic potential energy stored in the
spring, then it is all kinetic energy and finally it is all gravitational potential energy.
IDENTIFY: Use the work-energy theorem, Eq(7.7). The target variable μ k will be a factor in the work done by
friction.
SET UP: Let point 1 be where the block is released and let point 2 be where the block stops, as shown in
Figure 7.43.
K1 + U1 + Wother = K 2 + U 2
Work is done on the
block by the spring and
by friction, so Wother = W f and U = U el .
Figure 7.43 K1 = K 2 = 0 EXECUTE: U1 = U1,el = kx12 = 1 (100 N/m)(0.200 m)2 = 2.00 J
2
1
2 U 2 = U 2,el = 0, since after the block leaves the spring has given up all its stored energy
Wother = W f = ( f k cos φ ) s = μ k mg (cos φ ) s = − μ k mgs, since φ = 180° (The friction force is directed opposite to the
displacement and does negative work.) 7-14 Chapter 7 Putting all this into K1 + U1 + Wother = K 2 + U 2 gives
U1,el + W f = 0 μ k mgs = U1,el
μk = U1,el
mgs = EVALUATE:
7.44. 200 J
= 0.41.
(0.50 kg)(9.80 m/s 2 )(1.00 m)
U1,el + W f = 0 says that the potential energy originally stored in the spring is taken out of the system by the negative work done by friction.
IDENTIFY: Apply Eq.(7.14). Calculate f k from the fact that the crate slides a distance x = 5.60 m before coming
to rest. Then apply Eq.(7.14) again, with x = 2.00 m .
SET UP: U1 = U el = 360 J . U 2 = 0 . K1 = 0 . Wother = − f k x .
EXECUTE: Work done by friction against the crate brings it to a halt: U1 = −Wother . 360 J
= 64.29 N .
5.60 m
The friction force working over a 2.00-m distance does work equal to − f k x = −(64.29 N)(2.00 m) = −128.6 J. The
kinetic energy of the crate at this point is thus 360 J − 128.6 J = 231.4 J , and its speed is found from
f k x = potential energy of compressed spring , and f k = mv 2 / 2 = 231.4 J , so v = 7.45. 7.46. 2(231.4 J)
= 3.04 m/s .
50.0 kg EVALUATE: The energy of the compressed spring goes partly into kinetic energy of the crate and is partly
removed by the negative work done by friction. After the crate leaves the spring the crate slows down as friction
does negative work on it.
IDENTIFY: At its highest point between bounces all the mechanical energy of the ball is in the form of
gravitational potential energy.
SET UP: E = U = mgh , where h is the height at the highest point of the motion.
EXECUTE: (a) mgh = (0.650 kg)(9.80 m/s 2 )(2.50 m) = 15.9 J
(b) The second height is 0.75(2.50 m) = 1.875 m , so the second mgh = 11.9 J ; it loses 15.9 J − 11.9 J = 4.0 J on
first bounce. This energy is converted to thermal energy.
(c) The third height is 0.75(1.875 m) = 1.40 m , , so third mgh = 8.9 J ; it loses 11.9 J − 8.9 J = 3.0 J on second
bounce.
EVALUATE: In each bounce the ball loses 25% of its mechanical energy.
!
!
IDENTIFY: Apply Eq.(7.14) to relate h and vB . Apply ∑ F = ma at point B to find the minimum speed required at B for the car not to fall off the track.
2
2
SET UP: At B, a = vB / R , downward. The minimum speed is when n → 0 and mg = mvB / R . The minimum
speed required is vB = gR . K1 = 0 and Wother = 0 .
EXECUTE: 2
(a) Eq.(7.14) applied to points A and B gives U A − U B = 1 mvB . The speed at the top must be at least
2 1
5
gR . Thus, mg (h − 2 R) > mgR, or h > R.
2
2
(b) Apply Eq.(7.14) to points A and C. U A − U C = (2.50) Rmg = K C , so
vC = (5.00) gR = (5.00)(9.80 m/s 2 )(20.0 m) = 31.3 m/s.
2
vC
= 49.0 m/s 2 . The tangential direction is down, the normal force at point C is
R
horizontal, there is no friction, so the only downward force is gravity, and atan = g = 9.80 m/s 2 . The radial acceleration is arad = EVALUATE:
7.47. If h > 5 R , then the downward acceleration at B due to the circular motion is greater than g and the
2 track must exert a downward normal force n. n increases as h increases and hence vB increases.
(a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom.
SET UP: Let point 1 be where the piece of wood is released and point 2 be just before it enters the rough bottom.
Let y = 0 be at point 2.
EXECUTE: U1 = K 2 gives K 2 = mgy1 = 78.4 J.
IDENTIFY: Now apply work-energy relation to the motion along the rough bottom. Potential Energy and Energy Conservation SET UP: 7-15 Let point 1 be where it enters the rough bottom and point 2 be where it stops.
K1 + U1 + Wother = K 2 + U 2 EXECUTE: 7.48. Wother = W f = − μ k mgs, K 2 = U1 = U 2 = 0; K1 = 78.4 J 78.4 J − μ k mgs = 0; solving for s gives s = 20.0 m.
The wood stops after traveling 20.0 m along the rough bottom.
(b) Friction does −78.4 J of work.
EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical
energy is zero. The negative friction work takes away all the mechanical energy initially in the system.
IDENTIFY: Apply Eq.(7.14) to the rock. Wother = W fk .
SET UP: Let y = 0 at the foot of the hill, so U1 = 0 and U 2 = mgh , where h is the vertical height of the rock above
the foot of the hill when it stops.
EXECUTE: (a) At the maximum height, K 2 = 0 . Eq.(7.14) gives K Bottom + W fk = U Top . 12
12
h
= gh .
mv0 − μ k mg cos θ d = mgh . d = h sin θ , so v0 − μ k g cosθ
2
2
sin θ
1
cos 40°
(15 m/s) 2 − (0.20)(9.8 m/s 2 )
h = (9.8 m/s 2 )h and h = 9.3 m .
2
sin 40°
(b) Compare maximum static friction force to the weight component down the plane.
f s = μs mg cosθ = (0.75)(28 kg)(9.8 m/s 2 )cos 40° = 158 N . mg sinθ = (28 kg)(9.8 m/s 2 )(sin 40°) = 176 N > f s , so
the rock will slide down.
(c) Use same procedure as in part (a), with h = 9.3 m and vB being the speed at the bottom of the hill.
U Top + W fk = K B . mgh − μk mg cosθ h
12
= mvB and
sin θ 2 vB = 2 gh − 2 μ k gh cosθ sin θ = 11.8 m/s . 7.49. EVALUATE: For the round trip up the hill and back down, there is negative work done by friction and the speed
of the rock when it returns to the bottom of the hill is less than the speed it had when it started up the hill.
IDENTIFY: Apply Eq.(7.7) to the motion of the stone.
SET UP: K1 + U1 + Wother = K 2 + U 2
Let point 1 be point A and point 2 be point B. Take y = 0 at point B.
EXECUTE: 2
mgy1 + 1 mv12 = 1 mv2 , with h = 20.0 m and v1 = 10.0 m/s
2
2 v2 = v12 + 2 gh = 22.2 m/s
EVALUATE: The loss of gravitational potential energy equals the gain of kinetic energy.
(b) IDENTIFY: Apply Eq.(7.8) to the motion of the stone from point B to where it comes to rest against the
spring.
SET UP: Use K1 + U1 + Wother = K 2 + U 2 , with point 1 at B and point 2 where the spring has its maximum
compression x.
EXECUTE: U1 = U 2 = K 2 = 0; K1 = 1 mv12 with v1 = 22.2 m/s
2 Wother = W f + Wel = − μ k mgs − 1 kx 2 , with s = 100 m + x
2
The work-energy relation gives K1 + Wother = 0.
1
2 mv12 − μ k mgs − 1 kx 2 = 0
2 Putting in the numerical values gives x 2 + 29.4 x − 750 = 0. The positive root to this equation is x = 16.4 m.
EVALUATE: Part of the initial mechanical (kinetic) energy is removed by friction work and the rest goes into the
potential energy stored in the spring.
(c) IDENTIFY and SET UP: Consider the forces.
EXECUTE: When the spring is compressed x = 16.4 m the force it exerts on the stone is Fel = kx = 32.8 N. The
maximum possible static friction force is
max f s = μs mg = (0.80)(15.0 kg)(9.80 m/s 2 ) = 118 N.
EVALUATE: The spring force is less than the maximum possible static friction force so the stone remains at rest. 7-16 7.50. Chapter 7 IDENTIFY: Once the block leaves the top of the hill it moves in projectile motion. Use Eq.(7.14) to relate the
speed vB at the bottom of the hill to the speed vTop at the top and the 70 m height of the hill.
SET UP: For the projectile motion, take + y to be downward. ax = 0 , a y = g . v0 x = vTop , v0 y = 0 . For the motion up the hill only gravity does work. Take y = 0 at the base of the hill.
EXECUTE: First get speed at the top of the hill for the block to clear the pit. y = 12
1
gt . 20 m = (9.8 m/s 2 )t 2 .
2
2 40 m
= 20 m/s .
2.0 s
Energy conservation applied to the motion up the hill: K Bottom = U Top + K Top gives
t = 2.0 s . Then vTopt = 40 m gives vTop = 7.51. 7.52. 12
12
2
mvB = mgh + mvTop . vB = vTop + 2 gh = (20 m/s) 2 + 2(9.8 m/s 2 )(70 m) = 42 m/s .
2
2
EVALUATE: The result does not depend on the mass of the block.
IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the person.
SET UP: Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at point 2. y1 = 41.0 m. Wother = − 1 kx 2 , where x = 11.0 m is the amount the cord is stretched at point 2. The cord
2
does negative work.
EXECUTE: K1 = K 2 = U 2 = 0, so mgy1 − 1 kx 2 = 0 and k = 631 N/m.
2
Now apply F = kx to the test pulls:
F = kx so x = F/k = 0.602 m.
EVALUATE: All his initial gravitational potential energy is taken away by the negative work done by the force
exerted by the cord, and this amount of energy is stored as elastic potential energy in the stretched cord.
IDENTIFY: Apply Eq.(7.14) to the motion of the skier from the gate to the bottom of the ramp.
SET UP: Wother = −4000 J . Let y = 0 at the bottom of the ramp.
EXECUTE: For the skier to be moving at no more than 30.0 m/s ; his kinetic energy at the bottom of the ramp can be
mv 2 (85.0 kg)(30.0 m/s) 2
no bigger than
=
= 38,250 J . Friction does −4000 J of work on him during his run, which
2
2
means his combined U and K at the top of the ramp must be no more than 38,250 J + 4000 J = 42,250 J. His K at the
mv 2 (85.0 kg)(2.0 m/s) 2
=
= 170 J . His U at the top should thus be no more than 42,250 J − 170 J = 42,080 J ,
2
2
42,080 J
42,080 J
=
= 50.5 m.
which gives a height above the bottom of the ramp of h =
mg
(85.0 kg)(9.80 m/s 2 )
EVALUATE: In the absence of air resistance, for this h his speed at the bottom of the ramp would be 31.5 m/s.
The work done by air resistance is small compared to the kinetic and potential energies that enter into the
calculation.
IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for K 2 and then for v2 .
SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel,
as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by the 4400 N force.
K1 + U1 + Wother = K 2 + U 2
top is 7.53. Take y = 0 at his initial position.
EXECUTE: 2
K1 = 0, K 2 = 1 mv2
2 Wother = Wfric = − fs
Wother = −(40 N)(4.0 m) = −160 J
Figure 7.53 U1,grav = 0, U1,el = 1 kd 2 , where d is the distance the spring is initially compressed.
2
F
4400 N
=
= 4.00 m
k 1100 N/m
= 1 (1100 N/m)(4.00 m)2 = 8800 J
2 F = kd so d =
and U1,el U 2,grav = mgy2 = (60 kg)(9.80 m/s 2 )(2.5 m) = 1470 J, U 2, el = 0 Potential Energy and Energy Conservation 7-17 Then K1 + U1 + Wother = K 2 + U 2 gives
2
8800 J − 160 J = 1 mv2 + 1470 J
2 2(7170 J)
= 15.5 m/s
60 kg
EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by
friction. The rest goes partly into gravitational potential energy and partly into kinetic energy.
IDENTIFY: To be at equilibrium at the bottom, with the spring compressed a distance x0 , the spring force must
balance the component of the weight down the ramp plus the largest value of the static friction, or
kx0 = w sin θ + f . Apply Eq.(7.14) to the motion down the ramp.
1
2 7.54. 2
mv2 = 7170 J and v2 = SET UP: K 2 = 0 , K1 = 1 mv 2 , where v is the speed at the top of the ramp. Let U 2 = 0 , so U1 = wL sin θ , where L
2
is the total length traveled down the ramp.
12
1
2
EXECUTE: Eq.(7.14) gives kx0 = ( w sin θ − f ) L + mv 2 . With the given parameters, 1 kx0 = 248 J and
2
2
2
kx0 = 1.10 × 103 N. Solving for k gives k = 2440 N/m.
EVALUATE: 7.55. x0 = 0.451 m . w sin θ = 551 N . The decrease in gravitational potential energy is only slightly larger than the amount of mechanical energy removed by the negative work done by friction. 1 mv 2 = 243 J . The energy
2
stored in the spring is only slightly larger than the initial kinetic energy of the crate at the top of the ramp.
IDENTIFY: Apply Eq.(7.7) to the system consisting of the two buckets. If we ignore the inertia of the pulley we
ignore the kinetic energy it has.
SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.55. Figure 7.55 The tension force does positive work on the 4.0 kg bucket and an equal amount of negative work on the 12.0 kg
bucket, so the net work done by the tension is zero.
Work is done on the system only by gravity, so Wother = 0 and U = U grav
EXECUTE: K2 = m v
1
2 2
A A, 2 K1 = 0
2
+ 1 mB vB , 2 But since the two buckets are connected by a rope they move together and have the same
2 speed: v A, 2 = vB , 2 = v2 .
2
2
Thus K 2 = 1 ( mA + mB )v2 = (8.00 kg)v2 .
2 U1 = mA gy A,1 = (12.0 kg)(9.80 m/s 2 )(2.00 m) = 235.2 J.
U 2 = mB gyB , 2 = (4.0 kg)(9.80 m/s 2 )(2.00 m) = 78.4 J.
Putting all this into K1 + U1 + Wother = K 2 + U 2 gives
U1 = K 2 + U 2
2
235.2 J = (8.00 kg)v2 + 78.4 J 235.2 J − 78.4 J
= 4.4 m/s
8.00 kg
EVALUATE: The gravitational potential energy decreases and the kinetic energy increases by the same amount.
We could apply Eq.(7.7) to one bucket, but then we would have to include in Wother the work done on the bucket by
the tension T.
v2 = 7-18 7.56. Chapter 7 IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the rocket from the starting point to the base of the ramp. Wother is the work done by the thrust and by friction.
Let point 1 be at the starting point and let point 2 be at the base of the ramp. v1 = 0 , v2 = 50.0 m/s . Let SET UP: y = 0 at the base and take + y upward. Then y2 = 0 and y1 = d sin 53° , where d is the distance along the ramp
from the base to the starting point. Friction does negative work.
EXECUTE: K1 = 0 , U 2 = 0 . U1 + Wother = K 2 . Wother = (2000 N)d − (500 N)d = (1500 N)d .
2
mgd sin 53° + (1500 N) d = 1 mv2 .
2 d= 2
mv2
(1500 kg)(50.0 m/s)2
=
= 142 m .
2[mg sin 53° + 1500 N] 2[(1500 kg)(9.80 m/s 2 )sin 53° + 1500 N] EVALUATE: The initial height is y1 = (142 m)sin 53° = 113 m . An object free-falling from this distance attains a speed v = 2 gy1 = 47.1 m/s . The rocket attains a greater speed than this because the forward thrust is greater than
7.57. the friction force.
IDENTIFY: The force exerted by a spring is Fx = − kx . The acceleration of the object is given by Fx = max . Apply
Eq.(7.14) to relate position and speed.
SET UP: Let + x be when the spring is stretched.
EXECUTE: (a) U = 1 kx 2 . Let point 1 be when the spring is initially compressed a distance x0 , so x1 = − x0 .
2 K1 = 0 . Wother = 0 . 1
2 2
kx0 = U 2 + K 2 . The speed is maximum when x = 0 , so U 2 = 0 . Then 1
2 2
2
kx0 = 1 mv2 and
2 v2 = x0 k / m is this maximum speed.
k
k
x . a is maximum when x is maximum, so a = x0 .
m
m
(c) The speed is maximum when x = 0 , when the spring has returned to its natural length, and the acceleration is
maximum when x = − x0 , at the initial compression of the spring.
(b) Fx = − kx and Fx = max give ax = − 7.58. 2
(d) When the spring has maximum extension, v2 = 0 . 1 kx0 = 1 kx 2 and x = x0 .The magnitude of the maximum
2
2
extension equals the magnitude of the maximum compression.
(e) The machine part oscillates between x = − x0 and x = + x0 and never stops permanently.
EVALUATE: In any real system there are mechanical energy losses, for example due to negative work done by
friction, and the object eventually comes to rest.
IDENTIFY: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy.
SET UP: Since the two animals are equidistant from the axis, they each have the same speed v.
EXECUTE: One mass rises while the other falls, so the net loss of potential energy is
(0.500 kg − 0.200 kg)(9.80 m/s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the animals and is 2(1.176 J)
= 1.83 m/s.
(0.700 kg)
EVALUATE: The mouse gains both gravitational potential energy and kinetic energy. The rat’s gain in kinetic
energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse.
(a) IDENTIFY and SET UP: Apply Eq.(7.7) to the motion of the potato.
Let point 1 be where the potato is released and point 2 be at the lowest point in its motion, as shown in
Figure 7.59a.
K1 + U1 + Wother = K 2 + U 2
equal to 7.59. 1
2 mtot v 2 , and v = y1 = 2.50 m
y2 = 0
The tension in the string is at all points in
the motion perpendicular to the
displacement, so WT = 0
The only force that does work on the
potato is gravity, so Wother = 0.
Figure 7.59a Potential Energy and Energy Conservation EXECUTE: 7-19 2
K1 = 0, K 2 = 1 mv2 , U1 = mgy1 , U 2 = 0
2 Thus U1 = K 2 .
2
mgy1 = 1 mv2
2 v2 = 2 gy1 = 2(9.80 m/s 2 )(2.50 m) = 7.00 m/s
v2 is the same as if the potato fell through 2.50 m.
!
!
!
(b) IDENTIFY: Apply ∑ F = ma to the potato. The potato moves in an arc of a circle so its acceleration is arad ,
EVALUATE: where arad = v 2 / R and is directed toward the center of the circle. Solve for one of the forces, the tension T in the
string.
SET UP: The free-body diagram for the potato as it swings through its lowest point is given in Figure 7.59b. !
The acceleration arad is directed in toward
the center of the circular path, so at this
point it is upward. Figure 7.59b
EXECUTE: ∑F y = ma y T − mg = marad
⎛
v2 ⎞
T = m( g + arad ) = m ⎜ g + 2 ⎟ , where the radius R for the circular motion is the length L of the string.
R⎠
⎝
It is instructive to use the algebraic expression for v2 from part (a) rather than just putting in the numerical value:
2
v2 = 2 gy1 = 2 gL , so v2 = 2 gL ⎛
v2 ⎞
2 gL ⎞
⎛
Then T = m ⎜ g + 2 ⎟ = m ⎜ g +
⎟ = 3mg ; the tension at this point is three times the weight of the potato.
L⎠
L⎠
⎝
⎝
T = 3mg = 3(0.100 kg)(9.80 m/s 2 ) = 2.94 N
7.60. EVALUATE: The tension is greater than the weight; the acceleration is upward so the net force must be upward.
IDENTIFY: Eq.(7.14) says Wother = K 2 + U 2 − ( K1 + U1 ) . Wother is the work done on the baseball by the force
exerted by the air.
2
2
SET UP: U = mgy . K = 1 mv 2 , where v 2 = vx + v y .
2
(a) The change in total energy is the work done by the air,
⎛1 2
⎞
= ( K 2 + U 2 ) − ( K1 + U1 ) = m ⎜ (v2 − v12 ) + gy2 ⎟ .
2
⎝
⎠ EXECUTE: Wother ( ) Wother = (0.145 kg) (1/ 2 ⎡(18.6 m/s) 2 − (30.0 m/s) 2 − (40.0 m/s) 2 ⎤ + (9.80 m/s 2 )(53.6 m) .
⎣
⎦
Wother = −80.0 J .
(b) Similarly, Wother = ( K 3 + U 3 ) − ( K 2 + U 2 ) . ( ) Wother = (0.145 kg) (1/ 2) ⎡(11.9 m/s) 2 + (−28.7 m/s) 2 − (18.6 m/s) 2 ⎤ − (9.80 m/s 2 )(53.6 m) .
⎣
⎦
Wother = −31.3 J.
(c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by
the air is smaller in magnitude.
EVALUATE: The initial kinetic energy of the baseball is 1 (0.145 kg)(50.0 m/s) 2 = 181 J . For the total motion
2
from the ground, up to the maximum height, and back down the total work done by the air is 111 J. The ball
returns to the ground with 181 J − 111 J = 70 J of kinetic energy and a speed of 31 m/s, less than its initial speed of
50 m/s. 7-20 7.61. Chapter 7 IDENTIFY and SET UP: There are two situations to compare: stepping off a platform and sliding down a pole.
Apply the work-energy theorem to each.
(a) EXECUTE: Speed at ground if steps off platform at height h:
K1 + U1 + Wother = K 2 + U 2
2
2
mgh = 1 mv2 , so v2 = 2 gh
2
Motion from top to bottom of pole: (take y = 0 at bottom) K1 + U1 + Wother = K 2 + U 2
2
mgd − fd = 1 mv2
2
2
Use v2 = 2 gh and get mgd − fd = mgh
fd = mg ( d − h)
f = mg (d − h) / d = mg (1 − h / d )
EVALUATE: For h = d this gives f = 0 as it should (friction has no effect). For h = 0, v2 = 0 (no motion). The equation for f gives f = mg in this special case. When f = mg the forces on
him cancel and he doesn’t accelerate down the pole, which agrees with v2 = 0.
(b) EXECUTE: f = mg (1 − h / d ) = (75 kg)(9.80 m/s 2 )(1 − 1.0 m/2.5 m) = 441 N. (c) Take y = 0 at bottom of pole, so y1 = d and y2 = y. K1 + U1 + Wother = K 2 + U 2
0 + mgd − f ( d − y ) = 1 mv 2 + mgy
2
1
2 mv 2 = mg ( d − y ) − f ( d − y ) Using f = mg (1 − h / d ) gives 1
2 mv 2 = mg (d − y ) − mg (1 − h / d )(d − y ) mv 2 = mg ( h / d )(d − y ) and v = 2 gh(1 − y / d )
EVALUATE: This gives the correct results for y = 0 and for y = d .
IDENTIFY: Apply Eq.(7.14) to each stage of the motion.
SET UP: Let y = 0 at the bottom of the slope. In part (a), Wother is the work done by friction. In part (b), Wother is
1
2 7.62. the work done by friction and the air resistance force. In part (c), Wother is the work done by the force exerted by the
snowdrift.
EXECUTE: (a) The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus
the work done by friction, K1 = mgh − W f = (60.0 kg)(9.8 N/kg)(65.0 m) − 10,500 J, or K1 = 38,200 J − 10,500 J = 27,720 J . Then v1 = 2 K1
2(27,720 J)
=
= 30.4 m/s .
60 kg
m (b) K 2 = K1 − (W f + Wair ) = 27,720 J − ( μk mgd + f air d ). K 2 = 27,720 J − [(0.2)(588 N)(82 m) + (160 N)(82 m)] or K 2 = 27,720 J − 22,763 J = 4957 J . Then, v2 = 2K
2(4957 J)
=
= 12.9 m/s
60 kg
m (c) Use the Work-Energy Theorem to find the force. W = ΔK , F = K/d = (4957 J) (2.5 m) = 2000 N . In each case, Wother is negative and removes mechanical energy from the system.
!
!
IDENTIFY and SET UP: First apply ∑ F = ma to the skier.
EVALUATE: 7.63. Find the angle α where the normal force becomes zero, in terms of the speed v2 at this point. Then apply the
work-energy theorem to the motion of the skier to obtain another equation that relates v2 and α . Solve these two
equations for α . Let point 2 be where the skier loses contact
with the snowball, as sketched in Figure 7.63a
Loses contact implies n → 0.
y1 = R, y2 = R cos α Figure 7.63a Potential Energy and Energy Conservation 7-21 First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For
this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her
acceleration is arad = v 2 / R, directed in towards the center of the snowball.
EXECUTE: ∑F y = ma y mg cos α − n = mv / R
2
2 2
But n = 0 so mg cos α = mv2 / R
2
v2 = Rg cos α Figure 7.63b Now use conservation of energy to get another equation relating v2 to α :
K1 + U1 + Wother = K 2 + U 2
The only force that does work on the skier is gravity, so Wother = 0.
2
K1 = 0, K 2 = 1 mv2
2 U1 = mgy1 = mgR, U 2 = mgy2 = mgR cos α
2
Then mgR = 1 mv2 + mgR cos α
2
2
v2 = 2 gR(1 − cos α ) Combine this with the 7.64. ∑F y = ma y equation: Rg cos α = 2 gR (1 − cos α )
cos α = 2 − 2cos α
3 cos α = 2 so cos α = 2 / 3 and α = 48.2°
EVALUATE: She speeds up and her arad increases as she loses gravitational potential energy. She loses contact
when she is going so fast that the radially inward component of her weight isn’t large enough to keep her in the
circular path. Note that α where she loses contact does not depend on her mass or on the radius of the snowball.
IDENTIFY: Use conservation of energy to relate the speed at the lowest point to the speed at the highest point.
!
!
Use ∑ F = ma to calculate the tension.
SET UP: The rock has acceleration arad = v 2 / R , directed toward the center of the circle. EXECUTE: If the speed of the rock at the top is vt , then conservation of energy gives the speed vb at the bottom 2
from mv = 1 mvt2 + mg (2 R ) , R being the radius of the circle, and so vb = vt2 + 4 gR . The tension at the top and
2
1
2 2
b mvt2
mv 2
m2
and Tb − mg = b , so Tb − Tt = (vb − vt2 ) + 2mg = 6mg = 6w .
R
R
R
EVALUATE: The tensions Tt and Tb depend on the speed of the rock and on R, but the difference Tb − Tt is
independent of the speed of the rock and the radius of the circle.
IDENTIFY and SET UP:
bottom are found from Tt + mg = 7.65. yA = R
yB = yC = 0 Figure 7.65
(a) Apply conservation of energy to the motion from B to C:
K B + U B + Wother = K C + U C . The motion is described in Figure 7.65.
EXECUTE: The only force that does work on the package during this part of the motion is friction, so
Wother = W f = f k (cos φ ) s = μ k mg (cos180°) s = − μ k mgs
2
K B = 1 mvB , K C = 0
2 U B = 0, U C = 0 7-22 Chapter 7 Thus K B + W f = 0
1
2 2
mvB − μ k mgs = 0
2
μB (4.80 m/s) 2
= 0.392
2 gs 2(9.80 m/s 2 )(3.00 m)
EVALUATE: The negative friction work takes away all the kinetic energy.
(b) IDENTIFY and SET UP: Apply conservation of energy to the motion from A to B: μk = = K A + U A + Wother = K B + U B
Work is done by gravity and by friction, so Wother = W f . EXECUTE: 2
K A = 0, K B = 1 mvB = 1 (0.200 kg)(4.80 m/s) 2 = 2.304 J
2
2 U A = mgy A = mgR = (0.200 kg)(9.80 m/s 2 )(1.60 m) = 3.136 J, U B = 0
Thus U A + W f = K B
W f = K B − U A = 2.304 J − 3.136 J = −0.83 J
EVALUATE:
7.66. W f is negative as expected; the friction force does negative work since it is directed opposite to the displacement.
IDENTIFY: Apply Eq.(7.14) to the initial and final positions of the truck.
SET UP: Let y = 0 at the lowest point of the path of the truck. Wother is the work done by friction.
f r = μr n = μr mg cos β .
2
Denote the distance the truck moves up the ramp by x. K1 = 1 mv0 , U1 = mgL sin α , K 2 = 0 ,
2 EXECUTE: U 2 = mgx sin β and Wother = − μ r mgx cos β . From Wother = ( K 2 + U 2 ) − ( K1 + U1 ) , and solving for x,
x=
EVALUATE:
7.67. (v 2 /2 g ) + L sin α
K1 + mgL sin α
=0
.
mg (sin β + μ r cos β ) sin β + μ r cos β x increases when v0 increases and decreases when μ r increases. Fx = −α x − β x 2 , α = 60.0 N/m and β = 18.0 N/m 2
(a) IDENTIFY: Use Eq.(6.7) to calculate W and then use W = −ΔU to identify the potential energy function U ( x ).
SET UP: x2 WFx = U1 − U 2 = ∫ Fx ( x) dx
x1 Let x1 = 0 and U1 = 0. Let x2 be some arbitrary point x, so U 2 = U ( x).
EXECUTE: x x 0 0 x U ( x) = − ∫ Fx ( x) dx = − ∫ ( −α x − β x 2 ) dx = ∫ (α x + β x 2 ) dx = 1 α x 2 + 1 β x3 .
2
3
0 EVALUATE: If β = 0, the spring does obey Hooke’s law, with k = α , and our result reduces to
(b) IDENTIFY: Apply Eq.(7.15) to the motion of the object.
SET UP: The system at points 1 and 2 is sketched in Figure 7.67. 1
2 kx 2 . K1 + U1 + Wother = K 2 + U 2
The only force that does work on the
object is the spring force, so Wother = 0. Figure 7.67
EXECUTE: 2
K1 = 0, K 2 = 1 mv2
2 U1 = U ( x1 ) = 1 α x12 + 1 β x13 = 1 (60.0 N/m)(1.00 m) 2 + 1 (18.0 N/m 2 )(1.00 m)3 = 36.0 J
2
3
2
3
2
3
U 2 = U ( x2 ) = 1 α x2 + 1 β x2 = 1 (60.0 N/m)(0.500 m) 2 + 1 (18.0 N/m 2 )(0.500 m)3 = 8.25 J
2
3
2
3
2
Thus 36.0 J = 1 mv2 + 8.25 J
2 2(36.0 J − 8.25 J)
= 7.85 m/s
0.900 kg
EVALUATE: The elastic potential energy stored in the spring decreases and the kinetic energy of the object increases.
v2 = Potential Energy and Energy Conservation 7.68. 7.69. IDENTIFY: 7-23 Apply Eq.(7.14). Wother is the work done by F. SET UP: Wother = ΔK + ΔU . The distance the spring stretches is aθ . y2 − y1 = a sin θ .
EXECUTE: The force increases both the gravitational potential energy of the block and the potential energy of the
spring. If the block is moved slowly, the kinetic energy can be taken as constant, so the work done by the force is
the increase in potential energy, ΔU = mga sin θ + 1 k (aθ ) 2 .
2
EVALUATE: The force is kept tangent to the surface so the block will stay in contact with the surface.
IDENTIFY: Apply Eq.(7.14) to the motion of the block.
SET UP: Let y = 0 at the floor. Let point 1 be the initial position of the block against the compressed spring and
let point 2 be just before the block strikes the floor.
2
EXECUTE: With U 2 = 0, K1 = 0 , K 2 = U1 . 1 mv2 = 1 kx 2 + mgh . Solving for v2 ,
2
2 (1900 N/m)(0.045 m) 2
kx 2
+ 2 gh =
+ 2(9.80 m/s 2 )(1.20 m) = 7.01 m/s .
(0.150 kg)
m
EVALUATE: The potential energy stored in the spring and the initial gravitational potential energy all go into the
final kinetic energy of the block.
IDENTIFY: Apply Eq.(7.14). U is the total elastic potential energy of the two springs.
SET UP: Call the two points in the motion where Eq.(7.14) is applied A and B to avoid confusion with springs 1
and 2, that have force constants k1 and k2 . At any point in the motion the distance one spring is stretched equals
the distance the other spring is compressed. Let + x be to the right. Let point A be the initial position of the block,
where it is released from rest, so x1 A = +0.150 m and x2 A = −0.150 m .
v2 = 7.70. EXECUTE: (a) With no friction, Wother = 0 . K A = 0 and U A = K B + U B . The maximum speed is when U B = 0 and 2
2
this is at x1B = x2 B = 0 , when both springs are at their natural length. 1 k1 x12A + 1 k2 x2 A = 1 mvB .
2
2
2
2
x12A = x2 A = (0.150 m) 2 , so vB = k1 + k2
2500 N/m + 2000 N/m
(0.150 m) =
(0.150 m) = 5.81 m/s .
m
3.00 kg (b) At maximum compression of spring 1, spring 2 has its maximum extension and vB = 0 . Therefore, at this point U A = U B . The distance spring 1 is compressed equals the distance spring 2 is stretched, and vice versa: 7.71. x1 A = − x2 A and x1B = − x2 B . Then U A = U B gives 1 (k1 + k2 ) x12A = 1 (k1 + k2 ) x12B and x1B = − x1 A = −0.150 m . The
2
2
maximum compression of spring 1 is 15.0 cm.
EVALUATE: When friction is not present mechanical energy is conserved and is continually transformed between
kinetic energy of the block and potential energy in the springs. If friction is present, its work removes mechanical
energy from the system.
!
!
IDENTIFY: Apply conservation of energy to relate x and h. Apply ∑ F = ma to relate a and x.
SET UP: The first condition, that the maximum height above the release point is h, is expressed as 1 kx 2 = mgh .
2
The magnitude of the acceleration is largest when the spring is compressed to a distance x; at this point the net
upward force is kx − mg = ma , so the second condition is expressed as x = (m/k )( g + a ) .
EXECUTE: (a) Substituting the second expression into the first gives
2
1 ⎛m⎞
m( g + a ) 2
.
k ⎜ ⎟ ( g + a ) 2 = mgh, or k =
2 ⎝k⎠
2 gh
(b) Substituting this into the expression for x gives x = mg
and x = 2h . The initial spring force is kx = mg and the
2h
net upward force approaches zero. But 1 kx 2 = mgh and sufficient potential energy is stored in the spring to move
2
the mass to height h.
IDENTIFY: At equilibrium the upward spring force equals the weight mg of the object. Apply conservation of
energy to the motion of the fish.
SET UP: The distance that the mass descends equals the distance the spring is stretched. K1 = K 2 = 0 , so
EVALUATE: 7.72. 2 gh
.
g+a When a → 0 , our results become k = U1 (gravitational) = U 2 (spring)
EXECUTE: Following the hint, the force constant k is found from mg = kd , or k = mg / d . When the fish falls
from rest, its gravitational potential energy decreases by mgy; this becomes the potential energy of the spring,
1 mg 2
y = mgy, or y = 2d .
which is 1 ky 2 = 1 (mg / d ) y 2 . Equating these,
2
2
2d 7-24 7.73. Chapter 7 EVALUATE: At its lowest point the fish is not in equilibrium. The upward spring force at this point is ky = 2kd ,
and this is equal to twice the weight. At this point the net force is mg, upward, and the fish has an upward
acceleration equal to g.
IDENTIFY: Apply Eq.(7.15) to the motion of the block.
SET UP: The motion from A to B is described in Figure 7.73. Figure 7.73 The normal force is n = mg cosθ , so f k = μ k n = μ k mg cosθ .
y A = 0; yB = (60.0 m)sin 30.0° = 3.00 m
K A + U A + Wother = K B + U B
EXECUTE: Work is done by gravity, by the spring force, and by friction, so Wother = W f and U = U el + U grav 2
K A = 0, K B = 1 mvB = 1 (1.50 kg)(7.00 m/s) 2 = 36.75 J
2
2 U A = U el, A + U grav, A = U el, A , since U grav, A = 0
U B = U el, B + U grav, B = 0 + mgyB = (1.50 kg)(9.80 m/s 2 )(3.00 m) = 44.1 J
Wother = W f = ( f k cos φ ) s = μ k mg cosθ (cos180°) s = − μ k mg cosθ s
Wother = −(0.50)(1.50 kg)(9.80 m/s 2 )(cos30.0°)(6.00 m) = −38.19 J
Thus U el, A − 38.19 J = 36.75 J + 44.10 J
U el, A = 38.19 J + 36.75 J + 44.10 J = 119 J 7.74. EVALUATE: U el must always be positive. Part of the energy initially stored in the spring was taken away by
friction work; the rest went partly into kinetic energy and partly into an increase in gravitational potential energy.
IDENTIFY: Apply Eq.(7.14) to the motion of the package. Wother = W fk , the work done by the kinetic friction force.
SET UP: f k = μ k n = μ k mg cosθ , with θ = 53.1° . Let L = 4.00 m , the distance the package moves before
reaching the spring and let d be the maximum compression of the spring. Let point 1 be the initial position of the
package, point 2 be just as it contacts the spring, point 3 be at the maximum compression of the spring, and point 4
be the final position of the package after it rebounds.
EXECUTE: (a) K1 = 0 , U 2 = 0 , Wother = − f k L = − μ k L cosθ . U1 = mgL sin θ . K 2 = 1 mv 2 , where v is the speed
2
before the block hits the spring. Eq.(7.14) applied to points 1 and 2, with y2 = 0 , gives U1 + Wother = K 2 . Solving
for v,
v = 2 gL(sin θ − μ k cosθ ) = 2(9.80 m/s 2 )(4.00 m)(sin 53.1° − (0.20)cos53.1°) = 7.30 m/s.
(b) Apply Eq.(7.14) to points 1 and 3. Let y3 = 0 . K1 = K3 = 0 . U1 = mg ( L + d )sin θ . U 2 = 1 kd 2 .
2 Wother = − f k ( L + d ) . Eq.(7.14) gives mg ( L + d )sin θ − μk mg cosθ ( L + d ) = 1 kd 2 . This can be written as
2
k
− d − L = 0. The factor multiplying d 2 is 4.504 m −1 , and use of the quadratic formula
2mg (sin θ − μk cosθ )
gives d = 1.06 m .
(c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the
motion the spring has no potential energy, and the distance below the starting point is determined solely by how
much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has
moved a distance L + d down the incline and L + d − y up the incline. The magnitude of the friction force is the
d2 same in both directions, μ k mg cos θ , and so the work done by friction is − μk (2 L + 2d − y ) mg cos θ . This must be
equal to the change in gravitational potential energy, which is − mgy sin θ . Equating these and solving for y gives
2μk cosθ
2μk
= (L + d )
. Using the value of d found in part (b) and the given values for μ k
sin θ + μ k cosθ
tan θ + μk
and θ gives y = 1.32 m .
y = (L + d ) Potential Energy and Energy Conservation 7.75. 7-25 EVALUATE: Our expression for y gives the reasonable results that y = 0 when μ k = 0 ; in the absence of friction
the package returns to its starting point.
(a) IDENTIFY and SET UP: Apply K A + U A + Wother = K B + U B to the motion from A to B.
EXECUTE: 2
K A = 0, K B = 1 mvB
2 2
U A = 0, U B = U el, B = 1 kxB , where xB = 0.25 m
2 Wother = WF = FxB
2
2
Thus FxB = 1 mvB + 1 kxB . (The work done by F goes partly to the potential energy of the stretched spring and
2
2
partly to the kinetic energy of the block.)
2
FxB = (20.0 N)(0.25 m) = 5.0 J and 1 kxB = 1 (40.0 N/m)(0.25 m) 2 = 1.25 J
2
2 2(3.75 J)
= 3.87 m/s
0.500 kg
(b) IDENTIFY: Apply Eq.(7.15) to the motion of the block. Let point C be where the block is closest to the wall.
When the block is at point C the spring is compressed an amount xC , so the block is 0.60 m − xC from the wall, 2
Thus 5.0 J = 1 mvB + 1.25 J and vB =
2 and the distance between B and C is xB + xC .
SET UP: The motion from A to B to C is described in Figure 7.75.
K B + U B + Wother = K C + U C
EXECUTE: Wother = 0 K B = 1 mv = 5.0 J − 1.25 J = 3.75 J
2
(from part (a))
2
U B = 1 kxB = 1.25 J
2
2
B K C = 0 (instantaneously at rest at
point closest to wall)
2
1
U C = 2 k xC
Figure 7.75 Thus 3.75 J + 1.25 J = 1 k xC
2 2 2(5.0 J)
= 0.50 m
40.0 N/m
The distance of the block from the wall is 0.60 m − 0.50 m = 0.10 m.
EVALUATE: The work (20.0 N)(0.25 m) = 5.0 J done by F puts 5.0 J of mechanical energy into the system. No
mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J.
IDENTIFY: Apply Eq.(7.14) to the motion of the student.
SET UP: Let x0 = 0.18 m , x1 = 0.71 m . The spring constants (assumed identical) are then known in terms of the
xC = 7.76. unknown weight w, 4kx0 = w . Let y = 0 at the initial position of the student.
EXECUTE: (a) The speed of the brother at a given height h above the point of maximum compression is then
⎛ x2
⎞
1
1⎛ w⎞
(4k ) g 2
found from (4k ) x12 = ⎜ ⎟ v 2 + mgh, or v 2 =
x1 − 2 gh = g ⎜ 1 − 2h ⎟ . Therefore,
2
2⎝ g ⎠
w
⎝ x0
⎠
v = (9.80 m/s 2 )((0.71 m) 2 (0.18 m) − 2(0.90 m)) = 3.13 m/s , or 3.1 m/s to two figures.
(b) Setting v = 0 and solving for h, h = 2kx12
x2
= 1 = 1.40 m, or 1.4 m to two figures.
mg 2 x0
2 (c) No; the distance x0 will be different, and the ratio ⎛ 0.53 m ⎞
x12 ( x1 + 0.53 m) 2
=
= x1 ⎜1 +
⎟ will be different.
x0
x1
x1 ⎠
⎝ Note that on a planet with lower g, x1 will be smaller and h will be larger.
EVALUATE: We are able to solve the problem without knowing either the mass of the student or the force
constant of the spring. 7-26 7.77. Chapter 7 ax = d 2 x/dt 2 , a y = d 2 y/dt 2 . Fx = max , Fy = ma y . U = ∫ Fx dx + ∫ Fy dy . IDENTIFY:
SET UP: 1
1
d
d
(cos ω0t ) = −ω0 sin ω0t .
(sin ω0t ) = ω0 cos ω0t . ∫ cos ω0t dt = sin ω0t , ∫ sin ω0t dt = − cos ω0t .
ω0
ω0
dt
dt vx = dx / dt , v y = dy / dt . E = K + U .
2
2
2
2
(a) ax = d 2 x/dt 2 = −ω0 x, Fx = max = − mω0 x. a y = d 2 y/dt 2 = −ω0 y, Fy = ma y = − mω0 y EXECUTE: 1
2
2
(b) U = − ⎡ ∫ Fx dx + ∫ Fy dy ⎤ = mω0 ⎡ ∫ xdx + ∫ ydy ⎤ = mω0 ( x 2 + y 2 )
⎣
⎦
⎣
⎦2
(c) vx = dx/dt = − x0ω0 sin ω0t = − x0ω0 ( y/y0 ). v y = dy/dt = + y0ω0 cos ω0t = + y0ω0 ( x/x0 ). (i) When x = x0 and y = 0, vx = 0 and v y = y0ω0 ,
1
1 22
1
1
2
2
2
2
2
2
K = m(vx + v y ) = my0 ω0 , U = ω02 mx0 and E = K + U = mω0 ( x0 + y0 )
2
2
2
2
(ii) When x = 0 and y = y0 , vx = − x0ω0 and v y = 0 , 7.78. 12 2
1
1
22
2
2
2
K = ω0 mx0 , U = mω0 y0 and E = K + U = mω0 ( x0 + y0 )
2
2
2
EVALUATE: The total energy is the same at the two points in part (c); the total energy of the system is constant.
IDENTIFY: Calculate the increase in kinetic energy for the car.
SET UP: The car gets (0.15)(1.3 × 108 J) of energy from one gallon of gasoline.
(a) The mechanical energy increase of the car is K 2 − K1 = 1 (1500 kg)(37 m/s) 2 = 1.027 × 106 J. Let
2 EXECUTE: α be the number of gallons of gasoline consumed. α (1.3 × 108 J)(0.15) = 1.027 × 106 J and α = 0.053gallons .
(b) (1.00 gallons) α = 19 accelerations
7.79. EVALUATE: The time over which the increase in velocity occurs doesn't enter into the calculation.
IDENTIFY: U = mgh . Use h = 150 m for all the water that passes through the dam.
SET UP: m = ρV and V = AΔh is the volume of water in a height Δh of water in the lake.
EXECUTE: (a) Stored energy = mgh = ( ρ V ) gh = ρ A(1 m) gh . stored energy = (1000 kg/m3 )(3.0 × 106 m 2 )(1 m)(9.8 m/s 2 )(150 m) = 4.4 × 1012 J.
(b) 90% of the stored energy is converted to electrical energy, so (0.90)(mgh) = 1000 kWh .
(0.90) ρVgh = 1000 kWh . V = (1000 kWh)((3600 s) (1 h))
= 2.7 × 103 m3 .
(0.90)(1000 kg/m3 )(150 m)(9.8 m/s 2 ) V 2.7 × 103 m3
=
= 9.0 × 10−4 m .
A 3.0 × 106 m 2
EVALUATE: Δh is much less than 150 m, so using h = 150 m for all the water that passed through the dam was a
very good approximation.
IDENTIFY and SET UP: The potential energy of a horizontal layer of thickness dy, area A, and height y is
dU = (dm) gy. Let ρ be the density of water.
EXECUTE: dm = ρ dV = ρ A dy, so dU = ρ Agy dy.
The total potential energy U is
Change in level of the lake: AΔh = Vwater . Δh = 7.80. h h 0 0 U = ∫ dU = ρ Ag ∫ y dy = 1 ρ Agh 2 .
2
A = 3.0 × 106 m 2 and h = 150 m, so U = 3.3 × 1014 J = 9.2 × 107 kWh
EVALUATE: The volume is Ah and the mass of water is ρV = ρ Ah. The average depth is hav = h/2, so
U = mghav .
7.81. IDENTIFY:
SET UP: Apply Fx = − ∂U
∂U
∂U
, Fy = −
and Fz = −
.
∂x
∂y
∂z r = ( x 2 + y 2 + z 2 )1/ 2 . ∂ (1/r )
x
∂ (1/r )
y
∂ (1/r )
z
=− 2
,
=− 2
and
=− 2
.
( x + y 2 )3/ 2
( x + y 2 )3/ 2
( x + y 2 )3/ 2
∂x
∂y
∂z Potential Energy and Energy Conservation EXECUTE: Fy = − (a) U (r ) = − 7-27 Gm1m2 x
∂U
Gm1m2
⎡ ∂ (1/r ) ⎤
. Fx = −
= +Gm1m2 ⎢
⎥ = − ( x 2 + y 2 + z 2 )3/ 2 . Similarly,
r
∂x
⎣ ∂x ⎦ Gm1m2 y
Gm1m2 z
and Fz = − 2
.
( x 2 + y 2 + z 2 )3/ 2
( x + y 2 + z 2 )3/ 2 Gm1m2 x
Gm1m2 y
Gm1m2 z
, Fy = −
and Fz = −
.
r3
r3
r3
Gm1m2 2
Gm1m2
F = Fx2 + Fy2 + Fz2 =
x + y2 + z2 =
.
3
r
r2
!
!
(c) Fx , Fy and Fz are negative. Fx = α x , Fy = α y and Fz = α z , where α is a constant, so F and the vector r from
!
!
m1 to m2 are in the same direction. Therefore, F is directed toward m1 at the origin and F is attractive.
(b) ( x 2 + y 2 + z 2 )3/ 2 = r 3 so Fx = − 7.82. EVALUATE: When m2 moves to larger r, the work done on it by the attractive gravity force is negative. Since
W = −ΔU , negative work done by gravity means the gravitational potential energy increases.
Gm1m2
U (r ) = −
does increase (becomes less negative) as r increases. For an object near the surface of the earth,
r
Gm1m2
U (r ) = −
will be shown in Chapter 12 to be equivalent to U grav = mgy .
r
IDENTIFY: Calculate the work W done by this force. If the force is conservative, the work is path independent.
!
P2 !
SET UP: W = ∫ F ⋅ dl .
P
1 EXECUTE: P2 P2 P
1 P
1 (a) W = ∫ Fy dy = C ∫ y 2 dy . W doesn't depend on x, so it is the same for all paths between P and
1 P2 . The force is conservative.
P2 P2 P
1 P
1 (b) W = ∫ Fx dx = C ∫ y 2 dx . W will be different for paths between points P and P2 for which y has different
1 values. For example, if y has the constant value y0 along the path, then W = Cy0 ( x2 − x1 ) . W depends on the value 7.83. of y0 . The force is not conservative.
!
Cy 3
EVALUATE: F = Cy 2 ˆ has the potential energy function U ( y ) = −
j
. We cannot find a potential energy
3
!
ˆ
function for F = Cy 2 i .
!
F = −α xy 2 ˆ, α = 2.50 N/m3
j
!
!
IDENTIFY: F is not constant so use Eq.(6.14) to calculate W. F must be evaluated along the path.
(a) SET UP: The path is sketched in Figure 7.83a.
!
ˆ
j
dl = dxi + dyˆ
!!
F ⋅ dl = −α xy 2 dy
!!
On the path, x = y so F ⋅ dl = −α y 3 dy EXECUTE: Figure 7.83a
!
2!
y2
y2
4
W = ∫ F ⋅ dl = ∫ (−α y 3 ) dy = −(α / 4) y 4| = −(α / 4)( y2 − y14 )
1 ()
y1 y1 y1 = 0, y2 = 3.00 m, so W = − (2.50 N/m )(3.00 m) = −50.6 J
(b) SET UP: The path is sketched in Figure 7.83b.
1
4 3 4 Figure 7.83b
!
!!
ˆ
For the displacement from point 1 to point 2, dl = dxi , so F ⋅ dl = 0 and W = 0. (The force is perpendicular to
the displacement at each point along the path, so W = 0.) 7-28 Chapter 7 !
!!
For the displacement from point 2 to point 3, dl = dyˆ, so F ⋅ dl = −α xy 2 dy. On this path, x = 3.00 m, so
j
!!
F ⋅ dl = −(2.50 N/m3 )(3.00 m)y 2 dy = −(7.50 N/m 2 ) y 2 dy.
!
3!
y3
3
3
EXECUTE: W = ∫ F ⋅ dl = − (7.50 N/m 2 ) ∫ y 2 dy = −(7.50 N/m 2 ) 1 ( y3 − y2 )
3
2 W = −(7.50 N/m ) ( ) (3.00 m) = −67.5 J
2 7.84. y2 3 1
3 (c) EVALUATE: For these two paths between the same starting and ending points the work is different, so the
force is nonconservative.
!
P2 !
IDENTIFY: Use W = ∫ F ⋅ dl to calculate W for each segment of the path.
P
1
!!
SET UP: F ⋅ dl = Fx dx = α xy dx
EXECUTE: (a) The path is sketched in Figure 7.84.
!!
!
(b) (1): x = 0 along this leg, so F = 0 and W = 0 . (2): Along this leg, y = 1.50 m , so F ⋅ dl = (3.00 N m) xdx ,
!
!!
and W = (1.50 N m)((1.50 m) 2 − 0) = 3.38 J (3) F ⋅ dl = 0 , so W = 0 (4) y = 0 , so F = 0 and W = 0 . The work
done in moving around the closed path is 3.38 J.
(c) The work done in moving around a closed path is not zero, and the force is not conservative.
EVALUATE: There is no potential energy function for this force. Figure 7.84
7.85. IDENTIFY: Use Eq.(7.16) to relate Fx and U ( x) . The equilibrium is stable where U ( x ) is a local minimum and
the equilibrium is unstable where U ( x) is a local maximum.
SET UP: The maximum and minimum values of x are those for which U ( x) = E . K = E − U , so the maximum
speed is where U is a minimum.
dU
EXECUTE: (a) For the given proposed potential U ( x), −
= −kx + F , so this is a possible potential function.
dx
For this potential, U (0) = − F 2 2k , not zero. Setting the zero of potential is equivalent to adding a constant to the
potential; any additive constant will not change the derivative, and will correspond to the same force.
(b) At equilibrium, the force is zero; solving − kx + F = 0 for x gives x0 = F/k . U ( x0 ) = − F 2 /k , and this is a
minimum of U, and hence a stable point.
(c) The graph is given in Figure 7.85.
(d) No; Ftot = 0 at only one point, and this is a stable point.
(e) The extreme values of x correspond to zero velocity, hence zero kinetic energy, so U ( x± ) = E , where x± are the extreme points of the motion. Rather than solve a quadratic, note that 1
2 k ( x − F/k ) 2 − F 2 /k , so U ( x± ) = E 2 becomes 1⎛
F⎞
F2
F
F
F
F
x− = − .
. x± − = ±2 , so x+ = 3
k ⎜ x± − ⎟ F 2 /k =
k
k
k
k
2⎝
k⎠
k (f) The maximum kinetic energy occurs when U ( x ) is a minimum, the point x0 = F/k found in part (b). At this point K = E − U = ( F 2 /k ) − (− F 2 /k ) = 2 F 2 /k , so v = 2 F mk . Potential Energy and Energy Conservation 7-29 EVALUATE: As E increases, the magnitudes of x+ and x− increase. The particle cannot reach values of x for
which E < U ( x) because K cannot be negative. Figure 7.85
7.86. 7.87. IDENTIFY: Use Eq.(7.16) to relate Fx and U ( x) . The equilibrium is stable where U ( x ) is a local minimum and
the equilibrium is unstable where U ( x) is a local maximum.
SET UP: dU/dx is the slope of the graph of U versus x. K = E − U , so K is a maximum when U is a minimum.
The maximum x is where E = U .
EXECUTE: (a) The slope of the U vs. x curve is negative at point A, so Fx is positive (Eq. (7.16)).
(b) The slope of the curve at point B is positive, so the force is negative.
(c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m.
(d) The curve at point C looks pretty close to flat, so the force is zero.
(e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than
U ( A) , the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at
any point where the potential energy is larger than U ( A) . On the graph, that looks to be at about 2.2 m.
(f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m.
(g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C.
EVALUATE: If E is less than U at point C, the particle is trapped in one or the other of the potential "wells" and
cannot move from one allowed region of x to the other.
IDENTIFY: K = E − U determines v( x) .
SET UP: v is a maximum when U is a minimum and v is a minimum when U is a maximum. Fx = − dU/dx . The
extreme values of x are where E = U ( x ) .
EXECUTE: (a) Eliminating β in favor of α and x0 ( β = α /x0 ) , U ( x) = α
x 2 − β
x = 2
α x0
2
0 xx 2 − α
x0 x = 2
α ⎡⎛ x0 ⎞ ⎛ x0 ⎞ ⎤ ⎟ − ⎜ ⎟⎥ .
2 ⎢⎜
x0 ⎢⎝ x ⎠ ⎝ x ⎠ ⎥
⎣
⎦ ⎛α ⎞
U ( x0 ) = ⎜ 2 ⎟ (1 − 1) = 0 . U ( x) is positive for x < x0 and negative for x > x0 ( α and β must be taken as
⎝ x0 ⎠
positive). The graph of U ( x) is sketched in Figure 7.87a.
2
⎛ 2α ⎞ ⎛ ⎛ x ⎞ ⎛ x ⎞ ⎞
2
U = ⎜ 2 ⎟ ⎜ ⎜ 0 ⎟ − ⎜ 0 ⎟ ⎟ . The proton moves in the positive x-direction, speeding up until it
⎜
⎟
m
⎝ mx0 ⎠ ⎝ ⎝ x ⎠ ⎝ x ⎠ ⎠
reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the
square root in the expression for v ( x ) indicates that the particle will be found only in the region where U < 0 , that (b) v( x) = − is, x > x0 . The graph of v( x) is sketched in Figure 7.87b.
(c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy.
3
2
dU α ⎡ ⎛ x0 ⎞ ⎛ x0 ⎞ ⎤
This minimum occurs when dU = 0 , or
= 3 ⎢ −2 ⎜ ⎟ + ⎜ ⎟ ⎥ = 0,
dx
dx x0 ⎣ ⎝ x ⎠ ⎝ x ⎠ ⎦
⎢
⎥
which has the solution x = 2 x0 . U (2 x0 ) = − α
2
4 x0 , so v = α
2
2mx0 . (d) The maximum speed occurs at a point where dU = 0 , and from Eq. (7.15), the force at this point is zero.
dx 7-30 Chapter 7 (e) x1 = 3 x0 , and U (3x0 ) = − 2α
.
2
9 x0 2
⎛ ⎛ x ⎞ ⎛ x ⎞2
⎞
2
2 ⎡⎛ −2 α ⎞ α ⎛ ⎛ x0 ⎞ x0 ⎞ ⎤
− 2 ⎜ ⎜ ⎟ − ⎟ ⎥ = 2α2 ⎜ ⎜ 0 ⎟ − ⎜ 0 ⎟ − 2 9 ⎟ .
(U ( x1 ) − U ( x)) =
⎢⎜
2⎟
⎟
mx0 ⎜ ⎝ x ⎠ ⎝ x ⎠
m
m ⎢⎝ 9 x0 ⎠ x0 ⎜ ⎝ x ⎠
x ⎟⎥
⎝
⎠
⎝
⎠⎦
⎣
The particle is confined to the region where U ( x) < U ( x1 ) . The maximum speed still occurs at x = 2 x0 , but now v( x) = the particle will oscillate between x1 and some minimum value (see part (f)).
(f) Note that U ( x) − U ( x1 ) can be written as
2
α ⎡⎛ x0 ⎞ ⎛ x0 ⎞ ⎛ 2 ⎞ ⎤ α ⎡⎛ x0 ⎞ 1 ⎤ ⎡⎛ x0 ⎞ 2 ⎤
,
⎟ − ⎜ ⎟ + ⎜ ⎟⎥ = 2 ⎜ ⎟ −
⎜ ⎟−
2 ⎢⎜
x0 ⎢⎝ x ⎠ ⎝ x ⎠ ⎝ 9 ⎠ ⎥ x0 ⎢⎝ x ⎠ 3 ⎥ ⎢⎝ x ⎠ 3 ⎥
⎣
⎦⎣
⎦
⎣
⎦
which is zero (and hence the kinetic energy is zero) at x = 3 x0 = x1 and x = 3 x0 . Thus, when the particle is
2
released from x0 , it goes on to infinity, and doesn’t reach any maximum distance. When released from x1 , it
oscillates between
EVALUATE: 3
2 x0 and 3 x0 . In each case the proton is released from rest and E = U ( xi ) , where xi is the point where it is released. When xi = x0 the total energy is zero. When xi = x1 the total energy is negative. U ( x ) → 0 as x → ∞ , so
for this case the proton can't reach x → ∞ and the maximum x it can have is limited. Figure 7.87 8 MOMENTUM, IMPULSE, AND COLLISIONS 8.1. IDENTIFY and SET UP: (a) p = (10,000 kg)(12.0 m/s) = 1.20 × 105 kg ⋅ m/s EXECUTE:
(b) (i) v = p = mv. K = 1 mv 2 .
2 p 1.20 × 105 kg ⋅ m/s
=
= 60.0 m/s . (ii)
m
2000 kg vSUV = 8.2. 1
2 2
2
mT vT = 1 mSUV vSUV , so
2 10,000 kg
mT
(12.0 m/s) = 26.8 m/s
vT =
2000 kg
mSUV EVALUATE: The SUV must have less speed to have the same kinetic energy as the truck than to have the same
momentum as the truck.
IDENTIFY: Example 8.1 shows that the two iceboats have the same kinetic energy at the finish line. K = 1 mv 2 .
2
p = mv .
SET UP: Let A be the iceboat with mass m and let B be the iceboat with mass 2m, so mB = 2mA . EXECUTE: K A = K B gives 1
2 2
2
mv A = 1 mvB . v A =
2 mB
vB = 2vB .
mA ( ) p A = mAv A . pB = mB vB = (2mA ) v A / 2 = 2mAv A = 2 p A . 8.3. EVALUATE: The more massive boat must have less speed but greater momentum than the other boat in order to
have the same kinetic energy.
IDENTIFY and SET UP: p = mv . K = 1 mv 2 .
2
2 EXECUTE: (a) v = p2
p
⎛ p⎞
and K = 1 m ⎜ ⎟ =
.
2
m
⎝ m ⎠ 2m (b) K c = K b and the result from part (a) gives pc2
p2
mb
0.145 kg
= b . pb =
pc =
pc = 1.90 pc . The baseball
mc
2mc 2mb
0.040 kg has the greater magnitude of momentum. pc / pb = 0.526 .
(c) p 2 = 2mK so pm = pw gives 2mm K m = 2mw K w . w = mg , so wm K m = ww K w . 8.4. ⎛w ⎞
⎛ 700 N ⎞
Kw = ⎜ m ⎟ Km = ⎜
⎟ K m = 1.56 K m .
⎝ 450 N ⎠
⎝ ww ⎠
The woman has greater kinetic energy. K m / K w = 0.641 .
EVALUATE: For equal kinetic energy, the more massive object has the greater momentum. For equal momenta,
the less massive object has the greater kinetic energy.
IDENTIFY: Each momentum component is the mass times the corresponding velocity component.
SET UP: Let +x be along the horizontal motion of the shotput. Let +y be vertically upward. vx = v cosθ ,
v y = v sin θ .
EXECUTE: The horizontal component of the initial momentum is
px = mvx = mv cosθ = (7.30 kg)(15.0 m/s)cos 40.0° = 83.9 kg ⋅ m/s . The vertical component of the initial momentum is p y = mv y = mv sin θ = (7.30 kg)(15.0 m/s)sin40.0° = 70.4 kg ⋅ m/s
EVALUATE: The initial momentum is directed at 40.0° above the horizontal. 8-1 8-2 8.5. Chapter 8 !
!
IDENTIFY: For each object, p = mv and K = 1 mv 2 . The total momentum is the vector sum of the momenta of
2
each object. The total kinetic energy is the scalar sum of the kinetic energies of each object.
SET UP: Let object A be the 110 kg lineman and object B the 125 kg lineman. Let +x be the object to the right, so
v Ax = +2.75 m/s and vBx = −2.60 m/s .
EXECUTE: (a) Px = mAv Ax + mB vBx = (110 kg)(2.75 m/s) + (125 kg)(−2.60 m/s) = −22.5 kg ⋅ m/s . The net
momentum has magnitude 22.5 kg ⋅ m/s and is directed to the left. 8.6. 2
2
(b) K = 1 m Av A + 1 mB vB = 1 (110 kg)(2.75 m/s) 2 + 1 (125 kg)(2.60 m/s) 2 = 838 J
2
2
2
2
EVALUATE: The kinetic energy of an object is a scalar and is never negative. It depends only on the magnitude of
the velocity of the object, not on its direction. The momentum of an object is a vector and has both magnitude and
direction. When two objects are in motion, their total kinetic energy is greater than the kinetic energy of either one.
But if they are moving in opposite directions, the net momentum of the system has a smaller magnitude than the
magnitude of the momentum of either object.
!!
!
!
!
IDENTIFY: For each object p = mv and the net momentum of the system is P = pA + pB . The momentum
vectors are added by adding components. The magnitude and direction of the net momentum is calculated from its
x and y components.
SET UP: Let object A be the pickup and object B be the sedan. v Ax = −14.0 m/s , v Ay = 0 . vBx = 0 , vBy = +23.0 m/s . (a) Px = p Ax + pBx = mAv Ax + mB vBx = (2500 kg)(−14.0 m/s) + 0 = −3.50 × 104 kg ⋅ m/s EXECUTE: Py = p Ay + pBy = mAv Ay + mB vBy = (1500 kg)(+23.0 m/s) = +3.45 × 104 kg ⋅ m/s
(b) P = Px2 + Py2 = 4.91 × 104 kg ⋅ m/s . From Figure 8.6, tan θ = Px
Py = 3.50 × 104 kg ⋅ m/s
and θ = 45.4° . The net
3.45 × 104 kg ⋅ m/s momentum has magnitude 4.91 × 104 kg ⋅ m/s and is directed at 45.4° west of north.
EVALUATE: The momenta of the two objects must be added as vectors. The momentum of one object is west and
the other is north. The momenta of the two objects are nearly equal in magnitude, so the net momentum is directed
approximately midway between west and north. Figure 8.6
8.7. IDENTIFY: The average force on an object and the object’s change in momentum are related by Eq. 8.9. The
weight of the ball is w = mg .
SET UP: Let +x be in the direction of the final velocity of the ball, so v1x = 0 and v2 x = 25.0 m/s . EXECUTE: 8.8. ( Fav ) x (t2 − t1 ) = mv2 x − mv1x gives ( Fav ) x = mv2 x − mv1x (0.0450 kg)(25.0 m/s)
=
= 562 N .
t2 − t1
2.00 × 10−3 s w = (0.0450 kg)(9.80 m/s 2 ) = 0.441 N . The force exerted by the club is much greater than the weight of the ball,
so the effect of the weight of the ball during the time of contact is not significant.
EVALUATE: Forces exerted during collisions typically are very large but act for a short time.
IDENTIFY: The change in momentum, the impulse and the average force are related by Eq. 8.9.
SET UP: Let the direction in which the batted ball is traveling be the +x direction, so v1x = −45.0 m/s and
v2 x = 55.0 m/s .
EXECUTE: (a) Δpx = p2 x − p1x = m(v2 x − v1x ) = (0.145 kg)(55.0 m/s − [ −45.0 m/s]) = 14.5 kg ⋅ m/s . J x = Δpx , so J x = 14.5 kg ⋅ m/s . Both the change in momentum and the impulse have magnitude 14.5 kg ⋅ m/s .
J x 14.5 kg ⋅ m/s
=
= 7250 N .
Δt 2.00 × 10−3 s
EVALUATE: The force is in the direction of the momentum change.
IDENTIFY: Use Eq. 8.9. We know the intial momentum and the impluse so can solve for the final momentum and
then the final velocity.
(b) ( Fav ) x =
8.9. Momentum, Impulse, and Collisions 8-3 SET UP: Take the x-axis to be toward the right, so v1x = +3.00 m / s. Use Eq. 8.5 to calculate the impulse, since
the force is constant.
EXECUTE: (a) J x = p2 x − p1x J x = Fx (t2 − t1 ) = (+25.0 N)(0.050 s) = +1.25 kg ⋅ m/s
Thus p2 x = J x + p1x = +1.25 kg ⋅ m/s + (0.160 kg)( +3.00 m/s) = +1.73 kg ⋅ m/s v2 x = p2 x 1.73 kg ⋅ m/s
=
= +10.8 kg ⋅ m/s (to the right)
0.160 kg
m (b) J x = Fx (t2 − t1 ) = (−12.0 N)(0.050 s) = −0.600 kg ⋅ m/s (negative since force is to left) p2 x = J x + p1x = −0.600 kg ⋅ m/s + (0.160 kg)(+3.00 m/s) = −0.120 kg ⋅ m/s
v2 x = 8.10. p2 x −0.120 kg ⋅ m/s
=
= −0.75 m/s (to the left)
0.160 kg
m EVALUATE: In part (a) the impulse and initial momentum are in the same direction and vx increases. In part (b) the
impulse and initial momentum are in opposite directions and the velocity decreases.
IDENTIFY: The impulse, change in momentum and change in velocity are related by Eq. 8.9.
SET UP: Fy = 26,700 N and Fx = 0 . The force is constant, so ( Fav ) y = Fy .
(a) J y = Fy Δt = (26,700 N)(3.90 s) = 1.04 × 105 N ⋅ s . EXECUTE: (b) Δp y = J y = 1.04 × 105 kg ⋅ m/s .
(c) Δp y = mΔv y . Δv y = 8.11. Δp y
m = 1.04 × 105 kg ⋅ m/s
= 1.09 m/s .
95,000 kg 2
(d) The initial velocity of the shuttle isn’t known. The change in kinetic energy is ΔK = K 2 − K1 = 1 m(v2 − v12 ) . It
2
depends on the initial and final speeds and isn’t determined solely by the change in speed.
EVALUATE: The force in the +y direction produces an increase of the velocity in the +y direction.
!
t2 !
IDENTIFY: The force is not constant so J = ∫ F dt . The impulse is related to the change in velocity by Eq. 8.9.
t1 !
t2
Only the x component of the force is nonzero, so J x = ∫ Fx dt is the only nonzero component of J . SET UP: t1 J x = m(v2 x − v1x ) . t1 = 2.00 s , t2 = 3.50 s .
EXECUTE: (a) A = Fx 781.25 N
=
= 500 N/s 2 .
(1.25 s) 2
t2 t2 3
(b) J x = ∫ At 2 dt = 1 A(t2 − t13 ) = 1 (500 N/s 2 )([3.50 s]3 − [2.00 s]3 ) = 5.81×103 N ⋅ s .
3
3
t1 (c) Δvx = v2 x − v1x = 8.12. J x 5.81× 103 N ⋅ s
=
= 2.70 m/s . The x component of the velocity of the rocket increases by
m
2150 kg 2.70 m/s.
EVALUATE: The change in velocity is in the same direction as the impulse, which in turn is in the direction of the net
force. In this problem the net force equals the force applied by the engine, since that is the only force on the rocket.
IDENTIFY: Apply Eq. 8.9 to relate the change in momentum of the momentum to the components of the average
force on it.
SET UP: Let +x be to the right and +y be upward.
EXECUTE: (a) J x = Δpx = mv2 x − mv1x = (0.145 kg)(−[65.0 m/s]cos30° − 50.0 m/s) = −15.4 kg ⋅ m/s .
J y = Δp y = mv2 y − mv1 y = (0.145 kg)([65.0 m/s]sin 30° − 0) = 4.71 kg ⋅ m/s
The horizontal component is 15.4 kg ⋅ m/s , to the left and the vertical component is 4.71 kg ⋅ m/s , upward.
J
4.71 kg ⋅ m/s
J x −15.4 kg ⋅ m/s
=
= −8800 N . Fav-y = y =
= 2690 N .
−3
Δt
1.75 × 10 s
Δt 1.75 × 10−3 s
The horizontal component is 8800 N, to the left, and the vertical component is 2690 N, upward.
EVALUATE: The ball gains momentum to the left and upward and the force components are in these directions.
!!
!!
!
!
!
IDENTIFY: The force is constant during the 1.0 ms interval that it acts, so J = F Δt . J 5 p2 2 p1 5 m(v 2 2 v1 ) .
!
SET UP: Let +x be to the right, so v1x = +5.00 m/s . Only the x component of J is nonzero, and
(b) Fav-x = 8.13. J x = m(v2 x − v1x ) . 8-4 Chapter 8 EXECUTE: (a) The magnitude of the impulse is J = F Δt = (2.50 × 103 N)(1.00 × 10−3 s) = 2.50 N ⋅ s . The direction
of the impulse is the direction of the force.
+2.50 N ⋅ s
J
(b) (i) v2 x = x + v1x . J x = +2.50 N ⋅ s . v2 x =
+ 5.00 m/s = 6.25 m/s . The stone’s velocity has magnitude
2.00 kg
m 6.25 m/s and is directed to the right. (ii) Now J x = −2.50 N ⋅ s and v2 x = 8.14. stone’s velocity has magnitude 3.75 m/s and is directed to the right.
EVALUATE: When the force and initial velocity are in the same direction the speed increases and when they are
in opposite directions the speed decreases.
IDENTIFY: Apply conservation of momentum to the system of the astronaut and tool.
SET UP: Let A be the astronaut and B be the tool. Let +x be the direction in which she throws the tool, so
vB 2 x = +3.20 m/s . Assume she is initially at rest, so v A1x = vB1x = 0 . Solve for v A 2 x .
EXECUTE: 8.15. −2.50 N ⋅ s
+ 5.00 m/s = 3.75 m/s . The
2.00 kg P x = P2 x . P x = mAv A1x + mB vB1x = 0 . P2 x = mAv A 2 x + mB vB 2 x = 0 and
1
1 mv
(2.25 kg)(3.20 m/s)
vA2 x = − B A2 x = −
= −0.105 m/s . Her speed is 0.105 m/s and she moves opposite to the
mA
68.5 kg
direction in which she throws the tool.
EVALUATE: Her mass is much larger than that of the tool so to have the same magnitude of momentum as the
tool her speed is much less.
IDENTIFY: Since drag effects are neglected there is no net external force on the system of squid plus expelled
water and the total momentum of the system is conserved. Since the squid is initially at rest, with the water in its
cavity, the initial momentum of the system is zero. For each object, K = 1 mv 2 .
2
SET UP: Let A be the squid and B be the water it expels, so mA = 6.50 kg − 1.75 kg = 4.75 kg . Let +x be the direction in which the water is expelled. v A 2 x = −2.50 m/s . Solve for vB 2 x .
EXECUTE: (a) P x = 0 . P2 x = P x , so 0 = mAv A 2 x + mB vB 2 x . vB 2 x = −
1
1 m Av A 2 x
(4.75 kg)(−2.50 m/s)
=−
= +6.79 m/s .
mB
1.75 kg 2
2
(b) K 2 = K A 2 + K B 2 = 1 mAv A 2 + 1 mB vB 2 = 1 (4.75 kg)(2.50 m/s) 2 + 1 (1.75 kg)(6.79 m/s) 2 = 55.2 J The initial kinetic
2
2
2
2 8.16. energy is zero, so the kinetic energy produced is K 2 = 55.2 J .
EVALUATE: The two objects end up with momenta that are equal in magnitude and opposite in direction, so the
total momentum of the system remains zero. The kinetic energy is created by the work done by the squid as it
expels the water.
IDENTIFY: Apply conservation of momentum to the system of you and the ball. In part (a) both objects have the
same final velocity.
SET UP: Let +x be in the direction the ball is traveling initially. m A = 0.400 kg (ball). mB = 70.0 kg (you).
EXECUTE: 8.17. (a) P x = P2 x gives (0.400 kg)(10.0 m/s) = (0.400 kg + 70.0 kg)v2 and v2 = 0.0568 m/s .
1 (b) P x = P2 x gives (0.400 kg)(10.0 m/s) = (0.400 kg)( − 8.00 m/s) + (70.0 kg)vB 2 and vB 2 = 0.103 m/s .
1
EVALUATE: When the ball bounces off it has a greater change in momentum and you acquire a greater final speed.
IDENTIFY: Apply conservation of momentum to the system of the two pucks.
SET UP: Let +x be to the right.
EXECUTE: (a) P x = P2 x says (0.250)v A1 = (0.250 kg)(−0.120 m/s) + (0.350 kg)(0.650 m/s) and v A1 = 0.790 m/s .
1
(b) K1 = 1 (0.250 kg)(0.790 m/s) 2 = 0.0780 J .
2 8.18. K 2 = 1 (0.250 kg)(0.120 m/s) 2 + 1 (0.350 kg)(0.650 m/s) 2 = 0.0757 J and ΔK = K 2 − K1 = −0.0023 J .
2
2
EVALUATE: The total momentum of the system is conserved but the total kinetic energy decreases.
IDENTIFY: Since road friction is neglected, there is no net external force on the system of the two cars and the
total momentum of the system is conserved. For each object, K = 1 mv 2 .
2
SET UP: Let A be the 1750 kg car and B be the 1450 kg car. Let +x be to the right, so v A1x = +1.50 m/s , vB1x = −1.10 m/s , and v A 2 x = +0.250 m/s . Solve for vB 2 x .
EXECUTE: (a) P x = P2 x . mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x . vB 2 x =
1 vB 2 x = mAv A1x + mB vB1x − mAv A 2 x
.
mB (1750 kg)(1.50 m/s) + (1450 kg)(−1.10 m/s) − (1750 kg)(0.250 m/s)
= 0.409 m/s .
1450 kg After the collision the lighter car is moving to the right with a speed of 0.409 m/s. Momentum, Impulse, and Collisions 8-5 2
2
(b) K1 = 1 mAv A1 + 1 mB vB1 = 1 (1750 kg)(1.50 m/s)2 + 1 (1450 kg)(1.10 m/s) 2 = 2846 J .
2
2
2
2
2
2
K 2 = 1 mAv A 2 + 1 mB vB 2 = 1 (1750 kg)(0.250 m/s) 2 + 1 (1450 kg)(0.409 m/s) 2 = 176 J .
2
2
2
2 8.19. The change in kinetic energy is ΔK = K 2 − K1 = 176 J − 2846 J = −2670 J .
EVALUATE: The total momentum of the system is constant because there is no net external force during the
collision. The kinetic energy of the system decreases because of negative work done by the forces the cars exert on
each other during the collision.
IDENTIFY: Since the rifle is loosely held there is no net external force on the system consisting of the rifle, bullet
and propellant gases and the momentum of this system is conserved. Before the rifle is fired everything in the
system is at rest and the initial momentum of the system is zero.
SET UP: Let +x be in the direction of the bullet’s motion. The bullet has speed 601 m/s − 1.85 m/s = 599 m/s
relative to the earth. P2 x = prx + pbx + pgx , the momenta of the rifle, bullet and gases. vrx = −1.85 m/s and
vbx = +599 m/s .
EXECUTE: P2 x = P x = 0 . prx + pbx + pgx = 0 . pgx = − prx − pbx = −(2.80 kg)( −1.85 m/s) − (0.00720 kg)(599 m/s)
1 and pgx = +5.18 kg ⋅ m/s − 4.31 kg ⋅ m/s = 0.87 kg ⋅ m/s . The propellant gases have momentum 0.87 kg ⋅ m/s , in the 8.20. same direction as the bullet is traveling.
EVALUATE: The magnitude of the momentum of the recoiling rifle equals the magnitude of the momentum of the
bullet plus that of the gases as both exit the muzzle.
IDENTIFY: In part (a) no horizontal force implies Px is constant. In part (b) use the energy expression, Eq. 7.14,
to find the potential energy intially in the spring.
SET UP: Initially both blocks are at rest. Figure 8.20
EXECUTE: (a) mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x 0 = mAv A 2 x + mB vB 2 x
⎛m ⎞
⎛ 3.00 kg ⎞
v A 2 x = − ⎜ B ⎟ vB 2 x = − ⎜
⎟ (+1.20 m/s) = −3.60 m/s
⎝ 1.00 kg ⎠
⎝ mA ⎠
Block A has a final speed of 3.60 m/s, and moves off in the opposite direction to B.
(b) Use energy conservation: K1 + U1 + Wother = K 2 + U 2 .
Only the spring force does work so Wother = 0 and U = U el .
K1 = 0 (the blocks initially are at rest)
U 2 = 0 (no potential energy is left in the spring)
2
2
K 2 = 1 mAv A 2 + 1 mB vB 2 = 1 (1.00 kg)(3.60 m/s) 2 + 1 (3.00 kg)(1.20 m/s) 2 = 8.64 J
2
2
2
2 U1 = U1,el the potential energy stored in the compressed spring.
Thus U1,el = K 2 = 8.64 J 8.21. EVALUATE: The blocks have equal and opposite momenta as they move apart, since the total momentum is zero.
The kinetic energy of each block is positive and doesn’t depend on the direction of the block’s velocity, just on its
magnitude.
IDENTIFY: Since friction at the pond surface is neglected, there is no net external horizontal force and the
horizontal component of the momentum of the system of hunter plus bullet is conserved. Both objects are initially
at rest, so the initial momentum of the system is zero. Gravity and the normal force exerted by the ice together
produce a net vertical force while the rifle is firing, so the vertical component of momentum is not conserved.
SET UP: Let object A be the hunter and object B be the bullet. Let +x be the direction of the horizontal
component of velocity of the bullet. Solve for v A 2 x . 8-6 Chapter 8 (a) vB 2 x = +965 m/s . P x = P2 x = 0 . 0 = mAv A 2 x + mB vB 2 x and
1 EXECUTE: vA2 x = − 8.22. ⎛ 4.20 × 10−3 kg ⎞
mB
vB 2 x = − ⎜
⎟ (965 m/s) = −0.0559 m/s .
mA
⎝ 72.5 kg ⎠ ⎛ 4.20 × 10−3 kg ⎞
(b) vB 2 x = vB 2 cosθ = (965 m/s)cos56.0° = 540 m/s . v A 2 x = − ⎜
⎟ (540 m/s) = −0.0313 m/s .
⎝ 72.5 kg ⎠
EVALUATE: The mass of the bullet is much less than the mass of the hunter, so the final mass of the hunter plus
gun is still 72.5 kg, to three significant figures. Since the hunter has much larger mass, her final speed is much less
than the speed of the bullet.
IDENTIFY: Assume the nucleus is initially at rest. K = 1 mv 2 .
2
SET UP: Let +x be to the right. v A 2 x = −v A and vB 2 x = +vB . ⎛m ⎞
(a) P2 x = P x = 0 gives m Av A 2 x + mB vB 2 x = 0 . vB = ⎜ A ⎟ v A .
1
⎝ mB ⎠
2
2
1
mv
K
m Av A
m
2
(b) A = 1 A A =
= B.
2
K B 2 mB vB mB ( mAv A / mB )2 mA
EXECUTE: 8.23. EVALUATE: The lighter fragment has the greater kinetic energy.
IDENTIFY: Apply conservation of momentum to the nucleus and its fragments. The initial momentum is zero.
The 214 Po nucleus has mass 214(1.67 × 10−27 kg) = 3.57 × 10−25 kg , where 1.67 × 10−27 kg is the mass of a nucleon (proton or neutron). K = 1 mv 2 .
2
SET UP: Let +x be the direction in which the alpha particle is emitted. The nucleus that is left after the decay has
mass mn = 3.75 × 10−25 kg − mα = 3.57 × 10−25 kg − 6.65 × 10−27 kg = 3.50 × 10−25 kg .
EXECUTE: 8.24. P2 x = P x = 0 gives mα vα + mn vn = 0 . vn =
1 ⎛ 6.65 × 10−27 kg ⎞
7
5
vn = ⎜
⎟ (1.92 × 10 m/s) = 3.65 × 10 m/s .
−25
⎝ 3.50 × 10 kg ⎠
EVALUATE: The recoil velocity of the more massive nucleus is much less than the speed of the emitted alpha
particle.
IDENTIFY and SET UP: Let the +x-direction be horizontal, along the direction the rock is thrown. There is no net
horizontal force, so Px is constant. Let object A be you and object B be the rock.
EXECUTE: 0 = −mAv A + mB vB cos35.0°
vA = EVALUATE:
8.25. mα
2 Kα
2(1.23 × 10−12 J)
vα . vα =
=
= 1.92 × 107 m/s .
mn
6.65 × 10−27 kg
mα mB vB cos35.0°
= 2.11 m/s
mA Py is not conserved because there is a net external force in the vertical direction; as you throw the rock the normal force exerted on you by the ice is larger than the total weight of the system.
IDENTIFY: Each horizontal component of momentum is conserved. K = 1 mv 2 .
2
SET UP: Let +x be the direction of Rebecca’s initial velocity and let the +y axis make an angle of 36.9° with
respect to the direction of her final velocity. vD1x = vD1 y = 0 . vR1x = 13.0 m/s ; vR1 y = 0 . vR 2 x = (8.00 m/s)cos53.1° = 4.80 m/s ; vR 2 y = (8.00 m/s)sin 53.1° = 6.40 m/s . Solve for vD 2 x and vD 2 y .
EXECUTE: vD2 x = (a) P x = P2 x gives mR vR1x = mR vR 2 x + mD vD2 x .
1 mR (vR1x − vR 2 x ) (45.0 kg)(13.0 m/s − 4.80 m/s)
=
= 5.68 m/s .
mD
65.0 kg
⎛ 45.0 kg ⎞
mR
vR 2 y = − ⎜
⎟ (6.40 m/s) = −4.43 m/s .
mD
⎝ 65.0 kg ⎠
v
4.43 m/s
are sketched in Figure 8.25. tan θ = D 2 y =
and θ = 38.0° .
vD2 x 5.68 m/s P y = P2 y gives 0 = mR vR 2 y + mDvD2 y . vD2 y = −
1 !
!
!
The directions of vR1 , vR2 and vD2
2
2
vD = vD2 x + vD2 y = 7.20 m/s . Momentum, Impulse, and Collisions 8-7 2
(b) K1 = 1 mR vR1 = 1 (45.0 kg)(13.0 m/s) 2 = 3.80 × 103 J .
2
2
2
2
K 2 = 1 mR vR2 + 1 mD vD2 = 1 (45.0 kg)(8.00 m/s) 2 + 1 (65.0 kg)(7.20 m/s) 2 = 3.12 × 103 J .
2
2
2
2 ΔK = K 2 − K1 = −680 J .
EVALUATE: Each component of momentum is separately conserved. The kinetic energy of the system increases.
y vR2 vR1
u x vD2 Figure 8.25
8.26. IDENTIFY: There is no net external force on the system of astronaut plus canister, so the momentum of the
system is conserved.
SET UP: Let object A be the astronaut and object B be the canister. Assume the astronaut is initially at rest. After
the collision she must be moving in the same direction as the canister. Let +x be the direction in which the canister
is traveling initially, so v A1x = 0 , v A 2 x = +2.40 m/s , vB1x = +3.50 m/s , and vB 2 x = +1.20 m/s . Solve for mB . mA (v A 2 x − v A1x ) (78.4 kg)(2.40 m/s − 0)
=
= 81.8 kg .
vB1x − vB 2 x
3.50 m/s − 1.20 m/s
EVALUATE: She must exert a force on the canister in the − x direction to reduce its velocity component in the
+x direction. By Newton’s third law, the canister exerts a force on her that is in the +x direction and she gains
velocity in that direction.
IDENTIFY: The horizontal component of the momentum of the system of the rain and freight car is conserved.
SET UP: Let +x be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the
x axis.
EXECUTE: (a) P x = P2 x says (24,000 kg)(4.00 m/s) = (27,000 kg)v2 x and v2 x = 3.56 m/s .
1
(b) After it lands in the car the water must gain horizontal momentum, so the car loses horizontal momentum.
EVALUATE: The vertical component of the momentum is not conserved, because of the vertical external force
exerted by the track.
IDENTIFY: The x and y components of the momentum of the system of the two asteroids are separately conserved.
SET UP: The before and after diagrams are given in Figure 8.28 and the choice of coordinates is indicated. Each
asteroid has mass m.
EXECUTE: (a) P x = P2 x gives mv A1 = mv A 2 cos30.0° + mvB 2 cos 45.0° . 40.0 m/s = 0.866v A 2 + 0.707vB 2 and
1
EXECUTE: 8.27. 8.28. P x = P2 x . mAv A1x + mB vB1x = m Av A 2 x + mB vB 2 x . mB =
1 0.707vB 2 = 40.0 m/s − 0.866v A 2 . P2 y = P2 y gives 0 = mv A 2 sin 30.0° − mvB 2 sin 45.0° and 0.500v A 2 = 0.707vB 2 .
Combining these two equations gives 0.500v A 2 = 40.0 m/s − 0.866v A 2 and v A 2 = 29.3 m/s . Then ⎛ 0.500 ⎞
vB 2 = ⎜
⎟ (29.3 m/s) = 20.7 m/s .
⎝ 0.707 ⎠
2
2
2
(b) K1 = 1 mv A1 . K 2 = 1 mv A 2 + 1 mvB 2 .
2
2
2 2
2
K 2 v A 2 + vB 2 (29.3 m/s) 2 + (20.7 m/s) 2
=
=
= 0.804 .
2
K1
v A1
(40.0 m/s) 2 ΔK K 2 − K1 K 2
=
=
− 1 = −0.196 .
K1
K1
K1
19.6% of the original kinetic energy is dissipated during the collision.
EVALUATE: We could use any directions we wish for the x and y coordinate directions, but the particular choice
we have made is especially convenient. 8-8 Chapter 8 Figure 8.28
8.29. IDENTIFY: Since drag effects are neglected there is no net external force on the system of two fish and the momentum
of the system is conserved. The mechanical energy equals the kinetic energy, which is K = 1 mv 2 for each object.
2
SET UP: Let object A be the 15.0 kg fish and B be the 4.50 kg fish. Let +x be the direction the large fish is
moving initially, so v A1x = 1.10 m/s and vB1x = 0 . After the collision the two objects are combined and move with
!
velocity v2 . Solve for v2 x .
EXECUTE: (a) P x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x .
1 v2 x = mAv A1x + mB vB1x (15.0 kg)(1.10 m/s) + 0
=
= 0.846 m/s .
15.0 kg + 4.50 kg
mA + mB 2
2
2
(b) K1 = 1 mAv A1 + 1 mB vB1 = 1 (15.0 kg)(1.10 m/s) 2 = 9.08 J . K 2 = 1 ( mA + mB )v2 = 1 (19.5 kg)(0.846 m/s) 2 = 6.98 J .
2
2
2
2
2 8.30. ΔK = K 2 − K1 = −2.10 J . 2.10 J of mechanical energy is dissipated.
EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined.
IDENTIFY: There is no net external force on the system of the two otters and the momentum of the system is
conserved. The mechanical energy equals the kinetic energy, which is K = 1 mv 2 for each object.
2
!
SET UP: Let A be the 7.50 kg otter and B be the 5.75 kg otter. After the collision their combined velocity is v2 .
Let +x be to the right, so v A1x = −5.00 m/s and vB1x = +6.00 m/s . Solve for v2 x .
EXECUTE: (a) P x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x .
1 v2 x = mAv A1x + mB vB1x (7.50 kg)(−5.00 m/s) + (5.75)(+6.00 m/s)
=
= −0.226 m/s .
7.50 kg + 5.75 kg
mA + mB 2
2
(b) K1 = 1 mAv A1 + 1 mB vB1 = 1 (7.50 kg)(5.00 m/s)2 + 1 (5.75 kg)(6.00 m/s) 2 = 197.2 J .
2
2
2
2
2
K 2 = 1 ( mA + mB )v2 = 1 (13.25 kg)(0.226 m/s) 2 = 0.338 J .
2
2 8.31. ΔK = K 2 − K1 = −197 J . 197 J of mechanical energy is dissipated.
EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined.
IDENTIFY: Treat the comet and probe as an isolated system for which momentum is conserved.
SET UP: In part (a) let object A be the probe and object B be the comet. Let − x be the direction the probe is
traveling just before the collision. After the collision the combined object moves with speed v2 . The change in
velocity is Δv = v2 x − vB1x . In part (a) the impact speed of 37,000 km/h is the speed of the probe relative to the
comet just before impact: v A1x − vB1x = −37,000 km/h . In part (b) let object A be the comet and object B be the
earth. Let − x be the direction the comet is traveling just before the collision. The impact speed is 40,000 km/h, so
v A1x − vB1x = −40,000 km/h .
EXECUTE: (a) P x = P2 x . v2 x =
1 mAv A1x + mB vB1x
.
mA + mB ⎛ mA ⎞
⎛ mB − mA − mB ⎞
⎛ mA ⎞
Δv = v2 x − vB1x = ⎜
⎟ v A1x + ⎜
⎟ vB1x = ⎜
⎟ ( v A1x − vB1x ) .
mA + mB ⎠
mA + mB ⎠
⎝
⎝
⎝ mA + mB ⎠
⎛
⎞
372 kg
−6
Δv = ⎜
⎟ (−37,000 km/h) = −1.4 × 10 km/h .
372 kg + 0.10 × 1014 kg ⎠
⎝
The speed of the comet decreased by 1.4 × 10−6 km/h . This change is not noticeable. Momentum, Impulse, and Collisions 8.32. 8-9 ⎛
⎞
0.10 × 1014 kg
−8
(b) Δv = ⎜
⎟ (−40,000 km/h) = −6.7 × 10 km/h . The speed of the earth would change
0.10 × 1014 kg + 5.97 × 1024 kg ⎠
⎝
by 6.7 × 10−8 km/h . This change is not noticeable.
EVALUATE: v A1x − vB1x is the velocity of the projectile (probe or comet) relative to the target (comet or earth).
The expression for Δv can be derived directly by applying momentum conservation in coordinates in which the
target is initially at rest.
IDENTIFY: The forces the two vehicles exert on each other during the collision are much larger than the
horizontal forces exerted by the road, and it is a good approximation to assume momentum conservation.
!
SET UP: Let +x be eastward. After the collision two vehicles move with a common velocity v2 .
(a) P x = P2 x gives mSCvSCx + mT vTx = (mSC + mT )v2 x .
1 EXECUTE: v2 x = mSCvSCx + mT vTx (1050 kg)(−15.0 m/s) + (6320 kg)(+10.0 m/s)
=
= 6.44 m/s .
1050 kg + 6320 kg
mSC + mT The final velocity is 6.44 m/s, eastward. ⎛m ⎞
⎛ 1050 kg ⎞
(b) P x = P2 x = 0 gives mSCvSCx + mT vTx = 0 . vTx = − ⎜ SC ⎟ vSCx = − ⎜
⎟ (−15.0 m/s) = 2.50 m/s . The truck
1
mT ⎠
⎝ 6320 kg ⎠
⎝
would need to have initial speed 2.50 m/s.
(c) part (a): ΔK = 1 (7370 kg)(6.44 m/s) 2 − 1 (1050 kg)(15.0 m/s) 2 − 1 (6320 kg)(10.0 m/s) 2 = −2.81 × 105 J
2
2
2 8.33. part (b): ΔK = 0 − 1 (1050 kg)(15.0 m/s) 2 − 1 (6320 kg)(2.50 m/s) 2 = −1.38 × 105 J . The change in kinetic energy
2
2
has the greater magnitude in part (a).
EVALUATE: In part (a) the eastward momentum of the truck has a greater magnitude than the westward
momentum of the car and the wreckage moves eastward after the collision. In part (b) the two vehicles have equal
magnitudes of momentum, the total momentum of the system is zero, and the wreckage is at rest after the collision.
IDENTIFY: The forces the two players exert on each other during the collision are much larger than the horizontal
forces exerted by the slippery ground and it is a good approximation to assume momentum conservation. Each
component of momentum is separately conserved.
!
SET UP: Let +x be east and +y be north. After the collision the two players have velocity v2 . Let the linebacker
be object A and the halfback be object B, so v A1x = 0 , v A1 y = 8.8 m/s , vB1x = 7.2 m/s and vB1 y = 0 . Solve for v2 x and v2 y .
P x = P2 x gives mAv A1x + mB vB1x = (mA + mB )v2 x .
1 EXECUTE: v2 x = mAv A1x + mB vB1x (85 kg)(7.2 m/s)
=
= 3.14 m/s .
110 kg + 85 kg
mA + mB P y = P2 y gives mAv A1 y + mB vB1 y = (mA + mB )v2 y .
1
v2 y =
v= v +v
2
2x 2
2y mAv A1 y + mB vB1 y
mA + mB (110 kg)(8.8 m/s)
= 4.96 m/s .
110 kg + 85 kg = 5.9 m/s .
tan θ = 8.34. = v2 y
v2 x = 4.96 m/s
and θ = 58° .
3.14 m/s The players move with a speed of 5.9 m/s and in a direction 58° north of east.
EVALUATE: Each component of momentum is separately conserved.
IDENTIFY: There is no net external force on the system of the two skaters and the momentum of the system is
conserved.
SET UP: Let object A be the skater with mass 70.0 kg and object B be the skater with mass 65.0 kg. Let +x be to
the right, so v A1x = +2.00 m/s and vB1x = −2.50 m/s . After the collision the two objects are combined and move with
!
velocity v2 . Solve for v2 x .
EXECUTE: P x = P2 x . mAv A1x + mB vB1x = (mA + mB )v2 x .
1
v2 x = mAv A1x + mB vB1x (70.0 kg)(2.00 m/s) + (65.0)(−2.50 m/s)
=
= −0.167 m/s .
70.0 kg + 65.0 kg
mA + mB The two skaters move to the left at 0.167 m/s.
EVALUATE: There is a large decrease in kinetic energy. 8-10 8.35. 8.36. Chapter 8 IDENTIFY: Neglect external forces during the collision. Then the momentum of the system of the two cars is
conserved.
SET UP: mS = 1200 kg , mL = 3000 kg . The small car has velocity vS and the large car has velocity vL .
EXECUTE: (a) The total momentum of the system is conserved, so the momentum lost by one car equals the
!
!
momentum gained by the other car. They have the same magnitude of change in momentum. Since p = mv and
!
Δp is the same, the car with the smaller mass has a greater change in velocity. ⎛m ⎞
⎛ 3000 kg ⎞
mSΔvS = mL ΔvL and ΔvS = ⎜ L ⎟ ΔvL = ⎜
⎟ Δv = 2.50Δv .
⎝ 1200 kg ⎠
⎝ mS ⎠
(b) The acceleration of the small car is greater, since it has a greater change in velocity during the collision. The
large acceleration means a large force on the occupants of the small car and they would sustain greater injuries.
EVALUATE: Each car exerts the same magnitude of force on the other car but the force on the compact has a
greater effect on its velocity since its mass is less.
IDENTIFY: The collision forces are large so gravity can be neglected during the collision. Therefore, the
horizontal and vertical components of the momentum of the system of the two birds are conserved.
SET UP: The system before and after the collision is sketched in Figure 8.36. Use the coordinates shown. Figure 8.36
EXECUTE: There is no external force on the system so P x = P2 x and P y = P2 y .
1
1 P x = P2 x gives (1.5 kg)(9.0 m/s) = (1.5 kg)vraven-2 cos φ and vraven-2 cos φ = 9.0 m/s .
1
P y = P2 y gives (0.600 kg)(20.0 m/s) = (0.600 kg)( −5.0 m/s) + (1.5 kg)vraven-2 sin φ and vraven-2 sin φ = 10.0 m/s .
1
10.0 m/s
and φ = 48° .
9.0 m/s
EVALUATE: Due to its large initial speed the lighter falcon was able to produce a large change in the raven’s
direction of motion.
IDENTIFY: Since friction forces from the road are ignored, the x and y components of momentum are conserved.
SET UP: Let object A be the subcompact and object B be the truck. After the collision the two objects move
!
together with velocity v2 . Use the x and y coordinates given in the problem. v A1 y = vB1 y = 0 . Combining these two equations gives tan φ = 8.37. v2 x = (16.0 m/s)sin 24.0° = 6.5 m/s ; v2 y = (16.0 m/s)cos 24.0° = 14.6 m/s .
EXECUTE: P x = P2 x gives mAv A1x = (mA + mB )v2 x .
1
⎛ m + mB ⎞
⎛ 950 kg + 1900 kg ⎞
v A1x = ⎜ A
⎟ v2 x = ⎜
⎟ (6.5 m/s) = 19.5 m/s .
950 kg
mA ⎠
⎝
⎠
⎝ P y = P2 y gives mAvB1 y = (mA + mB )v2 y .
1
⎛ m + mB ⎞
⎛ 950 kg + 1900 kg ⎞
vB1 y = ⎜ A
⎟ v2 y = ⎜
⎟ (14.6 m/s) = 21.9 m/s .
1900 kg
mA ⎠
⎝
⎠
⎝ 8.38. Before the collision the subcompact car has speed 19.5 m/s and the truck has speed 21.9 m/s.
EVALUATE: Each component of momentum is independently conserved.
IDENTIFY: Apply conservation of momentum to the collision. Apply conservation of energy to the motion of the
block after the collision. Momentum, Impulse, and Collisions 8-11 SET UP: Conservation of momentum applied to the collision between the bullet and the block: Let object A be
the bullet and object B be the block. Let v A be the speed of the bullet before the collision and let V be the speed of
the block with the bullet inside just after the collision. Figure 8.38a Px is constant gives mAv A = (mA + mB )V .
Conservation of energy applied to the motion of the block after the collision:
V v50 #2 #1 y A1B x 0.230 m Figure 8.38b K1 + U1 + Wother = K 2 + U 2
EXECUTE: Work is done by friction so Wother = W f = ( f k cos φ ) s = − f k s = − μ k mgs U1 = U 2 = 0 (no work done by gravity)
K1 = 1 mV 2 ; K 2 = 0 (block has come to rest)
2
Thus 1
2 mV 2 − μ k mgs = 0 V = 2μ k gs = 2(0.20)(9.80 m/s 2 )(0.230 m) = 0.9495 m/s
Use this in the conservation of momentum equation
⎛ m + mB ⎞
⎛ 5.00 × 10−3 kg + 1.20 kg ⎞
vA = ⎜ A
V =⎜
⎟
⎟ (0.9495 m/s) = 229 m/s
5.00 × 10−3 kg
⎝
⎠
⎝ mA ⎠ 8.39. EVALUATE: When we apply conservation of momentum to the collision we are ignoring the impulse of the
friction force exerted by the surface during the collision. This is reasonable since this force is much smaller than
the forces the bullet and block exert on each other during the collision. This force does work as the block moves
after the collision, and takes away all the kinetic energy.
IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the
collision. After the collision the kinetic energy of the combined object is converted to gravitational potential
energy.
SET UP: Immediately after the collision the combined object has speed V. Let h be the vertical height through
which the pendulum rises.
EXECUTE: (a) Conservation of momentum applied to the collision gives
(12.0 ×10−3 kg)(380 m/s) = (6.00 kg + 12.0 × 10−3 kg)V and V = 0.758 m/s . Conservation of energy applied to the motion after the collision gives
h= 1
2 mtotV 2 = mtot gh and V 2 (0.758 m/s) 2
=
= 0.0293 m = 2.93 cm .
2 g 2(9.80 m/s 2 ) 2
(b) K = 1 mbvb = 1 (12.0 × 10−3 kg)(380 m/s) 2 = 866 J .
2
2 8.40. (c) K = 1 mtotV 2 = 1 (6.00 kg + 12.0 × 10−3 kg)(0.758 m/s) 2 = 1.73 J .
2
2
EVALUATE: Most of the initial kinetic energy of the bullet is dissipated in the collision.
IDENTIFY: Each component of horizontal momentum is conserved.
SET UP: Let +x be east and +y be north. vS1 y = vA1x = 0 . vS 2 x = (6.00 m/s)cos37.0° = 4.79 m/s , vS 2 y = (6.00 m/s)sin 37.0° = 3.61 m/s , v A 2 x = (9.00 m/s)cos 23.0° = 8.28 m/s and
v A 2 y = −(9.00 m/s)sin 23.0° = −3.52 m/s .
EXECUTE: P x = P2 x gives mSvS1x = mSvS2x + mA vA 2 x .
1
vS1x = mSvS2x + mA vA 2 x (80.0 kg)(4.79 m/s) + (50.0 kg)(8.28 m/s)
=
= 9.97 m/s .
mS
80.0 kg Sam’s speed before the collision was 9.97 m/s. 8-12 Chapter 8 P y = P2 y gives mA vA1 y = mSvS2y + mA vA2 y .
1
vA1 y = 8.41. mSvS2y + mA vA2 y
mS = (80.0 kg)(3.61 m/s) + (50.0 kg)(−3.52 m/s)
= 2.26 m/s .
50.0 kg Abigail’s speed before the collision was 2.26 m/s.
(b) ΔK = 1 (80.0 kg)(6.00 m/s)2 + 1 (50.0 kg)(9.00 m/s)2 − 1 (80.0 kg)(9.97 m/s)2 − 1 (50.0 kg)(2.26 m/s)2 . ΔK = −639 J .
2
2
2
2
EVALUATE: The total momentum is conserved because there is no net external horizontal force. The kinetic
energy decreases because the forces between the objects do negative work during the collision.
IDENTIFY: When the spring is compressed the maximum amount the two blocks aren’t moving relative to each
!
other and have the same velocity V relative to the surface. Apply conservation of momentum to find V and
conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give
the final velocity of each block after the collision.
SET UP: Let +x be the direction of the initial motion of A.
EXECUTE: (a) Momentum conservation gives (2.00 kg)(2.00 m/s) = (12.0 kg)V and V = 0.333 m/s . Both
blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the
initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy U b
stored in the bumpers: 1 (2.00 kg)(2.00 m/s) 2 = U b + 1 (12.0 kg)(0.333 m/s) 2 and U b = 3.33 J .
2
2 8.42. ⎛ m − mB ⎞
⎛ 2.00 kg − 10.0 kg ⎞
(b) v A 2 x = ⎜ A
⎟ v A1x = ⎜
⎟ (2.00 m/s) = −1.33 m/s . Block A is moving in the − x direction at
mA + mB ⎠
12.0 kg
⎝
⎠
⎝
1.33 m/s.
⎛ 2mA ⎞
2(2.00 kg)
vB 2 x = ⎜
(2.00 m/s) = +0.667 m/s . Block B is moving in the +x direction at 0.667 m/s.
⎟ v A1x =
mA + mB ⎠
12.0 kg
⎝
EVALUATE: When the spring is compressed the maximum amount the system must still be moving in order to
conserve momentum.
IDENTIFY: No net external horizontal force so Px is conserved. Elastic collision so K1 = K 2 and can use Eq. 8.27.
SET UP: Figure 8.42
EXECUTE: From conservation of x-component of momentum:
mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x
mAv A1 − mB vB1 = mAv A 2 x + mB vB 2 x
(0.150 kg)(0.80 m/s) − (0.300 kg)(2.20 m/s) = (0.150 kg)v A 2 x + (0.300 kg)vB 2 x
−3.60 m/s = vA2x + 2vB 2 x From the relative velocity equation for an elastic collision Eq. 8.27:
vB 2 x − v A 2 x = −(vB1x − v A1x ) = −(−2.20 m/s − 0.80 m/s) = +3.00 m/s
3.00 m/s = −vA2x + vB 2 x 8.43. Adding the two equations gives −0.60 m/s = 3vB 2 x and vB 2 x = −0.20 m/s. Then v A 2 x = vB 2 x − 3.00 m/s = −3.20 m/s.
The 0.150 kg glider (A) is moving to the left at 3.20 m/s and the 0.300 kg glider (B) is moving to the left at
0.20 m/s.
EVALUATE: We can use our v A 2 x and vB 2 x to show that Px is constant and K1 = K 2
IDENTIFY: Since the collision is elastic, both momentum conservation and Eq. 8.27 apply.
SET UP: Let object A be the 30.0 kg marble and let object B be the 10.0 g marble. Let +x be to the right.
EXECUTE: (a) Conservation of momentum gives
(0.0300 kg)(0.200 m/s) + (0.0100 kg)(−0.400 m/s) = (0.0300 kg)v A 2 x + (0.0100 kg)vB 2 x .
3v A 2 x + vB 2 x = 0.200 m/s . Eq. 8.27 says vB 2 x − v A 2 x = −(−0.400 m/s − 0.200 m/s) = +0.600 m/s . Solving this pair of
equations gives v A 2 x = −0.100 m/s and vB 2 x = +0.500 m/s . The 30.0 g marble is moving to the left at 0.100 m/s
and the 10.0 g marble is moving to the right at 0.500 m/s. Momentum, Impulse, and Collisions 8-13 (b) For marble A, ΔPAx = mAv A 2 x − mAv A1x = (0.0300 kg)(−0.100 m/s − 0.200 m/s) = −0.00900 kg ⋅ m/s . For marble B, ΔPBx = mB vB 2 x − mB vB1x = (0.0100 kg)(0.500 m/s − [−0.400 m/s]) = +0.00900 kg ⋅ m/s .
The changes in momentum have the same magnitude and opposite sign.
2
2
(c) For marble A, ΔK A = 1 mAv A 2 − 1 mAv A1 = 1 (0.0300 kg)([0.100 m/s]2 − [0.200 m/s]2 ) = −4.5 × 10−4 J .
2
2
2 8.44. 2
2
For marble B, ΔK B = 1 mB vB 2 − 1 mB vB1 = 1 (0.0100 kg)([0.500 m/s]2 − [0.400 m/s]2 ) = +4.5 × 10−4 J .
2
2
2
The changes in kinetic energy have the same magnitude and opposite sign.
EVALUATE: The results of parts (b) and (c) show that momentum and kinetic energy are conserved in the
collision.
IDENTIFY and SET UP: Without rounding, the calculation in Example 8.12 gives vB 2 = 20 m/s .
EXECUTE: The two equations in Example 8.12 for α and β are (0.500 kg)(4.00 m/s) = (0.500 kg)(2.00 m/s)(cosα ) + (0.300 kg)( 20 m/s)(cos β ) Eq. 1
and
0 = (0.500 kg)(2.00 m/s)(sin α ) − (0.300 kg)( 20 m/s)sin β Eq. 2.
Dividing each equation by (0.500 kg)(1.00 m/s) gives
4.00 = 2.00cos α + 0.6 20 cos β Eq. 3
and
0 = 2.00sin α − 0.6 20 sin β Eq. 4.
4.00 − 2.00cos α
and cos 2 β = 2.222 − 2.222cos α + 0.5556cos 2 α .
0.6 20
Eq. 4 gives sin β = 0.7454sin α and sin 2 β = 0.5556sin 2 α = 0.5556 − 0.5556cos 2 α .
Eq. 3 gives cos β = 8.45. Adding the two equations and using sin 2 β + cos 2 β = 1 gives 1 = 2.778 − 2.222cos α and cos α = 0.8002 .
α = 36.9° . Then sin β = 0.7454sin α gives β = 26.6° .
EVALUATE: For these values of α and β , the x component of momentum, the y component of momentum and
the kinetic energy are all conserved in the collision.
IDENTIFY: Eqs. 8.24 and 8.25 apply, with object A being the neutron.
SET UP: Let +x be the direction of the initial momentum of the neutron. The mass of a neutron is mn = 1.0 u .
⎛ m − mB ⎞
1.0 u − 2.0 u
v A1x = −v A1x / 3.0 . The speed of the neutron after the collision
(a) v A 2 x = ⎜ A
⎟ v A1x =
1.0 u + 2.0 u
⎝ mA + mB ⎠
is one-third its initial speed.
1
2
(b) K 2 = 1 mn vn = 1 mn (v A1 / 3.0) 2 =
K1 .
2
2
9.0
EXECUTE: n 8.46. n 1
⎛1⎞
⎛1⎞
(c) After n collisions, v A 2 = ⎜
, so 3.0n = 59,000 . n log3.0 = log59,000 and n = 10 .
⎟=
⎟ v A1 . ⎜
⎝ 3.0 ⎠ 59,000
⎝ 3.0 ⎠
EVALUATE: Since the collision is elastic, in each collision the kinetic energy lost by the neutron equals the
kinetic energy gained by the deuteron.
IDENTIFY: Elastic collision. Solve for mass and speed of target nucleus.
SET UP: (a) Let A be the proton and B be the target nucleus. The collision is elastic, all velocities lie along a line,
and B is at rest before the collision. Hence the results of Eqs. 8.24 and 8.25 apply.
EXECUTE: Eq. 8.24: mB (vx + v Ax ) = mA (vx − v Ax ), where vx is the velocity component of A before the collision
and v Ax is the velocity component of A after the collision. Here, vx = 1.50 × 107 m/s (take direction of incident
beam to be positive) and v Ax = −1.20 × 107 m/s (negative since traveling in direction opposite to incident beam).
⎛v −v ⎞
⎛ 1.50 × 107 m/s + 1.20 × 107 m/s ⎞
⎛ 2.70 ⎞
mB = mA ⎜ x Ax ⎟ = m ⎜
⎟ = m⎜
⎟ = 9.00m.
vx + v Ax ⎠
1.50 × 107 m/s − 1.20 × 107 m/s ⎠
⎝ 0.30 ⎠
⎝
⎝
⎛ 2mA ⎞
2m
⎛
⎞
7
6
(b) Eq. 8.25: vBx = ⎜
⎟v = ⎜
⎟ (1.50 × 10 m/s) = 3.00 × 10 m/s.
mA + mB ⎠
m + 9.00m ⎠
⎝
⎝
EVALUATE: Can use our calculated vBx and mB to show that Px is constant and that K1 = K 2 . 8-14 8.47. Chapter 8 IDENTIFY: Apply Eq. 8.28.
SET UP: mA = 0.300 kg , mB = 0.400 kg , mC = 0.200 kg .
EXECUTE: xcm = mA x A + mB xB + mC xC
.
mA + mB + mC xcm = (0.300 kg)(0.200 m) + (0.400 kg)(0.100 m) + (0.200 kg)(−0.300 m)
= 0.0444 m .
0.300 kg + 0.400 kg + 0.200 kg
ycm = ycm = 8.48. mA y A + mB yB + mC yC
.
mA + mB + mC (0.300 kg)(0.300 m) + (0.400 kg)(−0.400 m) + (0.200 kg)(0.600 m)
= 0.0556 m .
0.300 kg + 0.400 kg + 0.200 kg EVALUATE: There is mass at both positive and negative x and at positive and negative y and therefore the center
of mass is close to the origin.
IDENTIFY: Calculate xcm .
SET UP: Apply Eq. 8.28 with the sun as mass 1 and Jupiter as mass 2. Take the origin at the sun and let Jupiter
lie on the positive x-axis. Figure 8.48 xcm =
EXECUTE: x1 = 0 and x2 = 7.78 × 1011 m
xcm = 8.49. m1 x1 + m2 x2
m1 + m2 (1.90 ×10 27 kg ) ( 7.78 × 1011 m ) 1.99 × 1030 kg + 1.90 × 1027 kg = 7.42 × 108 m The center of mass is 7.42 × 108 m from the center of the sun and is on the line connecting the centers of the sun
and Jupiter. The sun’s radius is 6.96 × 108 m so the center of mass lies just outside the sun.
EVALUATE: The mass of the sun is much greater than the mass of Jupiter so the center of mass is much closer to
the sun. For each object we have considered all the mass as being at the center of mass (geometrical center) of the
object.
IDENTIFY: The location of the center of mass is given by Eq. 8.48. The mass can be expressed in terms of the
diameter. Each object can be replaced by a point mass at its center.
SET UP: Use coordinates with the origin at the center of Pluto and the +x direction toward Charon, so xP = 0
xC = 19,700 km . m = ρV = ρ 4 π r 3 = 1 ρπ d 3 .
3
6
EXECUTE: xcm = 3
3
1
⎛
⎞
⎛ dC ⎞
ρπ d C
mP xP + mC xC ⎛ mC ⎞
6
x.
=⎜
⎟ xC = ⎜ 3
⎟ xC = ⎜ 1
3
3
3⎟ C
1
mP + mC
⎝ mP + mC ⎠
⎝ dP + dC ⎠
⎝ 6 ρπ d P + 6 ρπ d C ⎠ ⎛
⎞
[1250 km]3
3
xcm = ⎜
⎟ (19,700 km) = 2.52 × 10 km .
[2370 km]3 + [1250 km]3 ⎠
⎝ 8.50. The center of mass of the system is 2.52 × 103 km from the center of Pluto.
EVALUATE: The center of mass is closer to Pluto because Pluto has more mass than Charon.
IDENTIFY: Apply Eqs. 8.28, 8.30 and 8.32. There is only one component of position and velocity.
SET UP: mA = 1200 kg , mB = 1800 kg . M = mA + mB = 3000 kg . Let +x be to the right and let the origin be at
the center of mass of the station wagon.
m x + mB xB 0 + (1800 kg)(40.0 m)
=
= 24.0 m.
EXECUTE: (a) xcm = A A
mA + mB
1200 kg + 1800 kg
The center of mass is between the two cars, 24.0 m to the right of the station wagon and 16.0 m behind the lead
car. Momentum, Impulse, and Collisions 8-15 (b) Px = mAv A1 + mB vB1 = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 5.04 × 104 kg ⋅ m/s.
(c) vcm, x = 8.51. mAv A, x + mB vB , x
mA + mB = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s)
= 16.8 m/s.
1200 kg + 1800 kg (d) Px = Mvcm-x = (3000 kg)(16.8 m/s) = 5.04 × 104 kg ⋅ m/s , the same as in part (b).
EVALUATE: The total momentum can be calculated either as the vector sum of the momenta of the individual
objects in the system, or as the total mass of the system times the velocity of the center of mass.
IDENTIFY: Use Eq. 8.28 to find the x and y coordinates of the center of mass of the machine part for each
configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be represented
by a point mass at its geometrical center.
SET UP: Use coordinates with the axis at the hinge and the +x and +y axes along the horizontal and vertical bars
in the figure in the problem. Let ( xi , yi ) and ( xf , yf ) be the coordinates of the bar before and after the vertical bar
is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar and 3 be the ball.
m x + m2 x2 + m3 x3 (4.00 kg)(0.750 m) + 0 + 0
=
= 0.333 m .
EXECUTE: xi = 1 1
m1 + m2 + m3
4.00 kg + 3.00 kg + 2.00 kg yi =
xf = 8.52. m1 y1 + m2 y2 + m3 y3 0 + (3.00 kg)(0.900 m) + (2.00 kg)(1.80 m)
=
= 0.700 m .
m1 + m2 + m3
9.00 kg
(4.00 kg)(0.750 m) + (3.00 kg)( −0.900 m) + (2.00 kg)( −1.80 m)
= −0.366 m .
9.00 kg yf = 0 . xf − xi = −0.700 m and yf − yi = −0.700 m . The center of mass moves 0.700 m to the right and 0.700 m
upward.
EVALUATE: The vertical bar moves upward and to the right so it is sensible for the center of mass of the machine
part to move in these directions.
(a) IDENTIFY: Use Eq. 8.28.
SET UP: The target variable is m1.
EXECUTE: xcm = 2.0 m, ycm = 0
xcm = m1 x1 + m2 x2 m1 ( 0 ) + ( 0.10 kg ) ( 8.0 m ) 0.80 kg ⋅ m
=
=
.
m1 + m2
m1 + ( 0.10 kg )
m1 + 0.10 kg
xcm = 2.0 m gives 2.0 m =
m1 + 0.10 kg = 0.80 kg ⋅ m
.
m1 + 0.10 kg 0.80 kg ⋅ m
= 0.40 kg.
2.0 m m1 = 0.30 kg.
The cm is closer to m1 so its mass is larger then m2 .
!
(b) IDENTIFY: Use Eq. 8.32 to calculate P .
!
SET UP: v 5 ( 5.0 m/s ) ˆ.
j
EVALUATE: cm !
!
ˆ
ˆ
P 5 Mvcm 5 ( 0.10 kg + 0.30 kg )( 5.0 m/s ) i 5 ( 2.0 kg ⋅ m/s ) i .
(c) IDENTIFY: Use Eq. 8.31.
!
!
!
!
m v + m2v2
SET UP: vcm 5 1 1
. The target variable is v1. Particle 2 at rest says v2 = 0.
m1 + m2 ⎛ 0.30 kg + 0.10 kg ⎞
! ⎛ m + m2 ⎞ !
ˆ
ˆ
v1 5 ⎜ 1
⎟ vcm 5 ⎜
⎟ ( 5.00 m/s ) i 5 ( 6.7 m/s ) i .
m1 ⎠
0.30 kg
⎝
⎠
⎝
!
!
!
EVALUATE: Using the result of part (c) we can calculate p1 and p2 and show that P as calculated in part (b)
!
!
does equal p1 1 p2 .
IDENTIFY: There is no net external force on the system of James, Ramon and the rope and the momentum of the
system is conserved and the velocity of its center of mass is constant. Initially there is no motion, and the velocity
of the center of mass remains zero after Ramon has started to move.
EXECUTE: 8.53. 8-16 Chapter 8 SET UP: Let +x be in the direction of Ramon’s motion. Ramon has mass mR = 60.0 kg and James has mass mJ = 90.0 kg .
vcm-x = EXECUTE: 8.54. mR vRx + mJ vJx
=0.
mR + mJ ⎛m ⎞
⎛ 60.0 kg ⎞
vJx = − ⎜ R ⎟ vRx = − ⎜
⎟ (0.700 m/s) = −0.47 m/s . James’ speed is 0.47 m/s.
⎝ 90.0 kg ⎠
⎝ mJ ⎠
EVALUATE: As they move, the two men have momenta that are equal in magnitude and opposite in direction, and
the total momentum of the system is zero. Also, Example 8.14 shows that Ramon moves farther than James in the
same time interval. This is consistent with Ramon having a greater speed.
(a) IDENTIFY and SET UP: Apply Eq. 8.28 and solve for m1 and m2 .
ycm = EXECUTE: m1 + m2 = m1 y1 + m2 y2
m1 + m2 m1 y1 + m2 y2 m1 (0) + (0.50 kg)(6.0 m)
=
= 1.25 kg and m1 = 0.75 kg.
ycm
2.4 m EVALUATE: ycm is closer to m1 since m1 > m2 .
!
!
(b) IDENTIFY and SET UP: Apply a 5 d v / dt for the cm motion.
!
!
dv
ˆ
EXECUTE: acm 5 cm 5 (1.5 m/s3 ) ti .
dt
(c) IDENTIFY and SET UP: Apply Eq. 8.34.
!
!
ˆ
EXECUTE: ∑ Fext 5 Macm 5 (1.25 kg ) (1.5 m/s3 ) ti .
!
ˆ
ˆ
At t = 3.0 s, ∑ Fext 5 (1.25 kg ) (1.5 m/s3 ) ( 3.0 s ) i 5 ( 5.6 N ) i . 8.55. 8.56. !
EVALUATE: vcm- x is positive and increasing so acm − x is positive and Fext is in the + x -direction. There is no
motion and no force component in the y-direction.
!
! dP
to the airplane.
IDENTIFY: Apply ∑ F =
dt
dn
SET UP:
( t ) = nt n −1 . 1 N = 1 kg ⋅ m/s2 .
dt
!
!
!
dP
5 [ −(1.50 kg ⋅ m/s3 )t ]i 1 (0.25 kg ⋅ m/s 2 ) j . Fx = −(1.50 N/s)t , Fy = 0.25 N , Fz = 0 .
EXECUTE:
dt
EVALUATE: There is no momentum or change in momentum in the z direction and there is no force component
in this direction.
IDENTIFY: Use Eq. 8.38, applied to a finite time interval.
SET UP: vex = 1600 m/s
Δm
−0.0500 kg
= −(1600 m/s)
= +80.0 N .
Δt
1.00 s
(b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by
ejecting the gas in a direction with a component perpendicular to the rocket’s velocity and braked by ejecting it in a
direction parallel (as opposed to antiparallel) to the rocket’s velocity.
EVALUATE: The thrust depends on the speed of the ejected gas relative to the rocket and on the mass of gas
ejected per second.
v dm
IDENTIFY: a = − ex
. Assume that dm / dt is constant over the 5.0 s interval, since m doesn’t change much
m dt
dm
.
during that interval. The thrust is F = −vex
dt
SET UP: Take m to have the constant value 110 kg + 70 kg = 180 kg . dm / dt is negative since the mass of the
MMU decreases as gas is ejected.
dm
m
⎛ 180 kg ⎞
2
= − a = −⎜
EXECUTE: (a)
⎟ (0.029 m/s ) = −0.0106 kg/s . In 5.0 s the mass that is ejected is
dt
vex
⎝ 490 m/s ⎠
(0.0106 kg/s)(5.0 s) = 0.053 kg .
EXECUTE: 8.57. (a) F = −vex (b) F = −vex dm
= −(490 m/s)( −0.0106 kg/s) = 5.19 N .
dt Momentum, Impulse, and Collisions 8.58. 8-17 EVALUATE: The mass change in the 5.0 s is a very small fraction of the total mass m, so it is accurate to take m
to be constant.
v dm
IDENTIFY and SET UP: Apply Eq. 8.39: a = − ex
. Solve for dm / dt.
m dt
EXECUTE: ( 6000 kg ) ( 25.0 m/s 2 )
dm
ma
=−
=−
= −75.0 kg/s .
dt
vex
2000 m/s 8.59. So in 1 s the rocket must eject 75.0 kg of gas.
EVALUATE: We have approximated dm / dt by Δm / Δt. We have assumed that 25.0 m/s 2 is the average
acceleration for the first second.
IDENTIFY: Use Eq. 8.39, applied to a finite time interval. Solve for vex . 8.60. Δm
m
=−
.
160
Δt
a
−15.0 m/s 2
v Δm
=
= 2.40 × 103 m/s = 2.40 km/s
EXECUTE: a = − ex
. vex = −
⎛ Δm ⎞ ⎛ m ⎞
m Δt
/m⎟ ⎜−
⎜
⎟/m
⎝ Δt
⎠ ⎝ 160 ⎠
EVALUATE: The acceleration is proportional to the speed of the exhaust gas and to the rate at which mass is
ejected.
IDENTIFY and SET UP: ( Fav ) Δt = J relates the impulse J to the average thrust Fav . Eq. 8.38 applied to a finite time
SET UP: interval gives Fav = −vex
EXECUTE:
(b) vex = − 8.61. (a) F = Δm
⎛m ⎞
. v − v0 = vex ln ⎜ 0 ⎟ . The remaining mass m after 1.70 s is 0.0133 kg.
Δt
⎝m⎠ J 10.0 N ⋅ s
=
= 5.88 N . Fav / Fmax = 0.442 .
Δt
1.70 s Fav Δt
= 800 m/s .
−0.0125 kg ⎛ 0.0258 kg ⎞
⎛m ⎞
(c) v0 = 0 and v = vex ln ⎜ 0 ⎟ = (800 m/s)ln ⎜
⎟ = 530 m/s .
m⎠
⎝
⎝ 0.0133 kg ⎠
EVALUATE: The acceleration of the rocket is not constant. It increases as the mass remaining decreases.
⎛m ⎞
IDENTIFY: v − v0 = vex ln ⎜ 0 ⎟ .
⎝m⎠
SET UP: v0 = 0 .
3
m
⎛ m ⎞ v 8.00 × 10 m/s
ln ⎜ 0 ⎟ =
=
= 3.81 and 0 = e3.81 = 45.2 .
2100 m/s
m
⎝ m ⎠ vex
EVALUATE: Note that the final speed of the rocket is greater than the relative speed of the exhaust gas.
IDENTIFY and SET UP: Use Eq. 8.40: v − v0 = vex ln ( m0 / m ) . EXECUTE:
8.62. v0 = 0 (“fired from rest”), so v / vex = ln ( m0 / m ) .
Thus m0 / m = ev / vex , or m / m0 = e − v / vex .
If v is the final speed then m is the mass left when all the fuel has been expended; m / m0 is the fraction of the
initial mass that is not fuel.
(a) EXECUTE: v = 1.00 × 10−3 c = 3.00 × 105 m/s gives
5 m / m0 = e − (3.00×10 8.63. m/s) /(2000 m/s) = 7.2 × 10−66 . EVALUATE: This is clearly not feasible, for so little of the initial mass to not be fuel.
(b) EXECUTE: v = 3000 m/s gives m / m0 = e − ( 3000 m/s)/(2000 m/s) = 0.223 .
EVALUATE: 22.3% of the total initial mass not fuel, so 77.7% is fuel; this is possible.
IDENTIFY: Use the heights to find v1 y and v2 y , the velocity of the ball just before and just after it strikes the slab. Then apply J y = Fy Δt = Δp y .
SET UP: Let +y be downward. 8-18 Chapter 8 EXECUTE: (a) 1
2 mv 2 = mgh so v = ± 2 gh . v1 y = + 2(9.80 m/s 2 )(2.00 m) = 6.26 m/s . v2 y = − 2(9.80 m/s 2 )(1.60 m) = −5.60 m/s .
J y = Δp y = m(v2 y − v1 y ) = (40.0 × 10−3 kg)(−5.60 m/s − 6.26 m/s) = −0.474 kg ⋅ m/s .
The impulse is 0.474 kg ⋅ m/s , upward.
−0.474 kg ⋅ m/s
=
= −237 N . The average force on the ball is 237 N, upward.
Δt
2.00 × 10−3 s
EVALUATE: The upward force on the ball changes the direction of its momentum.
IDENTIFY: Momentum is conserved in the explosion. At the highest point the velocity of the boulder is zero.
Since one fragment moves horizontally the other fragment also moves horizontally. Use projectile motion to relate
the initial horizontal velocity of each fragment to its horizontal displacement.
SET UP: Use coordinates where +x is north. Since both fragments start at the same height with zero vertical
component of velocity, the time in the air, t, is the same for both. Call the fragments A and B, with A being the one
that lands to the north. Therefore, mB = 3mA .
(b) Fy =
8.64. Jy EXECUTE: Apply P x = P2 x to the collision: 0 = mAv Ax + mB vBx . vBx = −
1 to the motion after the collision: x − x0 = v0 xt . Since t is the same, 8.65. mA
v Ax = −v Ax / 3 . Apply projectile motion
mB ( x − x0 ) A ( x − x0 ) B
=
and
v Ax
vBx ⎛v ⎞
⎛ −v / 3 ⎞
( x − x0 ) B = ⎜ Bx ⎟ ( x − x0 ) A = ⎜ Ax ⎟ ( x − x0 ) A = −(274 m) / 3 = −91.3 m . The other fragment lands 91.3 m
v Ax ⎠
⎝
⎝ v Ax ⎠
directly south of the point of explosion.
EVALUATE: The fragment that has three times the mass travels one-third as far.
IDENTIFY: The impulse, force and change in velocity are related by Eq. 8.9
!!
SET UP: m = w / g = 0.0571 kg . Since the force is constant, F = Fav .
EXECUTE: (a) J x = Fx Δt = ( −380 N)(3.00 × 10−3 s) = −1.14 N ⋅ s . J y = Fy Δt = (110 N)(3.00 ×10−3 s) = 0.330 N ⋅ s . J
Jx
−1.14 N ⋅ s
0.330 N ⋅ s
+ v1x =
+ 20.0 m/s = 0.04 m/s . v2 y = y − v1 y =
+ (−4.0 m/s) = +1.8 m/s .
m
0.0571 kg
0.0571 kg
m
!
!
EVALUATE: The change in velocity Δv is in the same direction as the force, so Δv has a negative x component
and a positive y component.
IDENTIFY: The horizontal component of the momentum of the system of cars is conserved.
SET UP: Let +x be the direction the cars are traveling. Each car has mass m. Let v1 be the initial speed of the
(b) v2 x = 8.66. three cars. v2 = 1 v1 . Let N be the number of cars in the final collection.
5
3v1
v
= 3 1 = 15 .
v2
v1 / 5
EVALUATE: In the complete absence of friction or other external horizontal forces this process of adding cars and
slowing down continues forever.
IDENTIFY: Px = p Ax + pBx and Py = p Ay + pBy .
EXECUTE: 8.67. SET UP: P x = P2 x . (3m)v1 = ( Nm)v2 . N =
1 Let object A be the convertible and object B be the SUV. Let +x be west and +y be south, p Ax = 0 and pBy = 0 .
EXECUTE: vBx = Px = (8000 kg ⋅ m/s)sin 60.0° = 6928 kg ⋅ m/s , so pBx = 6928 kg ⋅ m/s and 6928 kg ⋅ m/s
= 3.46 m/s .
2000 kg Py = (8000 kg ⋅ m/s)cos 60.0° = 4000 kg ⋅ m/s , so pBx = 4000 kg ⋅ m/s and v Ay = 8.68. 4000 kg ⋅ m/s
= 2.67 m/s .
1500 kg The convertible has speed 2.67 m/s and the SUV has speed 3.46 m/s.
EVALUATE: Each component of the total momentum arises from a single vehicle.
IDENTIFY: The total momentum of the system is conserved and is equal to zero, since the pucks are released
from rest.
SET UP: Each puck has the same mass m. Let +x be east and +y be north. Let object A be the puck that moves
west. All three pucks have the same speed v. Momentum, Impulse, and Collisions EXECUTE: 8-19 P x = P2 x gives 0 = −mv + mvBx + mvCx and v = vBx + vCx . P y = P2 y gives 0 = mvBy + mvCy and
1
1 vBy = −vCy . Since vB = vC and the y components are equal in magnitude, the x components must also be equal:
vBx = vCx and v = vBx + vCx says vBx = vCx = v / 2 . If vBy is positive then vCy is negative. The angle θ that puck B
v/2
and θ = 60° . One puck moves in a direction 60° north of east and
v
the other puck moves in a direction 60° south of east.
EVALUATE: Each component of momentum is separately conserved.
IDENTIFY: The x and y components of the momentum of the system are conserved.
!
Set Up: After the collision the combined object with mass mtot = 0.100 kg moves with velocity v2 . Solve for
makes with the x axis is given by cosθ = 8.69. vCx and vCy .
EXECUTE: (a) P x = P2 x gives m Av Ax + mB vBx + mC vCx = mtot v2 x .
1 vCx = −
vCx = − mAv Ax + mB vBx − mtot v2 x
mC (0.020 kg)( −1.50 m/s) + (0.030 kg)( −0.50 m/s)cos60° − (0.100 kg)(0.50 m/s)
.
0.050 kg
vCx = 1.75 m/s . P y = P2 y gives mAv Ay + mB vBy + mC vCy = mtot v2 y .
1
vCy = − mAv Ay + mB vBy − mtot v2 y
mC =− (0.030 kg)( −0.50 m/s)sin 60°
= +0.260 m/s .
0.050 kg 2
2
(b) vC = vCx + vCy = 1.77 m/s . ΔK = K 2 − K1 . 8.70. ΔK = 1 (0.100 kg)(0.50 m/s) 2 − [ 1 (0.020 kg)(1.50 m/s) 2 + 1 (0.030)(0.50 m/s) 2 + 1 (0.050 kg)(1.77 m/s) 2 ]
2
2
2
2
ΔK = −0.092 J .
EVALUATE: Since there is no horizontal external force the vector momentum of the system is conserved. The
forces the spheres exert on each other do negative work during the collision and this reduces the kinetic energy of
the system.
IDENTIFY: Use a coordinate system attached to the ground. Take the x-axis to be east (along the tracks) and the
y-axis to be north (parallel to the ground and perpendicular to the tracks). Then Px is conserved and Py is not
conserved, due to the sideways force exerted by the tracks, the force that keeps the handcar on the tracks.
(a) SET UP: Let A be the 25.0 kg mass and B be the car (mass 175 kg). After the mass is thrown sideways relative
to the car it still has the same eastward component of velocity, 5.00 m/s, as it had before it was thrown. Figure 8.70a Px is conserved so ( mA + mB ) v1 = mAv A 2 x + mB vB 2 x
EXECUTE: ( 200 kg )( 5.00 m/s ) = ( 25.0 kg ) ( 5.00 m/s ) + (175 kg ) vB 2 x .
vB 2 x = 1000 kg ⋅ m/s − 125 kg ⋅ m/s
= 5.00 m/s.
175 kg The final velocity of the car is 5.00 m/s, east (unchanged).
EVALUATE: The thrower exerts a force on the mass in the y-direction and by Newton’s 3rd law the mass exerts
an equal and opposite force in the − y -direction on the thrower and car.
(b) SET UP: We are applying Px = constant in coordinates attached to the ground, so we need the final velocity of
A relative to the ground. Use the relative velocity addition equation. Then use Px = constant to find the final
velocity of the car. 8-20 Chapter 8 EXECUTE: !
!
!
v A / E = v A / B + vB / E vB / E = +5.00 m/s
v A / B = −5.00 m/s (minus since the mass is moving west relative to the car). This gives v A / E = 0; the mass is at rest
relative to the earth after it is thrown backwards from the car.
As in part (a), ( mA + mB ) v1 = mAv A 2 x + mB vB 2 x .
Now v A 2 x = 0, so ( mA + mB ) v1 = mB vB 2 x .
⎛ m + mB ⎞
⎛ 200 kg ⎞
vB 2 x = ⎜ A
⎟ v1 = ⎜
⎟ ( 5.00 m/s ) = 5.71 m/s.
mB ⎠
⎝ 175 kg ⎠
⎝
The final velocity of the car is 5.71 m/s, east.
EVALUATE: The thrower exerts a force in the − x-direction so the mass exerts a force on him in the + x -direction
and he and the car speed up.
(c) SET UP: Let A be the 25.0 kg mass and B be the car (mass mB = 200 kg). Figure 8.70b
Px is conserved so mAv A1x + mB vB1x = ( m A + mB ) v2 x . EXECUTE: − mAv A1 + mB vB1 = ( mA + mB ) v2 x .
v2 x = 8.71. 8.72. mB vB1 − mAv A1 ( 200 kg ) ( 5.00 m/s ) − ( 25.0 kg ) ( 6.00 m/s )
=
= 3.78 m/s.
mA + mB
200 kg + 25.0 kg The final velocity of the car is 3.78 m/s, east.
EVALUATE: The mass has negative px so reduces the total Px of the system and the car slows down.
IDENTIFY: The horizontal component of the momentum of the sand plus railroad system is conserved.
SET UP: As the sand leaks out it retains its horizontal velocity of 15.0 m/s.
EXECUTE: The horizontal component of the momentum of the sand doesn’t change when it leaks out so the
speed of the railroad car doesn’t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and
initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car.
Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down.
EVALUATE: The horizontal momentum of the sand does change when it strikes the ground, due to the force that
is external to the system of sand plus railroad car.
IDENTIFY: Kinetic energy is K = 1 mv 2 and the magnitude of the momentum is p = mv . The force and the time t
2
it acts are related to the change in momentum whereas the force and distance d it acts are related to the change in
kinetic energy.
SET UP: Assume the net forces are constant and let the forces and the motion be along the x axis. The impulsemomentum theorem then says Ft = Δp and the work-energy theorem says Fd = ΔK . EXECUTE: (a) K N = 1 (840 kg)(9.0 m/s) 2 = 3.40 × 104 J . K P = 1 (1620 kg)(5.0 m/s) 2 = 2.02 × 104 J . The Nash has
2
2
the greater kinetic energy and KN
= 1.68 .
KP (b) pN = (840 kg)(9.0 m/s) = 7.56 × 103 kg ⋅ m/s . pP = (1620 kg)(5.0 m/s) = 8.10 × 103 kg ⋅ m/s . The Packard has
the greater magnitude of momentum and pN
= 0.933 .
pP (c) Since the cars stop the magnitude of the change in momentum equals the initial momentum. Since pP > pN ,
FN pN
=
= 0.933 .
FP pP
(d) Since the cars stop the magnitude of the change in kinetic energy equals the initial kinetic energy. Since
F
K
K N > K P , FN > FP and N = N = 1.68 .
FP K P
EVALUATE: If the stopping forces were the same, the Packard would have a larger stopping time but would
travel a shorter distance while stopping. This consistent with it having a smaller initial speed.
FP > FN and Momentum, Impulse, and Collisions 8.73. 8.74. 8-21 IDENTIFY: Use the impulse-momentum theorem to relate the average force on the bullets to their rate of change
in momentum. By Newton’s third law, the average force the weapon exerts on the bullets is equal in magnitude
and opposite in direction to the recoil force the bullets exert on the weapon.
SET UP: Consider a time interval of 1.00 minute. Let +x be the direction of motion of the bullets and use
coordinated fixed to the ground. The bullets start from rest.
(1000)(7.45 × 10−3 kg)(293 m/s)
= 36.4 N . The recoil force is 36.4 N.
EXECUTE: Fav Δt = Δp gives Fav =
60.0 s
EVALUATE: The change in momentum for each bullet is small since the mass is small, but over 16 bullets are
fired per second.
IDENTIFY: Find k for the spring from the forces when the frame hangs at rest, use constant acceleration equations
to find the speed of the putty just before it strikes the frame, apply conservation of momentum to the collision
between the putty and the frame and then apply conservation of energy to the motion of the frame after the collision.
SET UP: Use the free-body diagram for the frame when it hangs at rest on the end of the spring to find the force
constant k of the spring. Let s be the amount the spring is stretched. EXECUTE: Figure 8.74a ∑ Fy = ma y . − mg + ks = 0 .
k= mg ( 0.150 kg ) ( 9.80 m/s
=
s
0.050 m 2 ) = 29.4 N/m . SET UP: Next find the speed of the putty when it reaches the frame. The putty falls with acceleration a = g ,
downward. Figure 8.74b
v0 = 0
y − y0 = 0.300 m
a = +9.80 m/s 2
v=?
2
v 2 = v0 + 2a ( y − y0 ) EXECUTE:
SET UP: v = 2a ( y − y0 ) = 2 ( 9.80 m/s 2 ) ( 0.300 m ) = 2.425 m/s .
Apply conservation of momentum to the collision between the putty (A) and the frame (B): Figure 8.74c Py is conserved, so − mAv A1 = − ( mA + mB ) v2 .
EXECUTE: ⎛ mA ⎞
⎛ 0.200 kg ⎞
v2 = ⎜
⎟ v A1 = ⎜
⎟ ( 2.425 m/s ) = 1.386 m/s .
⎝ 0.350 kg ⎠
⎝ mA + mB ⎠ 8-22 Chapter 8 SET UP: Apply conservation of energy to the motion of the frame on the end of the spring after the collision. Let
point 1 be just after the putty strikes and point 2 be when the frame has its maximum downward displacement. Let
d be the amount the frame moves downward. Figure 8.74d When the frame is at position 1 the spring is stretched a distance x1 = 0.050 m. When the frame is at position 2 the
spring is stretched a distance x2 = 0.050 m + d . Use coordinates with the y-direction upward and y = 0 at the
lowest point reached by the frame, so that y1 = d and y2 = 0. Work is done on the frame by gravity and by the
spring force, so Wother = 0, and U = U el + U gravity .
EXECUTE: K1 + U1 + Wother = K 2 + U 2 . Wother = 0 .
K1 = 1 mv12 = 1 ( 0.350 kg )(1.386 m/s ) = 0.3362 J .
2
2
2 U1 = U1, el + U1, grav = 1 kx12 + mgy1 = 1 ( 29.4 N/m )( 0.050 m ) + ( 0.350 kg ) ( 9.80 m/s 2 ) d .
2
2
2 U1 = 0.03675 J + ( 3.43 N ) d . ( 29.4 N/m )( 0.050 m + d ) .
U 2 = 0.03675 J + (1.47 N ) d + (14.7 N/m ) d 2 .
Thus 0.3362 J + 0.03675 J + ( 3.43 N ) d = 0.03675 J + (1.47 N ) d + (14.7 N/m ) d 2 .
(14.7 N/m ) d 2 − (1.96 N ) d − 0.3362 J = 0 .
2
U 2 = U 2, el + U 2, grav = 1 kx2 + mgy2 =
2 8.75. 2 1
2 2
d = (1/ 29.4 ) ⎡1.96 ± (1.96 ) − 4 (14.7 )( −0.3362 ) ⎤ m = 0.0667 m ± 0.1653 m.
⎢
⎥
⎣
⎦
The solution we want is a positive (downward) distance, so d = 0.0667 m + 0.1653 m = 0.232 m.
EVALUATE: The collision is inelastic and mechanical energy is lost. Thus the decrease in gravitational potential
energy is not equal to the increase in potential energy stored in the spring.
IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the
collision.
SET UP: Let +x be to the right. The total mass is m = mbullet + mblock = 1.00 kg . The spring has force constant k= F
0.750 N
=
= 300 N/m . Let V be the velocity of the block just after impact.
x 0.250 × 10−2 m EXECUTE: (a) Conservation of energy for the motion after the collision gives K1 = U el2 .
V =x 1
2 mV 2 = 1 kx 2 and
2 k
300 N/m
= (0.150 m)
= 2.60 m/s .
m
1.00 kg (b) Conservation of momentum applied to the collision gives mbullet v1 = mV .
v1 = 8.76. mV
(1.00 kg)(2.60 m/s)
=
= 325 m/s .
mbullet
8.00 × 10−3 kg EVALUATE: The initial kinetic energy of the bullet is 422 J. The energy stored in the spring at maximum
compression is 3.38 J. Most of the initial mechanical energy of the bullet is dissipated in the collision.
IDENTIFY: The horizontal components of momentum of the system of bullet plus stone are conserved. The
collision is elastic if K1 = K 2 . Momentum, Impulse, and Collisions 8-23 SET UP: Let A be the bullet and B be the stone.
(a) Figure 8.76
EXECUTE: Px is conserved so mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x . m Av A1 = mB vB 2 x .
⎛m ⎞
⎛ 6.00 × 10−3 kg ⎞
vB 2 x = ⎜ A ⎟ v A1 = ⎜
⎟ ( 350 m/s ) = 21.0 m/s
⎝ 0.100 kg ⎠
⎝ mB ⎠
Py is conserved so m Av A1 y + mB vB1 y = m Av A 2 y + mB vB 2 y .
0 = −mAv A 2 + mB vB 2 y .
⎛m ⎞
⎛ 6.00 × 10−3 kg ⎞
vB 2 y = ⎜ A ⎟ v A 2 = ⎜
⎟ ( 250 m/s ) = 15.0 m/s .
⎝ 0.100 kg ⎠
⎝ mB ⎠
2
2
vB 2 = vB 2 x + vB 2 y = tan θ = vB 2 y
vB 2 x = ( 21.0 m/s ) 2 + (15.0 m/s ) = 25.8 m/s .
2 15.0 m/s
= 0.7143; θ = 35.5° (defined in the sketch).
21.0 m/s (b) To answer this question compare K1 and K 2 for the system:
2
2
K1 = 1 mAv A1 + 1 mB vB1 =
2
2 1
2 ( 6.00 × 10 −3 kg ) ( 350 m/s ) = 368 J .
2 2
2
K 2 = 1 mAv A 2 + 1 mB vB 2 = 1 ( 6.00 × 10−3 kg ) ( 250 m/s ) + 1 ( 0.100 kg )( 25.8 m/s ) = 221 J .
2
2
2
2
2 2 ΔK = K 2 − K1 = 221 J − 368 J = −147 J . EVALUATE: The kinetic energy of the system decreases by 147 J as a result of the collision; the collision is not
elastic. Momentum is conserved because ∑ Fext, x = 0 and ∑ Fext, y = 0. But there are internal forces between the
8.77. bullet and the stone. These forces do negative work that reduces K.
IDENTIFY: Apply conservation of momentum to the collision between the two people. Apply conservation of
energy to the motion of the stuntman before the collision and to the entwined people after the collision.
SET UP: For the motion of the stuntman, y1 − y2 = 5.0 m . Let vS be the magnitude of his horizontal velocity just
before the collision. Let V be the speed of the entwined people just after the collision. Let d be the distance they
slide along the floor.
2
EXECUTE: (a) Motion before the collision: K1 + U1 = K 2 + U 2 . K1 = 0 and 1 mvS = mg ( y1 − y2 ) .
2
vS = 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(5.0 m) = 9.90 m/s .
Collision: mSvS = mtotV . V = ⎛ 80.0 kg ⎞
mS
vS = ⎜
⎟ (9.90 m/s) = 5.28 m/s .
mtot
⎝ 150.0 kg ⎠ (b) Motion after the collision: K1 + U1 + Wother = K 2 + U 2 gives mtotV 2 − μ k mtot gd = 0 . V2
(5.28 m/s) 2
=
= 5.7 m .
2 μ k g 2(0.250)(9.80 m/s 2 )
EVALUATE: Mechanical energy is dissipated in the inelastic collision, so the kinetic energy just after the collision
is less than the initial potential energy of the stuntman.
IDENTIFY: Apply conservation of energy to the motion before and after the collision and apply conservation of
momentum to the collision.
SET UP: Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object
have mass m.
EXECUTE: Conservation of energy says 1 mv 2 = mgR; v = 2 gR .
2
d= 8.78. 1
2 8-24 Chapter 8 SET UP: This is speed v1 for the collision. Let v2 be the speed of the combined object just after the collision.
EXECUTE: Conservation of momentum applied to the collision gives mv1 = 2mv2 so v2 = v1 / 2 = gR / 2
SET UP: Apply conservation of energy to the motion of the combined object after the collision. Let y3 be the
final height above the bottom of the bowl.
2
EXECUTE: 1 ( 2m ) v2 = ( 2m ) gy3 .
2
2
v2
1 ⎛ gR ⎞
=
⎜
⎟ = R/4.
2g 2g ⎝ 2 ⎠
EVALUATE: Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the
initial gravitational potential energy.
IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. Apply conservation of energy to the
motion of the block after the collision.
SET UP: Object B is the block, initially at rest. If L is the length of the wire and θ is the angle it makes with the
vertical, the height of the block is y = L (1 − cosθ ) . Initially, y1 = 0 . y3 = 8.79. 8.80. ⎛ 2m A ⎞
⎛ 2M ⎞
EXECUTE: Eq. 8.25 gives vB = ⎜
⎟ vA = ⎜
⎟ (5.00 m/s) = 2.50 m/s . Conservation of energy gives
m A + mB ⎠
⎝ M + 3M ⎠
⎝
v2
(2.50 m/s) 2
2
1
mB vB = mB gL(1 − cosθ ) . cosθ = 1 − B = 1 −
= 0.362 and θ = 68.8° .
2
2 gL
2(9.80 m/s 2 )(0.500 m)
EVALUATE: Only a portion of the initial kinetic energy of the ball is transferred to the block in the collision.
IDENTIFY: Apply conservation of energy to the motion before and after the collision. Apply conservation of
momentum to the collision.
SET UP: First consider the motion after the collision. The combined object has mass mtot = 25.0 kg. Apply
!
!
∑ F 5 ma to the object at the top of the circular loop, where the object has speed v3. The acceleration is
2
arad = v3 / R, downward. EXECUTE: T + mg = m 2
v3
.
R The minimum speed v3 for the object not to fall out of the circle is given by setting T = 0. This gives v3 = Rg ,
where R = 3.50 m.
SET UP: Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the
loop. Take y = 0 at point 2. Only gravity does work, so K 2 + U 2 = K 3 + U 3 EXECUTE: 1
2 2
2
mtot v2 = 1 mtot v3 + mtot g ( 2 R ) .
2 Use v3 = Rg and solve for v2 : v2 = 5 gR = 13.1 m/s . SET UP: Now apply conservation of momentum to the collision between the dart and the sphere. Let v1 be the
speed of the dart before the collision.
EXECUTE: ( 5.00 kg ) v1 = ( 25.0 kg ) (13.1 m/s ) . 8.81. v1 = 65.5 m/s .
EVALUATE: The collision is inelastic and mechanical energy is removed from the system by the negative work
done by the forces between the dart and the sphere.
!
!
IDENTIFY: Use Eq. 8.25 to find the speed of the hanging ball just after the collision. Apply ∑ F = ma to find
the tension in the wire. After the collision the hanging ball moves in an arc of a circle with radius R = 1.35 m and
acceleration arad = v 2 / R .
!
!
SET UP: Let A be the 2.00 kg ball and B be the 8.00 kg ball. For applying ∑ F = ma to the hanging ball, let +y
!
be upward, since arad is upward. The free-body force diagram for the 8.00 kg ball is given in Figure 8.81.
⎛ 2m A ⎞
⎛
⎞
2[2.00kg]
vB 2 x = ⎜
⎟ v A1x = ⎜
⎟ (5.00 m/s) = 2.00 m/s . Just after the collision the 8.00 kg
mA + mB ⎠
2.00 kg + 8.00 kg ⎠
⎝
⎝
ball has speed v = 2.00 m/s . Using the free-body diagram, ∑ Fy = ma y gives T − mg = marad . EXECUTE: ⎛
⎛
v2 ⎞
[2.00 m/s]2 ⎞
T = m ⎜ g + ⎟ = (8.00 kg) ⎜ 9.80 m/s 2 +
⎟ = 102 N .
1.35 m ⎠
R⎠
⎝
⎝ Momentum, Impulse, and Collisions 8-25 EVALUATE: The tension before the collision is the weight of the ball, 78.4 N. Just after the collision, when the
ball has started to move, the tension is greater than this.
y
T arad
x mg 8.82. Figure 8.81
IDENTIFY: The impulse applied to the ball equals its change in momentum. The height of the ball and its speed
are related by conservation of energy.
SET UP: Let +y be upward.
EXECUTE: Applying conservation of energy to the motion of the ball from its height h to the floor gives
1
mv12 = mgh , where v1 is its speed just before it hits the floor. Just before it hits, it is traveling downward, so the
2
velocity of the ball just before it hits the floor is v1 y = − 2 gh . Applying conservation of energy to the motion of the
ball from just after it bounces off the floor with speed v2 to its maximum height of 0.90h gives 1
2 2
mv2 = mg (0.90h) . It is moving upward, so v2 y = + 2 g (0.90h) . The impulse applied to the ball is J y = p2 y − p1 y = m(v2 y − v1 y ) = 8.83. m 2 g (0.90h) + m 2 gh = 2.76m gh . The floor exerts an upward impulse of 2.76m gh to the ball.
EVALUATE: The impulse increases when m increases and when h increases. The ball does not return to its initial
height because some mechanical energy is dissipated during the collision with the floor.
IDENTIFY: Apply conservation of momentum to the collision between the bullet and the block and apply
conservation of energy to the motion of the block after the collision.
(a) SET UP: Collision between the bullet and the block: Let object A be the bullet and object B be the block.
Apply momentum conservation to find the speed vB 2 of the block just after the collision. Figure 8.83a
EXECUTE: Px is conserved so mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x . m Av A1 = mAv A 2 + mB vB 2 x .
mA ( v A1 − v A 2 ) 4.00 × 10−3 kg ( 400 m/s − 120 m/s )
=
= 1.40 m/s .
mB
0.800 kg
SET UP: Motion of the block after the collision.
Let point 1 in the motion be just after the collision, where the block has the speed 1.40 m/s calculated above, and
let point 2 be where the block has come to rest.
vB 2 x = Figure 8.83b
K1 + U1 + Wother = K 2 + U 2 . EXECUTE: Work is done on the block by friction, so Wother = W f .
Wother = W f = ( f k cos φ ) s = − f k s = − μ k mgs, where s = 0.450 m
U1 = 0, U2 = 0 K1 = mv ,
1
2 Thus 1
2 2
1 K 2 = 0 (block has come to rest) mv − μ k mgs = 0.
2
1 v12
(1.40 m/s )
=
= 0.222 .
2 gs 2 ( 9.80 m/s 2 ) ( 0.450 m )
2 μk = 8-26 Chapter 8 (b) For the bullet, K1 = 1 mv12 = 1 ( 4.00 × 10−3 kg ) ( 400 m/s ) = 320 J .
2
2
2 2
K 2 = 1 mv2 = 1 ( 4.00 × 10−3 kg ) (120 m/s ) = 28.8 J .
2
2
2 ΔK = K 2 − K1 = 28.8 J − 320 J = −291 J .
The kinetic energy of the bullet decreases by 291 J.
(c) Immediately after the collision the speed of the block is 1.40 m/s so its kinetic energy is
K = 1 mv 2 =
2 8.84. 1
2 ( 0.800 kg )(1.40 m/s ) 2 = 0.784 J. EVALUATE: The collision is highly inelastic. The bullet loses 291 J of kinetic energy but only 0.784 J is gained
by the block. But momentum is conserved in the collision. All the momentum lost by the bullet is gained by the
block.
IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion of the
block after the collision.
SET UP: Let +x be to the right. Let the bullet be A and the block be B. Let V be the velocity of the block just after
the collision.
EXECUTE: Motion of block after the collision: K1 = U grav2 . 1 mBV 2 = mB gh .
2 V = 2 gh = 2(9.80 m/s 2 )(0.450 × 10−2 m) = 0.297 m/s .
Collision: vB 2 = 0.297 m/s . P x = P2 x gives mAv A1 = mAv A 2 + mB vB 2 .
1
vA2 = 8.85. mAv A1 − mB vB 2 (5.00 × 10−3 kg)(450 m/s) − (1.00 kg)(0.297 m/s)
=
= 391 m/s .
mA
5.00 × 10−3 kg EVALUATE: We assume the block moves very little during the time it takes the bullet to pass through it.
IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. The value of M where the kinetic energy
loss K loss of the neutron is a maximum satisfies dK loss / dM = 0 .
SET UP: Let object A be the neutron and object B be the nucleus. Let the initial speed of the neutron be v A1 . All
2
motion is along the x-axis. K 0 = 1 mv A1 .
2 EXECUTE: (a) v A 2 = ⎛ ⎡ m − M ⎤2 ⎞ 2
2m 2 M 2
4 K 0 mM
m−M
2
2
v A1 . K loss = 1 mv A1 − 1 mv A 2 = 1 m ⎜1 − ⎢
v A1 =
, as
⎟ v A1 =
2
2
2
⎜ ⎣m + M ⎥ ⎟
m+M
( M + m) 2
( M + m) 2
⎦⎠
⎝ was to be shown.
⎡
1
2M ⎤
dK loss
2M
(b)
= 1 and M = m . The incident neutron loses the most
= 4K0m ⎢
−
=0.
2
3⎥
M +m
dM
⎣ ( M + m ) ( M + m) ⎦
kinetic energy when the target has the same mass as the neutron.
(c) When mA = mB , Eq. 8.24 says v A 2 = 0 . The final speed of the neutron is zero and the neutron loses all of its
kinetic energy.
EVALUATE: When M >> m , v A 2 x ≈ −v A1x and the neutron rebounds with speed almost equal to its initial speed.
8.86. In this case very little kinetic energy is lost; K loss = 4 K 0 m / M , which is very small.
IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision.
SET UP: Let all the motion be along the x axis. v A1x = v0 .
⎛ m − mB ⎞
⎛ 2mA ⎞
2
1
EXECUTE: (a) v A 2 x = ⎜ A
⎟ v0 and vB 2 x = ⎜
⎟ v0 . K1 = 2 mAv0 .
⎝ mA + mB ⎠
⎝ mA + mB ⎠
2 2 2 ⎛ m − mB ⎞ 2 ⎛ mA − mB ⎞
K A 2 ⎛ mA − mB ⎞
2
=⎜
K A 2 = 1 m Av A 2 x = 1 m A ⎜ A
⎟.
⎟ v0 = ⎜
⎟ K1 and
2
2
K1 ⎝ mA + mB ⎠
⎝ mA + mB ⎠
⎝ m A + mB ⎠
2 ⎛ 2m A ⎞ 2
4mA mB
K
4mAmB
2
.
K B 2 = 1 mB vB 2 x = 1 mB ⎜
K1 and B 2 =
⎟ v0 =
2
2
2
K1 ( mA + mB )2
mA + mB ⎠
mA + mB )
(
⎝
K
K
K
4
K
5
(b) (i) For mA = mB , A 2 = 0 and B 2 = 1 . (ii) For mA = 5mB , A 2 = and B 2 = .
K1
K1
K1 9
K1 9 Momentum, Impulse, and Collisions 8-27 2 (c) Equal sharing of the kinetic energy means K A 2 K B 2 1 ⎛ mA − mB ⎞ 1
=
= .⎜
⎟= .
K1
K1 2 ⎝ mA + mB ⎠ 2 2
2
2
2
2
2
2mA + 2mB − 4mAmB = mA + 2mA mB + mB . mA − 6mAmB + mB = 0 . The quadratic formula gives mA
= 5.83 or
mB mA
K
1
= 0.172 . We can also verify that these values give B 2 = .
mB
K1 2
8.87. EVALUATE: When mA << mB or when mA >> mB , object A retains almost all of the original kinetic energy.
IDENTIFY: Apply conservation of energy to the motion of the package before the collision and apply
conservation of the horizontal component of momentum to the collision.
(a) SET UP: Apply conservation of energy to the motion of the package from point 1 as it leaves the chute to
point 2 just before it lands in the cart. Take y = 0 at point 2, so y1 = 4.00 m. Only gravity does work, so K1 + U1 = K 2 + U 2 .
EXECUTE: 1
2 2
mv12 + mgy1 = 1 mv2 .
2 v2 = v12 + 2 gy1 = 9.35 m/s .
(b) SET UP: In the collision between the package and the cart momentum is conserved in the horizontal direction.
(But not in the vertical direction, due to the vertical force the floor exerts on the cart.) Take + x to be to the right.
Let A be the package and B be the cart.
EXECUTE: Px is constant gives mAv A1x + mB vB1x = ( mA + mB ) v2 x . vB1x = −5.00 m/s . v A1x = ( 3.00 m/s ) cos37.0° . (The horizontal velocity of the package is constant during its free-fall.) 8.88. Solving for v2 x gives v2 x = −3.29 m/s. The cart is moving to the left at 3.29 m/s after the package lands in it.
EVALUATE: The cart is slowed by its collision with the package, whose horizontal component of momentum is in
the opposite direction to the motion of the cart.
IDENTIFY: Eqs. 8.24, 8.25, and 8.27 give the outcome of the elastic collision.
SET UP: The blue puck is object A and the red puck is object B. Let +x be the direction of the initial motion of A.
v A1x = 0.200 m/s , v A 2 x = 0.050 m/s and vB1x = 0
EXECUTE: (a) Eq. 8.27 gives vB 2 x = v A 2 x − vB1x + v A1x = 0.250 m/s . ⎛v
⎞
⎛ ⎡ 0.200 m/s ⎤ ⎞
(b) Eq. 8.25 gives mB = mA ⎜ 2 A1x − 1⎟ = (0.0400 kg) ⎜ 2 ⎢
⎥ − 1⎟ = 0.024 kg .
vB 2 x
⎝ ⎣ 0.250 m/s ⎦ ⎠
⎝
⎠
EVALUATE: We can verify that our results give K1 = K 2 and P x = P2 x , as required in an elastic collision.
1
8.89. 2
2
(a) IDENTIFY and SET UP: K = 1 mAv A + 1 mB vB .
2
2
!
!
!
!
!
!
Use v A = v′ + vcm and v B = v′ + vcm to replace v A and vB in this equation. Note v′ and v′ as defined in the
A
B
A
B
problem are the velocities of A and B in coordinates moving with the center of mass. Note also that
!
!
!
!
′
′
mAv′A + mB v′ = Mvcm where vcm is the velocity of the car in these coordinates. But that’s zero, so
B
!
!
mAv′A 1 mB v′ 5 0; we can use this in the proof.
B
!
In part (b), use that P is conserved in a collision.
!!
!
!
2
2
EXECUTE: v A = v′ + vcm , so v A = v′2 + vcm + 2v′A ⋅ vcm .
A
A !!
!
!
2
2
v B = v′ + vcm , so vB = v′2 + vcm + 2v′ ⋅ vcm .
B
B
B
!!
!
2
(We have used that for a vector A, A = A ⋅ A.)
!!
!!
2
2
Thus K = 1 mAv′2 + 1 mAvcm + mAv′ ⋅ vcm + 1 mB v′2 + 1 mB vcm + mB v′ ⋅ vcm .
A
A
B
B
2
2
2
2 K= 1
2 ! ! ! 2
( mA + mB ) vcm + 1 ( mAv′A2 + mBv′B2 ) + ( mAv′A + mBv′B ) ⋅ vcm .
2 !
!
2
But mA + mB = M and as noted earlier mAv′ + mB v′ = 0, so K = 1 Mvcm + 1 ( mAv′2 + mB v′2 ) . This is the result the
A
B
A
B
2
2 problem asked us to derive.
!
!
(b) EVALUATE: In the collision P = Mvcm is constant, so
relative kinetic energy but the 1
2 2
Mvcm must remain. 1
2 2
Mvcm stays constant. The asteroids can lose all their 8-28 8.90. Chapter 8 IDENTIFY: Eq. 8.27 describes the elastic collision, with x replaced by y. Speed and height are related by
conservation of energy.
SET UP: Let +y be upward. Let A be the large ball and B be the small ball, so vB1 y = −v and v A1 y = + v . If the
large ball has much greater mass than the small ball its speed is changed very little in the collision and v A 2 y = +v . EXECUTE: (a) vB 2 y − v A 2 y = −(vB1 y − v A1 y ) gives vB 2 y = + v A 2 y − vB1 y + v A1 y = v − ( −v) + v = +3v . The small ball
moves upward with speed 3v after the collision.
(b) Let h1 be the height the small ball fell before the collision. Conservation of energy applied to the motion from
the release point to the floor gives U1 = K 2 and mgh1 = 1 mv 2 . h1 =
2 v2
. Conservation of energy applied to the
2g motion of the small ball from immediately after the collision to its maximum height h2 (rebound distance) gives
9v 2
= 9h1 . The ball’s rebound distance is nine times the distance it fell.
2g
EVALUATE: The mechanical energy gained by the small ball comes from the energy of the large ball. But since
the large ball’s mass is much larger it can give up this energy with very little decrease in speed.
IDENTIFY: Apply conservation of momentum to the system consisting of Jack, Jill and the crate. The speed of
Jack or Jill relative to the ground will be different from 4.00 m/s.
SET UP: Use an inertial coordinate system attached to the ground. Let +x be the direction in which the people
jump. Let Jack be object A, Jill be B, and the crate be C.
EXECUTE: (a) If the final speed of the crate is v, vC 2 x = −v , and v A 2 x = vB 2 x = 4.00 m/s − v . P2 x = P x gives
1
K1 = U 2 and 8.91. 1
2 m(3v) 2 = mgh2 . h2 = mAv A 2 x + mB vBx 2 + mC vCx 2 = 0 . (75.0 kg)(4.00 m/s − v) + (45.0 kg)(4.00 m/s − v) + (15.0 kg)( −v) = 0 and
v= (75.0 kg + 45.0 kg)(4.00 m/s)
= 3.56 m/s .
75.0 kg + 45.0 kg + 15.0 kg (b) Let v′ be the speed of the crate after Jack jumps. Apply momentum conservation to Jack jumping:
(75.0 kg)(4.00 m/s)
(75.0 kg)(4.00 m/s − v′) + (60.0 kg)( −v′) = 0 and v′ =
= 2.22 m/s . Then apply momentum
135.0 kg
conservation to Jill jumping, with v being the final speed of the crate: P x = P2 x gives
1
(60.0 kg)(−v′) = (45.0 kg)(4.00 m/s − v) + (15.0 kg)(−v ) .
v= (45.0 kg)(4.00 m/s) + (60.0 kg)(2.22 m/s)
= 5.22 m/s .
60.0 kg (c) Repeat the calculation in (b), but now with Jill jumping first.
Jill jumps: (45.0 kg)(4.00 m/s − v′) + (90.0 kg)(−v′) = 0 and v′ = 1.33 m/s .
Jack jumps: (90.0 kg)(−v′) = (75.0 kg)(4.00 m/s − v ) + (15.0 kg)(−v) .
v= 8.92. (75.0 kg)(4.00 m/s) + (90.0 kg)(1.33 m/s)
= 4.66 m/s .
90.0 kg EVALUATE: The final speed of the crate is greater when Jack jumps first, then Jill. In this case Jack leaves with a
speed of 1.78 m/s relative to the ground, whereas when they both jump simultaneously Jack and Jill each leave
with a speed of only 0.44 m/s relative to the ground.
IDENTIFY: Momentum is conserved in the explosion. The total kinetic energy of the two fragments is Q.
SET UP: Let the final speed of the two fragments be v A and vB . They must move in opposite directions after the
explosion.
EXECUTE: (a) Since the initial momentum of the system is zero, conservation of momentum says mAv A = mB vB
⎛m ⎞
and vB = ⎜ A ⎟ v A . K A + K B = Q gives
⎝ mB ⎠ 2 1
2 ⎛m ⎞ 2
2
mAv A + 1 mB ⎜ A ⎟ v A = Q .
2
⎝ mB ⎠ 1
2 m⎞
2⎛
m Av A ⎜ 1 + A ⎟ = Q .
⎝ mB ⎠ ⎛ mB ⎞
⎛
Q
mB ⎞ ⎛ mA ⎞
=⎜
⎟ Q . K B = Q − K A = Q ⎜1 −
⎟=⎜
⎟Q .
1 + mA / mB ⎝ mA + mB ⎠
⎝ mA + mB ⎠ ⎝ mA + mB ⎠
4
1
(b) If mB = 4mA , then K A = Q and K B = Q . The lighter fragment gets 80% of the energy that is released.
5
5
EVALUATE: If mA = mB the fragments share the energy equally. In the limit that mB >> mA , the lighter fragment
gets almost all of the released energy.
KA = Momentum, Impulse, and Collisions 8.93. 8-29 IDENTIFY: Apply conservation of momentum to the system of the neutron and its decay products.
SET UP: Let the proton be moving in the +x direction with speed vp after the decay. The initial momentum of the
neutron is zero, so to conserve momentum the electron must be moving in the − x direction after the decay. Let the
speed of the electron be ve . EXECUTE: ⎛m ⎞
P x = P2 x gives 0 = mp vp − meve and ve = ⎜ p ⎟ vp . The total kinetic energy after the decay is
1
⎝ me ⎠
2 m⎞
⎛m ⎞ 2
2
2
2⎛
K tot = 1 meve2 + 1 mp vp = 1 me ⎜ p ⎟ vp + 1 mp vp = 1 mpvp ⎜1 + p ⎟ .
2
2
2
2
2
⎝ me ⎠
⎝ me ⎠
K
1
1
Thus, p =
=
= 5.44 × 10−4 = 0.0544% .
K tot 1 + mp / me 1 + 1836 8.94. EVALUATE: Most of the released energy goes to the electron, since it is much lighter than the proton.
IDENTIFY: Momentum is conserved in the decay. The results of Problem 8.92 give the kinetic energy of each
fragment.
SET UP: Let A be the alpha particle and let B be the radium nucleus, so mA / mB = 0.0176 . Q = 6.54 × 10−13 J .
Q
6.54 × 10−13 J
=
= 6.43 × 10−13 J and K B = 0.11 × 10−13 J .
1 + mA / mB
1 + 0.0176
EVALUATE: The lighter particle receives most of the released energy.
IDENTIFY: The momentum of the system is conserved.
SET UP: Let +x be to the right. P x = 0 . pex , pnx and panx are the momenta of the electron, polonium nucleus and
1
antineutrino, respectively.
EXECUTE: P x = P2 x gives pex + pnx + panx = 0 . panx = −( pex + pnx ) .
1 EXECUTE: 8.95. KA = panx = −(5.60 × 10−22 kg ⋅ m/s + [3.50 × 10−25 kg][−1.14 × 103 m/s]) = −1.66 × 10−22 kg ⋅ m/s . 8.96. The antineutrino has momentum to the left with magnitude 1.66 × 10−22 kg ⋅ m/s .
EVALUATE: The antineutrino interacts very weakly with matter and most easily shows its presence by the
momentum it carries away.
IDENTIFY: Momentum components in the x and y directions are separately conserved. For an elastic collision
K1 = K 2 . SET UP: v A1x = + v A1 , vB1x = 0 . v A 2 x = v A 2 cos α , v A 2 y = v A 2 sin α . vB 2 x = vB 2 cos α , vB 2 y = −vB 2 sin α . sin θ + cos 2 θ = 1 , for any angle θ . cos(α + β ) = cos α cos β − sin α sin β .
2 EXECUTE: (a) P x = P2 x gives mAv A1 = mAv A 2 cos α + mB vB 2 cos β .
1
P y = P2 y gives 0 = mAv A 2 sin α − mB vB 2 sin β .
1
22
22
22
(b) mAv A1 = mAv A 2 cos 2 α + mB vB 2 cos 2 β + 2mA mB v A 2vB 2 cosα cos β and
22
22
0 = mAv A 2 sin 2 α + mB vB 2 sin 2 β − 2mAmB v A 2vB 2 sin α sin β . Adding these two equations and using the trig identities in
22
22
22
the SET UP step gives mAv A1 = mAv A 2 + mB vB 2 + 2mA mB v A 2vB 2 cos(α + β ) . (c) K1 = K 2 says 1
2 2
2
2
mAv A1 = 1 mAv A 2 + 1 mB vB 2 . The result in part (b) agrees with this expression only if
2
2 cos(α + β ) = 0 . This requires that α + β = 90° = 8.97. π rad .
2
EVALUATE: The result of part (c) says that the two protons move in perpendicular directions after the collision.
IDENTIFY and SET UP: Figure 8.97
Px and Py are conserved in the collision since there is no external horizontal force. 8-30 Chapter 8 The result of Problem 8.96 part (d) applies here since the collision is elastic This says that 25.0° + θ B = 90°, so that θ B = 65.0°. (A and B move off in perpendicular directions.)
EXECUTE: Px is conserved so mAv A1x + mB vB1x = mAv A 2 x + mB vB 2 x . But mA = mB so v A1 = v A 2 cos 25.0° + vB 2 cos65.0° .
Py is conserved so mAv A1 y + mB vB1 y = m Av A 2 y + mB vB 2 y .
0 = v A 2 y + vB 2 y .
0 = v A 2 sin 25.0° − vB 2 sin 65.0° .
vB 2 = ( sin 25.0° / sin 65.0° ) v A 2 . ⎛ sin 25.0° cos65.0° ⎞
This result in the first equation gives v A1 = v A 2 cos 25.0° + ⎜
⎟ v A2 .
sin 65.0°
⎝
⎠
v A1 = 1.103v A 2 .
v A 2 = v A1 /1.103 = (15.0 m/s)/1.103 = 13.6 m/s . And then vB 2 = ( sin 25.0° / sin 65.0° )(13.6 m/s ) = 6.34 m/s. EVALUATE: We can use our numerical results to show that K1 = K 2 and that P x = P2 x and P y = P2 y .
1
1
8.98. IDENTIFY: Since there is no friction, the horizontal component of momentum of the system of Jonathan, Jane and
the sleigh is conserved.
SET UP: Let +x be to the right. wA = 800 N , wB = 600 N and wC = 1000 N .
EXECUTE: P x = P2 x gives 0 = mAv A 2 x + mB vB 2 x + mC vC 2 x . vC 2 x =
1 mAv A 2 x + mB vB 2 x wAv A 2 x + wB vB 2 x
=
.
mC
wC (800 N)(−[5.00 m/s]cos30.0°) + (600 N)(+[7.00 m/s]cos36.9°)
= −0.105 m/s .
1000 N
The sleigh’s velocity is 0.105 m/s, to the left.
EVALUATE: The vertical component of the momentum of the system consisting of the two people and the sleigh
is not conserved, because of the net force exerted on the sleigh by the ice while they jump.
IDENTIFY: In Eq. 8.28 treat each straight piece as an object in the system.
SET UP: The center of mass of each piece of length L is at its center.
EXECUTE: (a) From symmetry, the center of mass is on the vertical axis, a distance ( L / 2)cos(α / 2) below the
apex.
(b) The center of mass is on the vertical axis of symmetry, a distance 2( L / 2) / 3 = L / 3 above the center of the
horizontal segment.
(c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal and each is equal
to ( L / 2) / 2 = L / 4 . The center of mass is along the bisector of the angle, a distance L / 8 from the corner.
vC 2 x = 8.99. 8.100. (d) By symmetry, the center of mass is at the center of the equilateral triangle, a distance ( L / 3)sin 60° = L / 12
above the center of the horizontal segment.
EVALUATE: The center of mass need not lie on any point of the object, it can be in empty space.
IDENTIFY: There is no net horizontal external force so vcm is constant.
SET UP: Let +x be to the right, with the origin at the initial position of the left-hand end of the canoe.
mA = 45.0 kg , mB = 60.0 kg . The center of mass of the canoe is at its center.
EXECUTE: Initially, vcm = 0 , so the center of mass doesn’t move. Initially, xcm1 = mA x A1 + mB xB1
. After she
mA + mB mA x A 2 + mB xB 2
. xcm1 = xcm2 gives mA x A1 + mB xB1 = mA x A 2 + mB xB 2 . She walks to a point 1.00 m from
mA + mB
the right-hand end of the canoe, so she is 1.50 m to the right of the center of mass of the canoe and
x A 2 = xB 2 + 1.50 m . walks, xcm2 = (45.0 kg)(1.00 m) + (60.0 kg)(2.50 m) = (45.0 kg)( xB 2 + 1.50 m) + (60.0 kg) xB 2 .
(105.0 kg) xB 2 = 127.5 kg ⋅ m and xB 2 = 1.21 m . xB 2 − xB1 = 1.21 m − 2.50 m = −1.29 m . The canoe moves 1.29 m
to the left. Momentum, Impulse, and Collisions 8.101. 8-31 EVALUATE: When the woman walks to the right, the canoe moves to the left. The woman walks 3.00 m to the
right relative to the canoe and the canoe moves 1.29 m to the left, so she moves 3.00 m − 1.29 m = 1.71 m to the
right relative to the water. Note that this distance is (60.0 kg / 45.0 kg)(1.29 m) .
IDENTIFY: Take as the system you and the slab. There is no horizontal force, so horizontal momentum is
!
!
conserved. By Eq. 8.32, P is constant vcm is constant (for a system of constant mass). Use coordinates fixed to
!
!
the ice, with the direction you walk as the x-direction. vcm is constant and initially vcm = 0. Figure 8.101
!
vcm 5 !
!
mp vp 1 ms vs mp + ms
!
!
mpvcm 1 ms vs = 0 . 50 . mpvpx + ms vsx = 0 .
vsx = − ( mp / ms ) vpx = − ( mp / 5mp ) 2.00 m/s = −0.400 m/s . 8.102. The slab moves at 0.400 m/s, in the direction opposite to the direction you are walking.
EVALUATE: The initial momentum of the system is zero. You gain momentum in the + x -direction so the slab
gains momentum in the − x-direction. The slab exerts a force on you in the + x -direction so you exert a force on
the slab in the − x-direction.
IDENTIFY: Conservation of x and y components of momentum applies to the collision. At the highest point of the
trajectory the vertical component of the velocity of the projectile is zero.
SET UP: Let +y be upward and +x be horizontal and to the right. Let the two fragments be A and B, each with
mass m. For the projectile before the explosion and the fragments after the explosion. ax = 0 , a y = −9.80 m/s 2 .
2
2
EXECUTE: (a) v y = v0 y + 2a y ( y − y0 ) with v y = 0 gives that the maximum height of the projectile is h=− 2
v0 y 2a y =− ([80.0 m/s]sin 60.0°) 2
= 244.9 m . Just before the explosion the projectile is moving to the right with
2( −9.80 m/s 2 ) horizontal velocity vx = v0 x = v0 cos60.0° = 40.0 m/s . After the explosion v Ax = 0 since fragment A falls vertically.
Conservation of momentum applied to the explosion gives (2m)(40.0 m/s) = mvBx and vBx = 80.0 m/s . Fragment B
has zero initial vertical velocity so y − y0 = v0 yt + 1 a yt 2 gives a time of fall of
2
t= − 2h
2(244.9 m)
=−
= 7.07 s . During this time the fragment travels horizontally a distance
ay
−9.80 m/s 2 (80.0 m/s)(7.07 s) = 566 m . It also took the projectile 7.07 s to travel from launch to maximum height and during
this time it travels a horizontal distance of ([80.0 m/s]cos60.0°)(7.07 s) = 283 m . The second fragment lands
283 m + 566 m = 849 m from the firing point.
(b) For the explosion, K1 = 1 (20.0 kg)(40.0 m/s) 2 = 1.60 × 104 J . K 2 = 1 (10.0 kg)(80.0 m/s) 2 = 3.20 × 104 J . The
2
2
energy released in the explosion is 1.60 × 104 J .
EVALUATE: The kinetic energy of the projectile just after it is launched is 6.40 × 104 J . We can calculate the
speed of each fragment just before it strikes the ground and verify that the total kinetic energy of the fragments just
before they strike the ground is 6.40 × 104 J + 1.60 × 104 J = 8.00 × 104 J . Fragment A has speed 69.3 m/s just before
it strikes the ground, and hence has kinetic energy 2.40 × 104 J . Fragment B has speed
(80.0 m/s) 2 + (69.3 m/s) 2 = 105.8 m/s just before it strikes the ground, and hence has kinetic energy 5.60 × 104 J .
2
v0
sin(2α 0 ) = 565 m that the projectile
g
would have had if no explosion had occurred. One fragment lands at R / 2 so the other, equal mass fragment lands
at a distance 3R / 2 from the launch point.
IDENTIFY: The rocket moves in projectile motion before the explosion and its fragments move in projectile
motion after the explosion. Apply conservation of energy and conservation of momentum to the explosion. Also, the center of mass of the system has the same horizontal range R = 8.103. 8-32 Chapter 8 SET UP: Apply conservation of energy to the explosion. Just before the explosion the shell is at its maximum
height and has zero kinetic energy. Let A be the piece with mass 1.40 kg and B be the piece with mass 0.28 kg. Let
v A and vB be the speeds of the two pieces immediately after the collision.
2
2
EXECUTE: 1 mAv A + 1 mB vB = 860 J
2
2
SET UP: Since the two fragments reach the ground at the same time, their velocities just after the explosion must
be horizontal. The initial momentum of the shell before the explosion is zero, so after the explosion the pieces must
be moving in opposite horizontal directions and have equal magnitude of momentum: mAv A = mB vB . EXECUTE: Use this to eliminate v A in the first equation and solve for vB :
1
2 2
mB vB (1 + mB / mA ) = 860 J and vB = 71.6 m/s. Then v A = ( mB / mA ) vB = 14.3 m/s. (b) SET UP: Use the vertical motion from the maximum height to the ground to find the time it takes the pieces to
fall to the ground after the explosion. Take + y downward.
v0 y = 0, a y = +9.80 m/s 2 , y − y0 = 80.0 m, t = ? EXECUTE: y − y0 = v0 yt + 1 a yt 2 gives t = 4.04 s.
2 During this time the horizontal distance each piece moves is x A = v At = 57.8 m and xB = vBt = 289.1 m. They move 8.104. in opposite directions, so they are x A + xB = 347 m apart when they land.
EVALUATE: Fragment A has more mass so it is moving slower right after the collision, and it travels horizontally
a smaller distance as it falls to the ground.
IDENTIFY: Apply conservation of momentum to the collision. At the highest point of its trajectory the shell is
moving horizontally. If one fragment received some upward momentum in the collision, the other fragment would
have had to receive a downward component. Since they each the ground at the same time, each must have zero
vertical velocity immediately after the explosion.
SET UP: Let +x be horizontal, along the initial direction of motion of the projectile and let +y be upward. At its
maximum height the projectile has vx = v0 cos55.0° = 86.0 m/s . Let the heavier fragment be A and the lighter
fragment be B. mA = 9.00 kg and mB = 3.00 kg . EXECUTE: Since fragment A returns to the launch point, immediately after the explosion it has v Ax = −86.0 m/s .
Conservation of momentum applied to the explosion gives
(12.0 kg)(86.0 m/s) = (9.00 kg)(−86.0 m/s) + (3.00 kg)vBx and vBx = 602 m/s . The horizontal range of the
2
v0
sin(2α 0 ) = 2157 m . The horizontal distance each fragment
g
travels is proportional to its initial speed and the heavier fragment travels a horizontal distance R / 2 = 1078 m after
⎛ 602 m ⎞
the explosion, so the lighter fragment travels a horizontal distance ⎜
⎟ (1078 m) = 7546 m from the point of
⎝ 86 m ⎠
explosion and 1078 m + 7546 m = 8624 m from the launch point. The energy released in the explosion is
K 2 − K1 = 1 (9.00 kg)(86.0 m/s) 2 + 1 (3.00 kg)(602 m/s) 2 − 1 (12.0 kg)(86.0 m/s) 2 = 5.33 × 105 J .
2
2
2
EVALUATE: The center of mass of the system has the same horizontal range R = 2157 m as if the explosion
didn’t occur. This gives (12.0 kg)(2157 m) = (9.00 kg)(0) + (3.00 kg) d and d = 8630 m , where d is the distance
from the launch point to where the lighter fragment lands. This agrees with our calculation.
!
IDENTIFY: No external force, so P is conserved in the collision.
SET UP: Apply momentum conservation in the x and y directions: projectile, if no explosion occurred, would be R = 8.105. Figure 8.105
Solve for v1 and v2 . Momentum, Impulse, and Collisions EXECUTE: 8-33 Px is conserved so mv0 = m ( v1 cos 45° + vf cos10° + v2 cos30° ) . v0 − vf cos10° = v1 cos 45° + v2 cos30° .
1030.4 m/s = v1 cos 45° + v2 cos30° . Px is conserved so 0 = m ( v1 sin 45° − v2 sin 30° + vf sin10° ) . v1 sin 45° = v2 sin 30° − 347.3 m/s .
sin 45° = cos 45° so
1030.4 m/s = v2 sin 30° − 347.3 m/s + v2 cos30° .
v2 = 1030.4 m/s + 347.3 m/s
= 1010 m/s .
sin 30° + cos30.0° v2 sin 30° − 347.3 m/s
= 223 m/s. Then two emitted neutrons have speeds of 223 m/s and 1010 m/s.
sin 45°
The speeds of the Ba and Kr nuclei are related by Pz conservation. And then v1 = Pz is constant implies that 0 = mBa vBa − mKr vKr
⎛m ⎞
⎛ 2.3 × 10−25 kg ⎞
vKr = ⎜ Ba ⎟ vBa = ⎜
⎟ vBa = 1.5vBa .
−25
⎝ 1.5 × 10 kg ⎠
⎝ mKr ⎠
We can’t say what these speeds are but they must satisfy this relation. The value of vBa depends on energy
considerations. EVALUATE: K1 = 1 mn ( 3.0 × 103 m/s ) = ( 4.5 × 106 J/kg ) mn .
2
2 K 2 = 1 mn ( 2.0 × 103 m/s ) + 1 mn ( 223 m/s ) + 1 mn (1010 m/s ) + K Ba + K Kr = ( 2.5 × 106 J/kg ) mn + K Ba + K Kr .
2
2
2
2 2 2 We don’t know what K Ba and K Kr are, but they are positive. We will study such nuclear reactions further in
Chapter 43 and will find that energy is released in this process; K 2 > K1. Some of the potential energy stored in the
235 8.106. U nucleus is released as kinetic energy and shared by the collision fragments.
IDENTIFY: The velocity of the center of mass of the system of the two blocks is given by Eq. 8.30. Conservation
of momentum says the center of mass moves at constant speed.
!
!
SET UP: v A1x = v A1 , vB1x = 0 . The velocity u in the center of mass frame is related to the velocity v in the
!!!
p2
stationary frame by u = v − vcm . We can express kinetic energy as K =
.
2m
mAv A1
EXECUTE: (a) vcm-x =
.
mA + mB
(b) The center of mass moves with constant speed so this coordinate system is an inertial frame.
mB v A1
mv
(c) u A1x = v A1x − vcm-x =
. u B1x = vB1x − vcm-x = − A A1 . In this frame P x = mAu A1x + mBuB1x = 0 .
1
mA + mB
mA + mB (d) P2 x = P x = 0 gives p A1x + pB1x = 0 and p A 2 x + pB 2 x = 0 , so pB1x = − p A1x and pB 2 x = − p A 2 x . Conservation of
1
kinetic energy gives 2
2
2
2
p A 2 x pB 2 x p A1x pB1x
2
2
+
=
+
. Using pB 2 x = − p A 2 x and pB1x = − p A1x gives p A 2 x = p A1x and
2mA 2mB 2mA 2mB p A 2 x = ± p A1x . If a collision occurs p Ax changes and p A 2 x = − p A1x . But pB 2 x = − p A 2 x and pB1x = − p A1x , so
pB 2 x = − pB1x . In the center of mass frame the momentum and hence the velocity of each puck keeps the same
magnitude and reverses direction.
⎛ 0.400 kg ⎞
(e) vcm-x = ⎜
⎟ (6.00 m/s) = 4.00 m/s . u A1x = 6.00 m/s − 4.00 m/s = 2.00 m/s .
⎝ 0.600 kg ⎠
u B1x = 0 − 4.00 m/s = −4.00 m/s . u A 2 x = −2.00 m/s and u B 2 x = +4.00 m/s .
v A 2 x = u A 2 x + vcm-x = −2.00 m/s + 4.00 m/s = 2.00 m/s . vB 2 x = u B 2 x + vcm-x = 4.00 m/s + 4.00 m/s = 8.00 m/s .
⎛ 0.400 kg − 0.200 kg ⎞
Eq. 8.24 says v A 2 x = ⎜
⎟ (6.00 m/s) = 2.00 m/s . Eq. 8.25 says
⎝ 0.400 kg + 0.200 kg ⎠
⎛
⎞
2[0.400 kg]
vA2 x = ⎜
⎟ (6.00 m/s) = 8.00 m/s . Our result agrees with Eqs. 8.24 and 8.25.
0.400 kg + 0.200 kg ⎠
⎝ 8-34 8.107. Chapter 8 EVALUATE: Eqs. 8.24 and 8.25 apply only when vB1 = 0 . The result that the velocity of each puck in the center
of mass frame reverses direction and retains the same magnitude applies to all elastic collisions, even when both
are moving initially.
IDENTIFY and SET UP: Apply conservation of energy to find the total energy before and after the collision with
the floor from the initial and final maximum heights.
EXECUTE: (a) Objects stick together says that the relative speed after the collision is zero, so P = 0.
(b) In an elastic collision the relative velocity of the two bodies has the same magnitude before and after the
collision, so P = 1.
(c) Speed of ball just before collision: mgh = 1 mv12 .
2
v1 = 2 gh
2
Speed of ball just after collision: mgH1 = 1 mv2 .
2 v2 = 2 gH1
The second object (the surface) is stationary, so P = v2 / v1 = H1 / h . (d) P = H1 / h implies H1 = hP2 = (1.2 m )( 0.85 ) = 0.87 m .
2 (e) H1 = hP2 .
H 2 = H1P2 = hP4 . H 3 = H 2P2 = ( hP4 ) P2 = hP6 .
Generalize to H n = H n −1P2 = hP2( n −1)P2 = hP2 n .
(f) 8th bounce implies n = 8 .
H 8 = hP16 = 1.2 m ( 0.85 ) = 0.089 m .
16 8.108. EVALUATE: P is a measure of the kinetic energy lost in the collision. The collision here is between a ball and the
earth. Momentum lost by the ball is gained by the earth, but the velocity gained by the earth is very small and can
be taken to be zero.
IDENTIFY: Momentum is conserved in the collision. Conservation of energy says K 2 = K1 + Δ .
!
!
SET UP: For part (b) let v0 be the common speed of each atom before the collision and let V and v3 be the
velocities after the collision of the molecule and the atom that remains. m = 1.67 × 10−27 kg is the mass of one
hydrogen atom.
EXECUTE: (a) In the center of mass frame P x = 0 so P2 x = 0 and vcm2 = 0 . But in this frame the potential energy
1
2
decreases and the kinetic energy increases. This is inconsistent with K 2cm = 1 mtot vcm2 = 0 .
2 (b) Before the collision vcm = 0 . After the collision the molecule and remaining atom move in opposite directions
2
2
and (2m)V = mv3 ; v3 = 2V . Conservation of energy gives 1 (2m)V 2 + 1 mv3 = 3( 1 mv0 ) 2 + Δ . With v3 = 2V this
2
2
2
2
becomes V 2 = 1 v0 +
2 EVALUATE: 8.109. Δ
.V=
3m 1
2 (1.00 × 103 m/s) 2 + 7.23 × 10−19 J
= 1.20 × 104 m/s and v3 = 2V = 2.40 × 104 m/s .
3(1.67 × 10−27 ) 2
K = 3 ( 1 mv0 ) = 2.50 × 10−21 J , which is much less than the binding energy of the molecule. Other
2 initial conditions also lead to molecule formation; the one of zero initial momentum is just particularly simple to
analyze.
IDENTIFY: Apply conservation of energy to the motion of the wagon before the collision. After the collision the
combined object moves with constant speed on the level ground. In the collision the horizontal component of
momentum is conserved.
SET UP: Let the wagon be object A and treat the two people together as object B. Let +x be horizontal and to the
right. Let V be the speed of the combined object after the collision.
EXECUTE: (a) The speed v A1 of the wagon just before the collision is given by conservation of energy applied to
2
the motion of the wagon prior to the collision. U1 = K 2 says mA g ([50 m][sin 6.0°]) = 1 mAv A1 . v A1 = 10.12 m/s .
2 ⎛
⎞
300 kg
P x = P2 x for the collision says mAv A1 = (mA + mB )V and V = ⎜
⎟ (10.12 m/s) = 6.98 m/s .
1
⎝ 300 kg + 75.0 kg + 60.0 kg ⎠
In 5.0 s the wagon travels (6.98 m/s)(5.0 s) = 34.9 m , and the people will have time to jump out of the wagon
before it reaches the edge of the cliff. Momentum, Impulse, and Collisions 8-35 (b) For the wagon, K1 = 1 (300 kg)(10.12 m/s)2 = 1.54 × 104 J . Assume that the two heroes drop from a small
2
height, so their kinetic energy just before the wagon can be neglected compared to K1 of the wagon. 8.110. K 2 = 1 (435 kg)(6.98 m/s) 2 = 1.06 × 104 J . The kinetic energy of the system decreases by K1 − K 2 = 4.8 × 103 J .
2
EVALUATE: The wagon slows down when the two heroes drop into it. The mass that is moving horizontally
increases, so the speed decreases to maintain the same horizontal momentum. In the collision the vertical
momentum is not conserved, because of the net external force due to the ground.
IDENTIFY: Gravity gives a downward external force of magnitude mg. The impulse of this force equals the
change in momentum of the rocket.
SET UP: Let +y be upward. Consider an infinitesimal time interval dt. In Example 8.15, vex = 2400 m/s and
dm
m
= − 0 . In Example 8.16, m = m0 / 4 after t = 90 s .
dt
120 s
EXECUTE: (a) The impulse-momentum theorem gives − mgdt = (m + dm)(v + dv) + (dm)(v − vex ) − mv . This
simplifies to − mgdt = mdv + vex dm and m dv
dm
= −vex
− mg .
dt
dt dv
v dm
= − ex
−g.
dt
m dt
1⎞
v dm
⎛
2
2
(c) At t = 0 , a = − ex
− g = −(2400 m/s) ⎜ −
⎟ − 9.80 m/s = 10.2 m/s .
m0 dt
⎝ 120 s ⎠ (b) a = vex
m
dm − gdt . Integrating gives v − v0 = + vex ln 0 − gt . v0 = 0 and
m
m
v = + (2400 m/s)ln 4 − (9.80 m/s 2 )(90 s) = 2445 m/s .
EVALUATE: Both the initial acceleration in Example 8.15 and the final speed of the rocket in Example 8.16 are
reduced by the presence of gravity.
IDENTIFY and SET UP: Apply Eq. 8.40 to the single-stage rocket and to each stage of the two-stage rocket.
(a) EXECUTE: v − v0 = vex ln ( m0 / m ) ; v0 = 0 so v = vex ln ( m0 / m ) (d) dv = − 8.111. The total initial mass of the rocket is m0 = 12,000 kg + 1000 kg = 13,000 kg. Of this, 9000 kg + 700 kg = 9700 kg
is fuel, so the mass m left after all the fuel is burned is 13,000 kg − 9700 kg = 3300 kg.
v = vex ln (13,000 kg/3300 kg ) = 1.37vex . (b) First stage: v = vex ln ( m0 / m )
m0 = 13,000 kg
The first stage has 9000 kg of fuel, so the mass left after the first stage fuel has burned is
13,000 kg − 9000 kg = 4000 kg.
v = vex ln (13,000 kg/4000 kg ) = 1.18vex . (c) Second stage: m0 = 1000 kg, m = 1000 kg − 700 kg = 300 kg . v = v0 + vex ln ( m0 / m ) = 1.18vex + vex ln (1000 kg/300 kg ) = 2.38vex . (d) v = 7.00 km/s 8.112. vex = v / 2.38 = ( 7.00 km/s ) / 2.38 = 2.94 km/s . q EVALUATE: The two-stage rocket achieves a greater final speed because it jetisons the left-over mass of the first
stage before the second-state fires and this reduces the final m and increases m0 / m.
IDENTIFY: During an interval where the mass is constant the speed of the rocket is constant. During an interval
where the mass is changing at a constant rate, the equations of Section 8.6 apply.
dm
m
SET UP: For 0 ≤ t ≤ 90 s ,
= − 0 . From Example 8.15, vex = 2400 m/s .
dt
120 s
EXECUTE: (a) For t ≤ 0 , v = 0 . For 0 ≤ t ≤ 90 s , Eq. 8.40 says v = (2400 m/s)ln 4 = 3327 m/s . For t > 90 s , v
has the constant value 3327 m/s. The graph of v(t ) is given in Fig. 8.112a.
(b) For 0 ≤ t ≤ 90 s , Eq. 8.39 gives a = − vex dm
2400 m/s
20 m/s 2
⎛ m0 ⎞
=−
−
=
. a = 20 m/s 2 at t = 0
⎜
⎟
m dt
m0 (1 − t /[120 s]) ⎝ 120 s ⎠ 1 − t /[120 s] (as in Example 8.15) and a = 80 m/s 2 at t = 90 s . For t > 90 s , a = 0 . The graph of a (t ) is given in Fig. 8.112b.
(c) The astronaut has the same acceleration as the rocket. This is maximum at t = 90 s and
Fmax = mastronaut amax = (75 kg)(80 m/s 2 ) = 6.0 × 103 N . This is 8.2 times her weight on earth, since amax is 8.2 times g. 8-36 Chapter 8 EVALUATE: The acceleration increases because the mass decreases while the thrust F = −vex dm
remains constant.
dt a (m/s2)
100 v (m/s)
4000 80 3000 60 2000 40 1000 20 0 0 20 40 60 80 100 120 t (s) 0 0 20 (a) 40 60 80 100 120 t (s) (b) Figure 8.112
8.113. dm = ρ dV . dV = Adx . Since the thin rod lies along the x axis, ycm = 0 . The mass of the IDENTIFY and SET UP: rod is given by M = ∫ dm . EXECUTE: (a) xcm =
xcm = 1
M ∫ L
0 xdm = ρ
M L ρ A L2 0 M2 A∫ xdx = . The volume of the rod is AL and M = ρ AL . ρ AL2 L
= . The center of mass of the uniform rod is at its geometrical center, midway between its ends.
2 ρ AL 2 (b) xcm = 1
M ∫ L
0 xdm = 1
M ∫ L
0 xρ Adx = Aα
M ∫ L
0 x 2 dx = L
L
Aα L3
α AL2
. Therefore,
. M = ∫ dm = ∫ ρ Adx = α A∫ xdx =
0
0
2
3M ⎛ Aα L
xcm = ⎜
⎝3
EVALUATE: ⎞ ⎛ 2 ⎞ 2L
=
.
⎟⎜
2⎟
⎠ ⎝ α AL ⎠ 3
When the density increases with x, the center of mass is to the right of the center of the rod.
1
1
2
2
2
IDENTIFY: xcm =
∫ xdm and ycm = M ∫ ydm. At the upper surface of the plate, y + x = a .
M
SET UP: To find xcm , divide the plate into thin strips parallel to the y-axis, as shown in Fig. 8.114a. To find ycm ,
divide the plate into thin strips parallel to the x-axis as shown in Fig. 8.114b. The plate has volume one-half that of
a circular disk, so V = 1 π a 2t and M = 1 ρπ a 2t.
2
2
3 8.114. EXECUTE: In Fig.114a each strip has length y = a 2 − x 2 . xcm =
xcm = ρt M∫ a
−a 1
xdm, where dm = ρ tydx = ρ t a 2 − x 2 dx.
M∫ x a 2 − x 2 dx = 0, since the integrand is an odd function of x. xcm = 0 because of symmetry. In Fig.114b each strip has length 2 x = 2 a 2 − y 2 . ycm = 1
ydm, where dm = 2 ρ txdy = 2 ρ t a 2 − y 2 dy.
M∫ 2ρt a
y a 2 − y 2 dy . The integral can be evaluated using u = a 2 − y 2 , du = −2 ydy . This substitution gives
M ∫0
2 ρ t ⎛ 1 ⎞ 0 1/ 2
2 ρ ta 3 ⎛ 2 ρ ta 3 ⎞⎛ 2 ⎞ 4a
=
=⎜
=
.
⎟⎜
⎜ − ⎟ ∫ a 2 u du =
2⎟
3M
M ⎝ 2⎠
⎝ 3 ⎠ ⎝ ρπ a t ⎠ 3π ycm =
ycm EVALUATE: 4
= 0.424. ycm is less than a/2, as expected, since the plate becomes wider as y decreases.
3π
y y dy y
y
x x x dx 2x
(b) (a) Figure 8.114 Momentum, Impulse, and Collisions 8.115. 8-37 x2 IDENTIFY: The work is related to the force by W = ∫ Fdx . The force the person must apply equals the weight of
x1 the hanging portion. Since the rope is uniform, the center of mass of the hanging portion is at its geometrical
center.
SET UP: Let y be the length of the rope hanging over the edge and use coordinates where the origin is at the edge
of the table and +y is downward. When the rope is pulled onto the table, y goes from l / 4 to zero. A length y of the
rope has mass λ y .
EXECUTE: (a) When a length y hangs over the edge, the person must apply an upward force
0
0
λ gl 2
Fy = − m( y ) g = −λ yg . W = ∫ Fy ( y )dy = −λ g ∫ ydy =
.
l/4
l/4
32
(b) Initially, ycm = l /8 . The work done to raise an object of mass M a distance ycm is W = Mgycm . 8.116. 2
⎛ λl ⎞ ⎛ l ⎞ λ gl
.
W = ⎜ ⎟g⎜ ⎟ =
32
⎝ 4 ⎠ ⎝8⎠
EVALUATE: The answers from methods (a) and (b) agree. The change in gravitational potential energy of the
rope can be calculated by considering all its mass acting at its center of mass, and the work done by the person
equals the increase in gravitational potential energy of the rope.
IDENTIFY: From our analysis of motion with constant acceleration, if v = at and a is constant, then
x − x0 = v0t + 1 at 2 .
2 SET UP: Take v0 = 0 , x0 = 0 and let +x downward.
dv
dv
= a , v = at and x = 1 at 2 . Substituting into xg = x + v 2 gives
2
dt
dt
1
at 2 g = 1 at 2 a + a 2t 2 = 3 a 2t 2 . The nonzero solution is a = g / 3 .
2
2
2 EXECUTE: (a) (b) x = 1 at 2 = 1 gt 2 = 1 (9.80 m/s 2 )(3.00 s)2 = 14.7 m .
2
6
6
(c) m = kx = (2.00 g/m)(14.7 m) = 29.4 g .
EVALUATE: The acceleration is less than g because the small water droplets are initially at rest, before they
adhere to the falling drop. The small droplets are suspended by buoyant forces that we ignore for the raindrops. 9 ROTATION OF RIGID BODIES 9.1. 9.2. IDENTIFY: s = rθ , with θ in radians.
SET UP: π rad = 180° .
s 1.50 m
EXECUTE: (a) θ = =
= 0.600 rad = 34.4°
r 2.50 m
s
14.0 cm
(b) r = =
= 6.27 cm
θ (128°)(π rad /180°)
(c) s = rθ = (1.50 m)(0.700 rad) = 1.05 m
EVALUATE: An angle is the ratio of two lengths and is dimensionless. But, when s = rθ is used, θ must be in
radians. Or, if θ = s / r is used to calculate θ , the calculation gives θ in radians.
IDENTIFY: θ − θ 0 = ωt , since the angular velocity is constant.
SET UP: 1 rpm = (2π / 60) rad/s .
EXECUTE: (a) ω = (1900)(2π rad / 60 s) = 199 rad/s
(b) 35° = (35°)(π /180°) = 0.611 rad . t = In t = EVALUATE: θ − θ 0 0.611 rad
=
= 3.1× 10−3 s
ω
199 rad/s θ − θ0
we must use the same angular measure (radians, degrees or revolutions) for both
ω θ − θ 0 and ω .
9.3. t2
dω z
. Writing Eq.(2.16) in terms of angular quantities gives θ − θ = ∫ ω z dt .
t1
dt
dn
1 n +1
n −1
SET UP:
t
t = nt and ∫ t n dt =
n +1
dt
EXECUTE: (a) A must have units of rad/s and B must have units of rad/s3 .
(b) α z (t ) = 2 Bt = (3.00 rad/s3 )t . (i) For t = 0 , α z = 0 . (ii) For t = 5.00 s , α z = 15.0 rad/s 2 . IDENTIFY: α z (t ) = t2 3
(c) θ 2 − θ1 = ∫ ( A + Bt 2 )dt = A(t2 − t1 ) + 1 B (t2 − t13 ) . For t1 = 0 and t2 = 2.00 s ,
3 t1 9.4. θ 2 − θ1 = (2.75 rad/s)(2.00 s) + 1 (1.50 rad/s3 )(2.00 s)3 = 9.50 rad .
3
EVALUATE: Both α z and ω z are positive and the angular speed is increasing.
Δω z
IDENTIFY: α z = d ω z / dt . α av-z =
.
Δt SET UP: d2
(t ) = 2t
dt dωz
= −2 βt = ( − 1.60 rad s3 )t.
dt
(b) αz (3.0 s) = ( − 1.60 rad s3 )(3.0 s) = −4.80 rad s 2 .
EXECUTE: (a) αz (t ) = ωz (3.0 s) − ωz (0) −2.20 rad s − 5.00 rad s
=
= −2.40 rad s 2. ,
3.0 s
3.0 s
which is half as large (in magnitude) as the acceleration at t = 3.0 s.
α (0) + α z (3.0 s)
EVALUATE: α z (t ) increases linearly with time, so α av-z = z
. α z (0) = 0 .
2
αav-z = 9-1 9-2 9.5. Chapter 9 IDENTIFY and SET UP: Use Eq.(9.3) to calculate the angular velocity and Eq.(9.2) to calculate the average
angular velocity for the specified time interval.
EXECUTE: θ = γ t + β t 3 ; γ = 0.400 rad/s, β = 0.0120 rad/s3
dθ
(a) ω z =
= γ + 3β t 2
dt
(b) At t = 0, ω z = γ = 0.400 rad/s
(c) At t = 5.00 s, ω z = 0.400 rad/s + 3(0.0120 rad/s3 )(5.00 s) 2 = 1.30 rad/s ωav-z = Δθ θ 2 − θ1
=
t2 − t1
Δt For t1 = 0, θ1 = 0.
For θ 2 = 5.00 s, θ 2 = (0.400 rad/s)(5.00 s) + (0.012 rad/s3 )(5.00 s)3 = 3.50 rad
3.50 rad − 0
= 0.700 rad/s.
5.00 s − 0
EVALUATE: The average of the instantaneous angular velocities at the beginning and end of the time interval is
1
2 (0.400 rad/s + 1.30 rad/s) = 0.850 rad/s. This is larger than ωav-z , because ω z (t ) is increasing faster than linearly.
So ωav-z = 9.6. IDENTIFY:
SET UP: dθ
dω z
Δθ
. α z (t ) =
. ωav-z =
.
dt
dt
Δt
ωz = (250 rad s) − (40.0 rad s 2 )t − (4.50 rad s3 )t 2 . αz = −(40.0 rad s 2 ) − (9.00 rad s3 )t . EXECUTE: ω z (t ) = (a) Setting ωz = 0 results in a quadratic in t. The only positive root is t = 4.23 s . (b) At t = 4.23 s , αz = −78.1 rad s 2 .
(c) At t = 4.23 s , θ = 586 rad = 93.3 rev .
(d) At t = 0 , ωz = 250 rad/s .
(e) ωav-z = 586 rad = 138 rad s.
4.23 s
EVALUATE: Between t = 0 and t = 4.23 s , ω z decreases from 250 rad/s to zero. ω z is not linear in t, so ωav-z is not midway between the values of ω z at the beginning and end of the interval.
9.7. IDENTIFY: ω z (t ) = dθ
dω z
. α z (t ) =
. Use the values of θ and ω z at t = 0 and α z at 1.50 s to calculate a, b,
dt
dt and c. dn
t = nt n −1
dt
EXECUTE: (a) ω z (t ) = b − 3ct 2 . α z (t ) = −6ct . At t = 0 , θ = a = π / 4 rad and ω z = b = 2.00 rad/s . At t = 1.50 s ,
SET UP: α z = −6c(1.50 s) = 1.25 rad/s 2 and c = −0.139 rad/s3 .
(b) θ = π / 4 rad and α z = 0 at t = 0 .
(c) α z = 3.50 rad/s 2 at t = − θ= αz
6c =− 3.50 rad/s 2
= 4.20 s . At t = 4.20 s ,
6(−0.139 rad/s3 ) π rad + (2.00 rad/s)(4.20 s) − ( −0.139 rad/s 3 )(4.20 s) 3 = 19.5 rad .
4
ω z = 2.00 rad/s − 3( −0.139 rad/s3 )(4.20 s) 2 = 9.36 rad/s . EVALUATE:
9.8. IDENTIFY: ωav-z = θ , ω z and α z all increase as t increases.
dω
α z = z . θ − θ 0 = ωav-zt . When ω z is linear in t, ωav-z for the time interval t1 to t2 is ω z1 + ω z 2
t2 − t1 dt . SET UP: From the information given, ω z (t ) = −6.00 rad/s + (2.00 rad/s 2 )t
EXECUTE: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative
value to a positive value.
(b) It takes 3.00 seconds for the wheel to stop (ωz = 0) . During this time its speed is decreasing. For the next 4.00 s its speed is increasing from 0 rad s to + 8.00 rad s . Rotation of Rigid Bodies 9-3 −6.00 rad s + 8.00 rad s
= 1.00 rad s . θ − θ 0 = ωav-zt then leads to
2
displacement of 7.00 rad after 7.00 s.
EVALUATE: When α z and ω z have the same sign, the angular speed is increasing; this is the case for t = 3.00 s to
(c) The average angular velocity is 9.9. t = 7.00 s . When α z and ω z have opposite signs, the angular speed is decreasing; this is the case between
t = 0 and t = 3.00 s .
IDENTIFY: Apply the constant angular acceleration equations.
SET UP: Let the direction the wheel is rotating be positive.
EXECUTE: (a) ω z = ω0 z + α z t = 1.50 rad s + (0.300 rad s 2 )(2.50 s) = 2.25 rad s.
(b) θ − θ 0 = ω0 z t + 1 α zt 2 = (1.50 rad/s)(2.50 s) + 1 (0.300 rad/s 2 )(2.50 s) 2 = 4.69 rad .
2
2 ⎛ ω0 z + ω z ⎞ ⎛ 1.50 rad/s + 2.25 rad/s ⎞
⎟t = ⎜
⎟ (2.50 s) = 4.69 rad , the same as calculated with
2
2
⎝
⎠⎝
⎠
another equation in part (b).
IDENTIFY: Apply the constant angular acceleration equations to the motion of the fan.
(a) SET UP: ω0 z = (500 rev/min)(1 min/60 s) = 8.333 rev/s, ω z = ( 200 rev/min)(1 min/60 s) = 3.333 rev/s, θ − θ0 = ⎜ EVALUATE: 9.10. t = 4.00 s, α z = ? ω z = ω0 z + α z t
αz = EXECUTE: ω z − ω0 z
t = 3.333 rev/s − 8.333 rev/s
= −1.25 rev/s 2
4.00 s θ − θ0 = ?
θ − θ 0 = ω0 z t + 1 α zt 2 = (8.333 rev/s)(4.00 s) + 1 ( −1.25 rev/s 2 )(4.00 s) 2 = 23.3 rev
2
2
(b) SET UP: ω z = 0 (comes to rest); ω0 z = 3.333 rev/s; α z = −1.25 rev/s 2 ;
t =?
ω z = ω0 z + α z t t= EXECUTE: ω z − ω0 z 0 − 3.333 rev/s
=
= 2.67 s
αz
−1.25 rev/s 2 EVALUATE: The angular acceleration is negative because the angular velocity is decreasing. The average angular
velocity during the 4.00 s time interval is 350 rev/min and θ − θ 0 = ωav-z t gives θ − θ 0 = 23.3 rev, which checks.
9.11. IDENTIFY:
SET UP: Apply the constant angular acceleration equations to the motion. The target variables are t and θ − θ 0 .
(a) α z = 1.50 rad/s 2 ; ω0 z = 0 (starts from rest); ω z = 36.0 rad/s; t = ? ω z = ω0 z + α z t
EXECUTE: t= ω z − ω0 z 36.0 rad/s − 0
=
= 24.0 s
αz
1.50 rad/s 2 (b) θ − θ 0 = ? θ − θ 0 = ω0 z t + 1 α zt 2 = 0 + 1 (1.50 rad/s 2 )(2.40 s) 2 = 432 rad
2
2
θ − θ 0 = 432 rad(1 rev/2π rad) = 68.8 rev
EVALUATE: We could use θ − θ 0 = 1 (ω z + ω0 z )t to calculate θ − θ 0 = 1 (0 + 36.0 rad/s)(24.0 s) = 432 rad, which
2
2
9.12. checks.
IDENTIFY: In part (b) apply the equation derived in part (a).
SET UP: Let the direction the propeller is rotating be positive.
ω − ω0 z
. Rewriting Eq. (9.11) as θ − θ 0 = t (ω0 z + 1 ω z t ) and
EXECUTE: (a) Solving Eq. (9.7) for t gives t = z
2 αz substituting for t gives
⎛ ω − ω0 z ⎞ ⎛
1
1
⎞1
⎛ ω z + ω0 z ⎞
2
θ − θ0 = ⎜ z
(ω z2 − ω0 z ),
⎟ ω0 z + (ω z − ω0 z ) ⎟ = (ω z − ω0 z ) ⎜
⎟=
αz ⎠⎜
2
αz
2
2α z
⎝
⎠
⎝
⎠
⎝
which when rearranged gives Eq. (9.12).
⎛1⎞2
1
2
2
⎞
2
2
1⎛
(b) α z = 1 ⎜
⎟ ( ω − ω0 z ) = 2 ⎜
⎟ (16.0 rad s ) − (12.0 rad s ) = 8.00 rad/s
2
θ − θ0 ⎠ z
⎝ 7.00 rad ⎠
⎝ ( ) 9-4 Chapter 9 EVALUATE: ⎛ ω + ωz ⎞
We could also use θ − θ 0 = ⎜ 0 z
⎟ t to calculate t = 0.500 s . Then ω z = ω0 z + α zt gives
2
⎝
⎠ α z = 8.00 rad/s 2 , which agrees with our results in part (b).
9.13. IDENTIFY: Use a constant angular acceleration equation and solve for ω0 z .
SET UP: Let the direction of rotation of the flywheel be positive.
θ − θ0 1
60.0 rad 1
− 2αz =
− 2 (2.25 rad/s 2 )(4.00 s) = 10.5 rad/s .
EXECUTE: θ − θ 0 = ω0 z t + 1 α z2 gives ω0 z =
2
4.00 s
t
EVALUATE: At the end of the 4.00 s interval, ω z = ω0 z + α z t = 19.5 rad/s . ⎛ ω0 z + ω z ⎞ ⎛ 10.5 rad/s + 19.5 rad/s ⎞
⎟t = ⎜
⎟ (4.00 s) = 60.0 rad , which checks.
2
2
⎝
⎠⎝
⎠
IDENTIFY: Apply the constant angular acceleration equations.
SET UP: Let the direction of the rotation of the blade be positive. ω0 z = 0 . θ − θ0 = ⎜ 9.14. EXECUTE: ω z = ω0 z + α z gives α z = ω z − ω0 z
t = 140 rad/s − 0
= 23.3 rad/s 2 .
6.00 s ⎛ ω + ω z ⎞ ⎛ 0 + 140 rad/s ⎞
(θ − θ 0 ) = ⎜ 0 z
⎟t = ⎜
⎟ (6.00 s) = 420 rad
2
2
⎝
⎠
⎠
EVALUATE: We could also use θ − θ 0 = ω0 z t + 1 α z t 2 . This equation gives
2 θ − θ 0 = 1 (23.3 rad/s 2 )(6.00 s) 2 = 419 rad , in agreement with the result obtained above.
2
9.15. IDENTIFY: Apply constant angular acceleration equations.
SET UP: Let the direction the flywheel is rotating be positive.
θ − θ 0 = 200 rev, ω0 z = 500 rev min = 8.333 rev s, t = 30.0 s .
EXECUTE: ⎛ ω + ωz ⎞
(a) θ − θ 0 = ⎜ 0 z
⎟ t gives ω z = 5.00 rev s = 300 rpm
2
⎝
⎠ (b) Use the information in part (a) to find α z : ω z = ω0 z + α zt gives α z = −0.1111 rev s 2 . Then ω z = 0, ⎛ ω0 z + ω z ⎞
⎟ t gives
2
⎝
⎠ α z = −0.1111 rev s 2 , ω0 z = 8.333 rev s in ω z = ω0 z + α zt gives t = 75.0 s and θ − θ 0 = ⎜ 9.16. θ − θ 0 = 312 rev .
EVALUATE: The mass and diameter of the flywheel are not used in the calculation.
IDENTIFY: Use the constant angular acceleration equations, applied to the first revolution and to the first two
revolutions.
SET UP: Let the direction the disk is rotating be positive. 1 rev = 2π rad . Let t be the time for the first revolution.
The time for the first two revolutions is t + 0.750 s .
EXECUTE: (a) θ − θ 0 = ω0 zt + 1 α z t 2 applied to the first and to the first two revolutions gives 2π rad = 1 α zt 2 and
2
2
4π rad = 1 α z (t + 0.750 s) 2 . Eliminating α z between these equations gives 4π rad =
2
2t 2 = (t + 0.750 s) 2 . 2t = ± (t + 0.750 s) . The positive root is t = 2π rad
(t + 0.750 s) 2 .
t2 0.750 s
= 1.81 s .
2 −1 (b) 2π rad = 1 α zt 2 and t = 1.81 s gives α z = 3.84 rad/s 2
2
EVALUATE: 9.17. At the start of the second revolution, ω0 z = (3.84 rad/s 2 )(1.81 s) = 6.95 rad/s . The distance the disk rotates in the next 0.750 s is θ − θ 0 = ω0 z t + 1 α zt 2 = (6.95 rad/s)(0.750 s) + 1 (3.84 rad/s 2 )(0.750 s) 2 = 6.29 rad ,
2
2
which is two revolutions.
IDENTIFY: Apply Eq.(9.12) to relate ω z to θ − θ 0 .
SET UP: Establish a proportionality.
EXECUTE: From Eq.(9.12), with ω0 z = 0, the number of revolutions is proportional to the square of the initial
angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.00 rev.
EVALUATE: We don't have enough information to calculate α z ; all we need to know is that it is constant. Rotation of Rigid Bodies 9.18. IDENTIFY: In each case we apply constant acceleration equations to determine θ (t ) and ω z (t ) . SET UP: Let θ 0 = 0 . The following table gives the revolutions and the angle θ (in degrees) through which the
wheel has rotated for each instant in time (in seconds) and each of the three situations: t 0.05
0.10
0.15
0.20
EXECUTE:
EVALUATE: (a)
rev
0.50
1.00
1.50
2.00 (b) θ
180
360
540
720 rev
0.03
0.13
0.28
0.50 (c) θ
11.3
45
101
180 rev
0.44
0.75
0.94
1.00 θ
158
270
338
360 The θ and ω z graphs for each case are given in Figures 9.18 a–c.
The slope of the θ (t ) graph is ω z (t ) and the slope of the ω z (t ) graph is α z (t ) . Figure 9.18 9-5 9-6 9.19. Chapter 9 IDENTIFY: Apply the constant angular acceleration equations separately to the time intervals 0 to 2.00 s and
2.00 s until the wheel stops.
(a) SET UP: Consider the motion from t = 0 to t = 2.00 s:
θ − θ 0 = ?; ω0 z = 24.0 rad/s; α z = 30.0 rad/s 2 ; t = 2.00 s
EXECUTE: θ − θ 0 = ω0 z t + 1 α zt 2 = (24.0 rad/s)(2.00 s) + 1 (30.0 rad/s 2 )(2.00 s) 2
2
2 θ − θ 0 = 48.0 rad + 60.0 rad = 108 rad
Total angular displacement from t = 0 until stops: 108 rad + 432 rad = 540 rad
Note: At t = 2.00 s, ω z = ω0 z + α z t = 24.0 rad/s + (30.0 rad/s 2 )(2.00 s) = 84.0 rad/s; angular speed when breaker trips.
(b) SET UP: Consider the motion from when the circuit breaker trips until the wheel stops. For this calculation let
t = 0 when the breaker trips.
t = ?; θ − θ 0 = 432 rad; ω z = 0; ω0 z = 84.0 rad/s (from part (a))
⎛ ω0 z + ω z ⎞
⎟t
2
⎝
⎠
2(θ − θ 0 )
2(432 rad)
=
= 10.3 s
EXECUTE: t =
ω0 z + ω z 84.0 rad/s + 0
The wheel stops 10.3 s after the breaker trips so 2.00 s + 10.3 s = 12.3 s from the beginning.
(c) SET UP: α z = ?; consider the same motion as in part (b): θ − θ0 = ⎜ ω z = ω0 z + α z t 9.20. ω z − ω0 z 0 − 84.0 rad/s
=
= −8.16 rad/s 2
10.3 s
t
EVALUATE: The angular acceleration is positive while the wheel is speeding up and negative while it is slowing
ω 2 − ω02z 0 − (84.0 rad/s) 2
2
down. We could also use ω z2 = ω0 z + 2α z (θ − θ 0 ) to calculate α z = z
=
= −8.16 rad/s 2 for
2(θ − θ 0 )
2(432 rad)
the acceleration after the breaker trips.
IDENTIFY: The linear distance the elevator travels, its speed and the magnitude of its acceleration are equal to the
tangential displacement, speed and acceleration of a point on the rim of the disk. s = rθ , v = rω and a = rα . In
these equations the angular quantities must be in radians.
SET UP: 1 rev = 2π rad . 1 rpm = 0.1047 rad/s . π rad = 180° . For the disk, r = 1.25 m .
EXECUTE: αz = v 0.250 m/s
=
= 0.200 rad/s = 1.91 rpm .
1.25 m
r
a 1.225 m/s 2
(b) a = 1 g = 1.225 m/s 2 . α = =
= 0.980 rad/s 2 .
8
1.25 m
r
s 3.25 m
(c) s = 3.25 m . θ = =
= 2.60 rad = 149° .
r 1.25 m
EVALUATE: When we use s = rθ , v = rω and atan = rα to solve for θ , ω and α , the results are in rad, rad/s
EXECUTE: 9.21. (a) v = 0.250 m/s so ω = and rad/s 2 .
IDENTIFY: When the angular speed is constant, ω = θ / t . vtan = rω , atan = rα and arad = rω 2 . In these equations
radians must be used for the angular quantities.
SET UP: The radius of the earth is RE = 6.38×106 m and the earth rotates once in 1 day = 86,400 s . The orbit radius
of the earth is 1.50 ×1011 m and the earth completes one orbit in 1 y = 3.156 ×107 s . When ω is constant, ω = θ / t .
EXECUTE: (a) θ = 1 rev = 2π rad in t = 3.156 × 107 s . ω = (b) θ = 1 rev = 2π rad in t = 86,400 s . ω = 2π rad
= 1.99 ×10 −7 rad/s .
3.156 ×107 s 2π rad
= 7.27 ×10−5 rad/s
86,400 s (c) v = rω = (1.50 ×1011 m)(1.99 ×10−7 rad/s) = 2.98 ×104 m/s .
(d) v = rω = (6.38 × 106 m)(7.27 ×10−5 rad/s) = 464 m/s .
(e) arad = rω 2 = (6.38 ×106 m)(7.27 ×10−5 rad/s) 2 = 0.0337 m/s 2 . atan = rα = 0 . α = 0 since the angular velocity is
constant.
EVALUATE: The tangential speeds associated with these motions are large even though the angular speeds are
very small, because the radius for the circular path in each case is quite large. Rotation of Rigid Bodies 9.22. 9.23. 9-7 IDENTIFY: Linear and angular velocities are related by v = rω . Use ω z = ω0 z + α zt to calculate α z .
SET UP: ω = v / r gives ω in rad/s.
1.25 m/s
1.25 m/s
EXECUTE: (a)
= 50.0 rad/s,
= 21.6 rad/s.
−3
25.0 × 10 m
58.0 × 10−3 m
(b) (1.25 m/s) (74.0 min) (60 s/min) = 5.55 km.
(c) α z = 21.55 rad/s − 50.0 rad/s = −6.41 × 10−3 rad/s 2 .
(74.0 min) (60 s/min)
EVALUATE: The width of the tracks is very small, so the total track length on the disc is huge.
IDENTIFY: Use constant acceleration equations to calculate the angular velocity at the end of two revolutions.
v = rω .
SET UP: 2 rev = 4π rad. r = 0.200 m.
EXECUTE: 2
(a) ω z2 = ω0 z + 2α z (θ − θ 0 ). ω z = 2α z (θ − θ 0 ) = 2(3.00 rad/s 2 )(4π rad) = 8.68 rad/s. arad = rω 2 = (0.200 m)(8.68 rad/s) 2 = 15.1 m/s 2 .
v 2 (1.74 m/s) 2
=
= 15.1 m/s 2 .
0.200 m
r
rω 2 and v 2 / r are completely equivalent expressions for arad . (b) v = rω = (0.200 m)(8.68 rad/s) = 1.74 m/s. arad =
EVALUATE:
9.24. IDENTIFY: arad = rω 2 , with ω in rad/s. Solve for ω .
SET UP: 1 rpm = (2π / 60) rad/s
EXECUTE: 9.25. ω= arad
(400,000)(9.80 m/s 2 )
=
= 1.25 ×104 rad/s = 1.20 ×105 rpm
0.0250 m
r EVALUATE: In arad = rω 2 , ω must be in rad/s.
IDENTIFY and SET UP: Use constant acceleration equations to find ω and α after each displacement. The use
Eqs.(9.14) and (9.15) to find the components of the linear acceleration.
EXECUTE: (a) at the start t = 0
flywheel starts from rest so ω z = ω0 z = 0 atan = rα = (0.300 m)(0.600 rad/s 2 ) = 0.180 m/s 2
arad = rω 2 = 0
2
2
a = arad + atan = 0.180 m/s 2 (b) θ − θ 0 = 60° atan = rα = 0.180 m/s 2
Calculate ω:
θ − θ 0 = 60°(π rad/180°) = 1.047 rad; ω0 z = 0; α z = 0.600 rad/s 2 ; ω z = ? ω z2 = ω02z + 2α z (θ − θ 0 )
ω z = 2α z (θ − θ 0 ) = 2(0.600 rad/s 2 )(1.047 rad) = 1.121 rad/s and ω = ω z .
Then arad = rω 2 = (0.300 m)(1.121 rad/s) 2 = 0.377 m/s 2 .
2
2
a = arad + atan = (0.377 m/s 2 ) 2 + (0.180 m/s 2 ) 2 = 0.418 m/s 2 (c) θ − θ 0 = 120° atan = rα = 0.180 m/s 2
Calculate ω:
θ − θ 0 = 120°(π rad/180°) = 2.094 rad; ω0 z = 0; α z = 0.600 rad/s 2 ; ω z = ? ω z2 = ω02z + 2α z (θ − θ 0 )
ω z = 2α z (θ − θ 0 ) = 2(0.600 rad/s 2 )(2.094 rad) = 1.585 rad/s and ω = ω z .
Then arad = rω 2 = (0.300 m)(1.585 rad/s) 2 = 0.754 m/s 2 .
2
2
a = arad + atan = (0.754 m/s 2 )2 + (0.180 m/s 2 ) 2 = 0.775 m/s 2 EVALUATE: α is constant so α tan is constant. ω increases so arad increases. 9-8 9.26. Chapter 9 IDENTIFY: Apply constant angular acceleration equations. v = rω . A point on the rim has both tangential and
radial components of acceleration.
SET UP: atan = rα and arad = rω 2 .
(a) ω z = ω0 z + α z t = 0.250 rev/s + (0.900 rev/s 2 )(0.200 s) = 0.430 rev/s EXECUTE: (Note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to convert to radians).
(b) ωav-z Δt = (0.340 rev s)(0.2 s) = 0.068 rev .
(c) Here, the conversion to radians must be made to use Eq. (9.13), and ⎛ 0.750 m ⎞
v = rω = ⎜
⎟ ( 0.430 rev/s ) ( 2π rad rev ) = 1.01 m s.
2
⎝
⎠
(d) Combining equations (9.14) and (9.15),
2
2
a = arad + atan = (ω 2 r ) 2 + (α r ) 2 .
2 2 a = ⎡((0.430 rev/s)(2π rad/rev)) 2 (0.375 m) ⎤ + ⎡(0.900 rev/s 2 )(2π rad/rev)(0.375 m) ⎤ .
⎣
⎦⎣
⎦
a = 3.46 m s 2 .
9.27. EVALUATE: If the angular acceleration is constant, atan is constant but arad increases as ω increases.
IDENTIFY: Use Eq.(9.15) and solve for r.
SET UP: arad = rω 2 so r = arad / ω 2 , where ω must be in rad/s arad = 3000 g = 3000(9.80 m/s 2 ) = 29,400 m/s 2 EXECUTE: ⎛ 1 min ⎞⎛ 2π rad ⎞
⎟⎜
⎟ = 523.6 rad/s
⎝ 60 s ⎠⎝ 1 rev ⎠ ω = (5000 rev/min) ⎜ 29,400 m/s 2
= 0.107 m.
ω
(523.6 rad/s) 2
EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic.
IDENTIFY: In part (b) apply the result derived in part (a).
SET UP: arad = rω 2 and v = rω ; combine to eliminate r. Then r =
9.28. arad
2 = ⎛v⎞
(a) arad = ω 2 r = ω 2 ⎜ ⎟ = ωv.
⎝ω ⎠ EXECUTE: (b) From the result of part (a), ω =
EVALUATE:
9.29. arad 0.500 m s 2
=
= 0.250 rad s.
2.00 m s
v arad = rω 2 and v = rω both require that ω be in rad/s, so in arad = ωv , ω is in rad/s. IDENTIFY: v = rω and arad = rω 2 = v 2 / r .
SET UP: 2π rad = 1 rev , so π rad/s = 30 rev/min .
EXECUTE: (a) ω r = (1250 rev min ) rad/s
( 30πrev/min )⎛⎜⎝ 12.7 ×210 −3 m⎞
⎟ = 0.831 m s.
⎠ v2
(0.831 m s) 2
=
= 109 m s 2 .
r (12.7 ×10−3 m) 2
EVALUATE: In v = rω , ω must be in rad/s.
IDENTIFY: atan = rα , v = rω and arad = v 2 / r . θ − θ 0 = ωav-zt .
(b) 9.30. SET UP: When α z is constant, ωav-z = ω0 z + ω z
2 . Let the direction the wheel is rotating be positive. atan −10.0 m s 2
=
= −50.0 rad s 2
r
0.200 m
v 50.0 m s
(b) At t = 3.00 s , v = 50.0 m s and ω = =
= 250 rad s and at t = 0,
r 0.200 m
EXECUTE: (a) α = v = 50.0 m s + ( − 10.0 m s 2 )(0 − 3.00 s) = 80.0 m s , ω = 400 rad s.
(c) ωav-z t = (325 rad s)(3.00 s) = 975 rad = 155 rev . Rotation of Rigid Bodies 9-9 50.0 m/s − 1.40 m/s
= 4.86 s
10.0 m/s
after t = 3.00 s , or at t = 7.86 s . (There are many equivalent ways to do this calculation.)
EVALUATE: At t = 0 , arad = rω 2 = 3.20 ×104 m/s 2 . At t = 3.00 s , arad = 1.25 ×104 m/s 2 . For arad = g the wheel
must be rotating more slowly than at 3.00 s so it occurs some time after 3.00 s.
!
!
IDENTIFY and SET UP: Use Eq.(9.15) to relate ω to arad and ∑ F = ma to relate arad to Frad . Use Eq.(9.13) to
(d) v = arad r = (9.80 m/s 2 )(0.200 m) = 1.40 m/s. This speed will be reached at time 9.31. relate ω and v, where v is the tangential speed.
EXECUTE: (a) arad = rω 2 and Frad = marad = mrω 2
2 2 ⎛ ω ⎞ ⎛ 640 rev/min ⎞
=⎜ 2 ⎟ =⎜
⎟ = 2.29
Frad,1 ⎝ ω1 ⎠ ⎝ 423 rev/min ⎠
(b) v = rω
v2 ω2 640 rev/min
=
=
= 1.51
v1 ω1 423 rev/min
(c) v = rω
⎛ 1 min ⎞⎛ 2π rad ⎞
ω = (640 rev/min) ⎜
⎟⎜
⎟ = 67.0 rad/s
⎝ 60 s ⎠⎝ 1 rev ⎠
Then v = rω = (0.235 m)(67.0 rad/s) = 15.7 m/s.
Frad, 2 arad = rω 2 = (0.235 m)(67.0 rad/s) 2 = 1060 m/s 2
arad 1060 m/s 2
=
= 108; a = 108 g
g
9.80 m/s 2
EVALUATE: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert ω to rad/s.
In part (c), v and arad are calculated from ω , and ω must be in rad/s.
9.32. IDENTIFY:
SET UP: v = rω and atan = rα .
The linear acceleration of the bucket equals atan for a point on the rim of the axle. ⎛ 7.5 rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞
(a) v = Rω . 2.00 cm s = R ⎜
⎟⎜
⎟⎜
⎟ gives R = 2.55 cm .
⎝ min ⎠⎝ 60 s ⎠⎝ 1 rev ⎠
D = 2 R = 5.09 cm .
a
0.400 m s 2
(b) atan = Rα . α = tan =
= 15.7 rad s 2 .
0.0255 m
R
EVALUATE: In v = Rω and atan = Rα , ω and α must be in radians.
IDENTIFY: Apply v = rω .
SET UP: Points on the chain all move at the same speed, so rrωr = rf ωf .
EXECUTE: 9.33. vr 5.00 m s
=
= 15.15 rad s.
0.330 m
r
The angular velocity of the front wheel is ωf = 0.600 rev s = 3.77 rad s . rr = rf (ωf ωr ) = 2.99 cm .
EXECUTE: 9.34. The angular velocity of the rear wheel is ωr = EVALUATE: The rear sprocket and wheel have the same angular velocity and the front sprocket and wheel have
the same angular velocity. rω is the same for both, so the rear sprocket has a smaller radius since it has a larger
angular velocity. The speed of a point on the chain is v = rrωr = (2.99 ×10−2 m)(15.15 rad/s) = 0.453 m/s . The linear
speed of the bicycle is 5.50 m/s.
IDENTIFY and SET UP: Use Eq.(9.16). Treat the spheres as point masses and ignore I of the light rods.
EXECUTE: The object is shown in Figure 9.34a.
(a) r = (0.200 m) 2 + (0.200 m) 2 = 0.2828 m
I = ∑ mi ri 2 = 4(0.200 kg)(0.2828 m) 2
I = 0.0640 kg ⋅ m 2 Figure 9.34a 9-10 Chapter 9 (b) The object is shown in Figure 9.34b. r = 0.200 m
I = ∑ mi ri 2 = 4(0.200 kg)(0.200 m) 2
I = 0.0320 kg ⋅ m 2
Figure 9.34b
(c) The object is shown in Figure 9.34c. r = 0.2828 m
I = ∑ mi ri 2 = 2(0.200 kg)(0.2828 m) 2
I = 0.0320 kg ⋅ m 2 Figure 9.34c 9.35. EVALUATE: In general I depends on the axis and our answer for part (a) is larger than for parts (b) and (c). It just
happens that I is the same in parts (b) and (c).
IDENTIFY: Use Table 9.2. The correct expression to use in each case depends on the shape of the object and the
location of the axis.
SET UP: In each case express the mass in kg and the length in m, so the moment of inertia will be in kg ⋅ m 2 .
EXECUTE: (a) (i) I = 1 ML2 = 1 (2.50 kg)(0.750 m) 2 = 0.469 kg ⋅ m 2 .
3
3 1
(ii) I = 12 ML2 = 1 (0.469 kg ⋅ m 2 ) = 0.117 kg ⋅ m 2 . (iii) For a very thin rod, all of the mass is at the axis and I = 0 .
4
2
2
(b) (i) I = 5 MR 2 = 5 (3.00 kg)(0.190 m) 2 = 0.0433 kg ⋅ m 2 .
2
(ii) I = 3 MR 2 = 5 (0.0433 kg ⋅ m 2 ) = 0.0722 kg ⋅ m 2 .
3 (c) (i) I = MR 2 = (8.00 kg)(0.0600 m) 2 = 0.0288 kg ⋅ m 2 . 9.36. (ii) I = 1 MR 2 = 1 (8.00 kg)(0.0600 m) 2 = 0.0144 kg ⋅ m 2 .
2
2
EVALUATE: I depends on how the mass of the object is distributed relative to the axis.
IDENTIFY: Treat each block as a point mass, so for each block I = mr 2 , where r is the distance of the block from
the axis. The total I for the object is the sum of the I for each of its pieces.
SET UP: In part (a) two blocks are a distance L / 2 from the axis and the third block is on the axis. In part (b) two
blocks are a distance L / 4 from the axis and one is a distance 3L / 4 from the axis.
EXECUTE: (a) I = 2m( L / 2) 2 = 1 mL2 .
2
1
11
mL2 (2 + 9) = mL2 .
16
16
EVALUATE: For the same object I is in general different for different axes.
IDENTIFY: I for the object is the sum of the values of I for each part.
SET UP: For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2. For a
point mass, I = mr 2 , where r is the distance of the mass from the axis.
(b) I = 2m( L / 4) 2 + m(3L / 4) 2 = 9.37. 2 EXECUTE: (a) I = I bar + I balls = I= 1
⎛L⎞
M bar L2 + 2mballs ⎜ ⎟ .
12
⎝2⎠ 1
2
2
( 4.00 kg )( 2.00 m ) + 2 ( 0.500 kg )(1.00 m ) = 2.33 kg ⋅ m2
12 1
1
2
2
(b) I = mbar L2 + mball L2 = ( 4.00 kg )( 2.00 m ) + ( 0.500 kg ) ( 2.00 m ) = 7.33 kg ⋅ m 2
3
3
(c) I = 0 because all masses are on the axis.
(d) All the mass is a distance d = 0.500 m from the axis and
I = mbar d 2 + 2mball d 2 = M Total d 2 = (5.00 kg)(0.500 m) 2 = 1.25 kg ⋅ m 2 .
EVALUATE: I for an object depends on the location and direction of the axis. Rotation of Rigid Bodies 9.38. 9-11 IDENTIFY and SET UP: According to Eq.(9.16), I for the entire object equals the sum of I for each piece, the rod
plus the end caps. The object is shown in Figure 9.38.
EXECUTE: I = I rod + 2 I cap 1
1
I = 12 ML + 2(m)( L / 2) 2 = ( 12 M + 1 m ) L2
2
2 Figure 9.38
EVALUATE:
9.39. Table 9.2 was used for I rod and I = mr 2 for the end caps, since they are treated as point particles. IDENTIFY and SET UP: I = ∑ mi ri 2 implies I = I rim + I spokes EXECUTE: I rim = MR 2 = (1.40 kg)(0.300 m) 2 = 0.126 kg ⋅ m 2
Each spoke can be treated as a slender rod with the axis through one end, so
I spokes = 8 ( 1 ML2 ) = 8 (0.280 kg)(0.300 m) 2 = 0.0672 kg ⋅ m 2
3
3 I = I rim + I spokes = 0.126 kg ⋅ m 2 + 0.0672 kg ⋅ m 2 = 0.193 kg ⋅ m 2 9.40. 9.41. EVALUATE: Our result is smaller than mtot R 2 = (3.64 kg)(0.300 m) 2 = 0.328 kg ⋅ m 2 , since the mass of each
spoke is distributed between r = 0 and r = R.
IDENTIFY: Compare this object to a uniform disk of radius R and mass 2M.
SET UP: With an axis perpendicular to the round face of the object at its center, I for a uniform disk is the same
as for a solid cylinder.
EXECUTE: (a) The total I for a disk of mass 2M and radius R, I = 1 (2 M ) R 2 = MR 2 . Each half of the disk has the
2 same I, so for the half-disk, I = 1 MR 2 .
2
(b) The same mass M is distributed the same way as a function of distance from the axis.
(c) The same method as in part (a) says that I for a quarter-disk of radius R and mass M is half that of a half-disk of
radius R and mass 2M, so I = 1 ( 1 [2 M ]R 2 ) = 1 MR 2 .
22
2
EVALUATE: I depends on how the mass of the object is distributed relative to the axis, and this is the same for
any segment of a disk.
IDENTIFY: I for the compound disk is the sum of I of the solid disk and of the ring.
SET UP: For the solid disk, I = 1 md rd2 . For the ring, I r = 1 mr (r12 + r22 ) , where r1 = 50.0 cm, r2 = 70.0 cm . The
2
2
mass of the disk and ring is their area times their area density.
EXECUTE: I = I d + I r .
1
2
Disk: md = (3.00 g cm )π rd2 = 23.56 kg . I d = md rd2 = 2.945 kg ⋅ m 2 .
2
1
2
2
2
Ring: mr = (2.00 g cm )π (r2 − r1 ) = 15.08 kg . I r = mr (r12 + r22 ) = 5.580 kg ⋅ m 2 .
2
2
I = I d + I r = 8.52 kg ⋅ m .
EVALUATE: 9.42. IDENTIFY: Even though mr < md , I r > I d since the mass of the ring is farther from the axis. K = 1 I ω 2 . Use Table 9.2b to calculate I.
2
1
I = 12 ML2 . 1 rpm = 0.1047 rad/s SET UP:
EXECUTE: ⎛ 0.1047 rad/s ⎞
1
(a) I = 12 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 . ω = (2400 rev/min) ⎜
⎟ = 251 rad/s .
⎝ 1 rev/min ⎠ K = 1 I ω 2 = 1 (42.2 kg ⋅ m 2 )(251 rad/s) 2 = 1.33 ×106 J .
2
2
2
2
2
1
1
(b) K1 = 12 M 1L1ω12 , K 2 = 12 M 2 L2ω2 . L1 = L2 and K1 = K 2 , so M 1ω12 = M 2ω2 .
2 ω2 = ω1 9.43. M1
M1
= (2400 rpm)
= 2770 rpm
M2
0.750 M 1 EVALUATE: The rotational kinetic energy is proportional to the square of the angular speed and directly
proportional to the mass of the object.
IDENTIFY: K = 1 I ω 2 . Use Table 9.2 to calculate I.
2
SET UP: 2
I = 5 MR 2 . For the moon, M = 7.35 ×1022 kg and R = 1.74 ×106 m . The moon moves through 1 rev = 2π rad in 27.3 d. 1 d = 8.64 ×104 s . 9-12 Chapter 9 EXECUTE: ω= 2
(a) I = 5 (7.35 ×1022 kg)(1.74 ×106 m) 2 = 8.90 ×1034 kg ⋅ m 2 . 2π rad
= 2.66 ×10−6 rad/s .
(27.3 d)(8.64 ×10 4 s/d) K = 1 I ω 2 = 1 (8.90 ×1034 kg ⋅ m 2 )(2.66 × 10−6 rad/s) 2 = 3.15 ×1023 J .
2
2
3.15 ×1023 J
= 158 years . Considering the expense involved in tapping the moon’s rotational energy, this
5(4.0 ×1020 J)
does not seem like a worthwhile scheme for only 158 years worth of energy.
EVALUATE: The moon has a very large amount of kinetic energy due to its motion. The earth has even more, but
changing the rotation rate of the earth would change the length of a day.
IDENTIFY: K = 1 I ω 2 . Use Table 9.2 to relate I to the mass M of the disk.
2
(b) 9.44. SET UP:
EXECUTE: 45.0 rpm = 4.71 rad/s . For a uniform solid disk, I = 1 MR 2 .
2
(a) I = 2K ω2 = 2(0.250 J)
= 0.0225 kg ⋅ m 2 .
(4.71 rad/s) 2 2 I 2(0.0225 kg ⋅ m 2 )
=
= 0.500 kg .
(0.300 m) 2
R2
EVALUATE: No matter what the shape is, the rotational kinetic energy is proportional to the mass of the object.
IDENTIFY: K = 1 I ω 2 , with ω in rad/s. Solve for I.
2
SET UP: 1 rev/min = (2π / 60) rad/s . ΔK = −500 J
(b) I = 1 MR 2 and M =
2 9.45. EXECUTE: I= 9.46. ωi = 650 rev/min = 68.1 rad/s . ωf = 520 rev/min = 54.5 rad/s . ΔK = K f − K i = 1 I (ωf2 − ωi2 ) and
2 2( ΔK )
2(−500 J)
=
= 0.600 kg ⋅ m 2 .
2
2
ωf − ωi (54.5 rad/s) 2 − (68.1 rad/s) 2 EVALUATE: In K = 1 I ω 2 , ω must be in rad/s.
2
IDENTIFY: The work done on the cylinder equals its gain in kinetic energy.
SET UP: The work done on the cylinder is PL, where L is the length of the rope. K1 = 0 . K 2 = 1 I ω 2 .
2 ⎛ w⎞
I = 1 mr 2 = 1 ⎜ ⎟ r 2 .
2
2
⎝g⎠
1w 2
1 w v2
(40.0 N)(6.00 m s) 2
v , or P =
=
= 14.7 N.
2g
2 g L 2(9.80 m s 2 )(5.00 m)
EVALUATE: The linear speed v of the end of the rope equals the tangential speed of a point on the rim of the
cylinder. When K is expressed in terms of v, the radius r of the cylinder doesn't appear.
IDENTIFY and SET UP: Combine Eqs.(9.17) and (9.15) to solve for K. Use Table 9.2 to get I.
EXECUTE: K = 1 I ω 2
2
EXECUTE: 9.47. PL = arad = Rω 2 , so ω = arad / R = (3500 m/s 2 ) /1.20 m = 54.0 rad/s
For a disk, I = 1 MR 2 = 1 (70.0 kg)(1.20 m) 2 = 50.4 kg ⋅ m 2
2
2
Thus K = 1 I ω 2 = 1 (50.4 kg ⋅ m 2 )(54.0 rad/s) 2 = 7.35 ×104 J
2
2
9.48. EVALUATE: The limit on arad limits ω which in turn limits K.
IDENTIFY: Repeat the calculation in Example 9.9, but with a different expression for I.
SET UP: For the solid cylinder in Example 9.9, I = 1 MR 2 . For the thin-walled, hollow cylinder, I = MR 2 .
2
EXECUTE: (a) With I = MR 2 , the expression for v is v = 2 gh
.
1+ M m (b) This expression is smaller than that for the solid cylinder; more of the cylinder’s mass is concentrated at its
edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is
converted to the kinetic energy of the cylinder, and so less is available for the falling mass.
EVALUATE: When M is much larger than m, v is very small. When M is much less than m, v becomes v = 2 gh ,
the same as for a mass that falls freely from a height h. Rotation of Rigid Bodies 9.49. 9-13 IDENTIFY: Apply conservation of energy to the system of stone plus pulley. v = rω relates the motion of the
stone to the rotation of the pulley.
SET UP: For a uniform solid disk, I = 1 MR 2 . Let point 1 be when the stone is at its initial position and point 2 be
2
when it has descended the desired distance. Let + y be upward and take y = 0 at the initial position of the stone, so y1 = 0 and y2 = − h , where h is the distance the stone descends.
(a) K p = 1 I pω 2 . I p = 1 M p R 2 = 1 (2.50 kg)(0.200 m) 2 = 0.0500 kg ⋅ m 2 .
2
2
2 EXECUTE: ω= 2Kp Ip 2(4.50 J)
= 13.4 rad/s . The stone has speed v = Rω = (0.200 m)(13.4 rad/s) = 2.68 m/s . The
0.0500 kg ⋅ m 2 = stone has kinetic energy K s = 1 mv 2 = 1 (1.50 kg)(2.68 m/s) 2 = 5.39 J . K1 + U1 = K 2 + U 2 gives 0 = K 2 + U 2 .
2
2
0 = 4.50 J + 5.39 J + mg (− h) . h = 4.50 J
= 45.5% .
K tot 9.89 J
EVALUATE: The gravitational potential energy of the pulley doesn’t change as it rotates. The tension in the wire
does positive work on the pulley and negative work of the same magnitude on the stone, so no net work on the
system.
IDENTIFY: K p = 1 I ω 2 for the pulley and K b = 1 mv 2 for the bucket. The speed of the bucket and the rotational
2
2
(b) K tot = K p + K s = 9.89 J . 9.50. Kp 9.89 J
= 0.673 m .
(1.50 kg)(9.80 m/s 2 ) = speed of the pulley are related by v = Rω .
SET UP: K p = 1 K b
2 9.51. EXECUTE: 1 I ω 2 = 1 ( 1 mv 2 ) = 1 mR 2ω 2 . I = 1 mR 2 .
2
22
4
2
EVALUATE: The result is independent of the rotational speed of the pulley and the linear speed of the mass.
IDENTIFY: The general expression for I is Eq.(9.16). K = 1 I ω 2 .
2
SET UP: R will be multiplied by f.
EXECUTE: (a) In the expression of Eq. (9.16), each term will have the mass multiplied by f 3 and the distance multiplied by f , and so the moment of inertia is multiplied by f 3 ( f ) 2 = f 5 . 9.52. (b) (2.5 J)(48)5 = 6.37 × 108 J.
EVALUATE: Mass and volume are proportional to each other so both scale by the same factor.
IDENTIFY: The work the person does is the negative of the work done by gravity. Wgrav = U grav,1 − U grav,2 . U grav = Mgycm .
SET UP: The center of mass of the ladder is at its center, 1.00 m from each end.
ycm,1 = (1.00 m)sin 53.0° = 0.799 m . ycm,2 = 1.00 m .
EXECUTE: 9.53. Wgrav = (9.00 kg)(9.80 m/s 2 )(0.799 m − 1.00 m) = −17.7 J . The work done by the person is 17.7 J. EVALUATE: The gravity force is downward and the center of mass of the ladder moves upward, so gravity does
negative work. The person pushes upward and does positive work.
IDENTIFY: U = Mgycm . ΔU = U 2 − U1 .
SET UP: Half the rope has mass 1.50 kg and length 12.0 m. Let y = 0 at the top of the cliff and take + y to be upward. The center of mass of the hanging section of rope is at its center and ycm,2 = −6.00 m .
EXECUTE:
9.54. 9.55. ΔU = U 2 − U1 = mg ( ycm,2 − ycm,1 ) = (1.50 kg)(9.80 m/s 2 )(−6.00 m − 0) = −88.2 J . EVALUATE: The potential energy of the rope decreases when part of the rope moves downward.
IDENTIFY: Apply Eq.(9.19), the parallel-axis theorem.
SET UP: The center of mass of the hoop is at its geometrical center.
EXECUTE: In Eq. (9.19), I cm = MR 2 and d = R 2 , so I P = 2 MR 2 .
EVALUATE: I is larger for an axis at the edge than for an axis at the center. Some mass is closer than distance R
from the axis but some is also farther away. Since I for each piece of the hoop is proportional to the square of the
distance from the axis, the increase in distance has a larger effect.
IDENTIFY: Use Eq.(9.19) to relate I for the wood sphere about the desired axis to I for an axis along a diameter.
SET UP: For a thin-walled hollow sphere, axis along a diameter, I = 2 MR 2 .
3
2
For a solid sphere with mass M and radius R, I cm = 5 MR 2 , for an axis along a diameter. 9-14 Chapter 9 EXECUTE:
2
3 Find d such that I P = I cm + Md 2 with I P = 2 MR 2 :
3 2
MR 2 = 5 MR 2 + Md 2 The factors of M divide out and the equation becomes 2
( 2 − 5 ) R2 = d 2
3 d = (10 − 6) /15R = 2 R / 15 = 0.516 R.
The axis is parallel to a diameter and is 0.516R from the center.
EVALUATE: I cm (lead) > I cm (wood) even though M and R are the same since for a hollow sphere all the mass is a
9.56. distance R from the axis. Eq.(9.19) says I P > I cm , so there must be a d where I P (wood) = I cm (lead).
IDENTIFY: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its
end and perpendicular to the rod.
1
SET UP: The center of mass of the rod is at its center, and I cm = 12 ML2 .
2 M2
M2
⎛L⎞
L +M⎜ ⎟ =
L.
12
3
⎝2⎠
EVALUATE: I is larger when the axis is not at the center of mass.
IDENTIFY and SET UP: Use Eq.(9.19). The cm of the sheet is at its geometrical center. The object is sketched in
Figure 9.57.
EXECUTE: I P = I cm + Md 2 .
EXECUTE: 9.57. I p = I cm + Md 2 = From part (c) of Table 9.2,
1
I cm = 12 M (a 2 + b 2 ).
The distance d of P from
the cm is d = (a / 2) 2 + (b / 2) 2 .
Figure 9.57
1
1
Thus I P = I cm + Md 2 = 12 M (a 2 + b 2 ) + M ( 1 a 2 + 1 b 2 ) = ( 12 + 1 ) M (a 2 + b 2 ) =
4
4
4
1
3 9.58. 9.59. M (a 2 + b 2 ) EVALUATE: I P = 4 I cm . For an axis through P mass is farther from the axis.
IDENTIFY: Consider the plate as made of slender rods placed side-by-side.
SET UP: The expression in Table 9.2(a) gives I for a rod and an axis through the center of the rod.
1
EXECUTE: (a) I is the same as for a rod with length a: I = 12 Ma 2 .
1
(b) I is the same as for a rod with length b: I = 12 Mb 2 .
EVALUATE: I is smaller when the axis is through the center of the plate than when it is along one edge.
IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis
theorem says I p = I cm + Md 2 . SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length L / 2 and mass M / 2 .
EXECUTE: (a) For each segment the moment of inertia is for a rod with mass M / 2 , length L / 2 and the axis
2 1 ⎛ M ⎞⎛ L ⎞
1
1
ML2 . For the rod, I a = 2 I s = ML2 .
through one end. For one segment, I s = ⎜ ⎟⎜ ⎟ =
3 ⎝ 2 ⎠⎝ 2 ⎠ 24
12
(b) The center of mass of each segment is at the center of the segment, a distance of L / 4 from each end. For each
2 segment, I cm = 1 ⎛ M ⎞⎛ L ⎞
1
2
⎜ ⎟⎜ ⎟ = ML . Axis b is a distance L / 4 from the cm of each segment, so for each
12 ⎝ 2 ⎠⎝ 2 ⎠ 96 segment the parallel axis theorem gives I for axis b to be I s = 1
M
ML2 +
96
2 2 1
1
⎛L⎞
2
2
⎜ ⎟ = ML and I b = 2 I s = ML .
4 ⎠ 24
12
⎝ Rotation of Rigid Bodies EVALUATE: 9-15 I for these two axes are the same. Figure 9.59
9.60. 9.61. 9.62. Apply the parallel-axis theorem.
M2
L and d = ( L 2 − h ) .
SET UP: In Eq .( 9.19 ) , I cm =
12
2
⎡1
1
⎛L
⎞⎤
⎡1
⎤
⎡1
⎤
EXECUTE: I P = M ⎢ L2 + ⎜ − h ⎟ ⎥ = M ⎢ L2 + L2 − Lh + h 2 ⎥ = M ⎢ L2 − Lh + h 2 ⎥ ,
4
⎝2
⎠⎥
⎣12
⎦
⎣3
⎦
⎢12
⎣
⎦
which is the same as found in Example 9.11.
EVALUATE: Example 9.11 shows that this result gives the expected result for h = 0 , h = L and h = L / 2 .
IDENTIFY: Apply Eq.(9.20).
SET UP: dm = ρ dV = ρ (2π rL dr ) , where L is the thickness of the disk. M = π Lρ R 2 .
EXECUTE: The analysis is identical to that of Example 9.12, with the lower limit in the integral being zero and
the upper limit being R. The result is I = 1 MR 2 .
2
EVALUATE: Our result agrees with Table 9.2(f).
IDENTIFY: Eq.(9.20), I = ∫ r 2 dm
IDENTIFY: SET UP: Figure 9.62 Take the x-axis to lie along the rod, with the origin at the left end. Consider a thin slice at coordinate x and width
dx, as shown in Figure 9.62. The mass per unit length for this rod is M / L, so the mass of this slice is
dm = ( M / L) dx.
EXECUTE:
9.63. L L 0 0 I = ∫ x 2 ( M / L) dx = ( M / L) ∫ x 2 dx = ( M / L)( L3 / 3) = 1 ML2
3 EVALUATE: This result agrees with Table 9.2.
IDENTIFY: Apply Eq.(9.20).
SET UP: For this case, dm = γ dx.
L EXECUTE: (a) M = ∫ dm = ∫ γ x dx = γ
0 L (b) I = ∫ x 2 (γ x)dx = γ
0 4L x
4 0 = γ L4
4 x2
2 L =
0 γ L2
2 = M L2 . This is larger than the moment of inertia of a uniform rod of the same
2 mass and length, since the mass density is greater further away from the axis than nearer the axis.
L 9.64. L
L
⎛ x2
x3 x 4 ⎞
L4 M 2
=
L.
(c) I = ∫ (L − x) 2γ xdx = γ ∫ ( L2 x − 2 Lx 2 + x 3 )dx = γ ⎜ L2 − 2 L + ⎟ = γ
3 4 ⎠0
12 6
⎝2
0
0
This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end.
EVALUATE: For a uniform rod with an axis at one end, I = 1 ML2 . The result in (b) is larger than this and the
3
result in (a) is smaller than this.
!
!
IDENTIFY: We know that v = rω and v is tangential. We know that arad = rω 2 and arad is in toward the center of
the wheel. See if the vector product expressions give these results.
!
!
SET UP: A × B = AB sin φ , where φ is the angle between A and B .
!
EXECUTE: (a) For a counterclockwise rotation, ω will be out of the page. 9-16 9.65. Chapter 9 !
!
(b) The upward direction crossed into the radial direction is, by the right-hand rule, counterclockwise. ω and r are
!!
perpendicular, so the magnitude of ω × r is ω r = v .
!
!
!!
(c) ω is perpendicular to v and so ω × v has magnitude ωv = arad, and from the right-hand rule, the upward
!
direction crossed into the counterclockwise direction is inward, the direction of arad .
!
!
!
EVALUATE: If the wheel rotates clockwise, the directions of ω and v are reversed, but arad is still inward.
IDENTIFY: Apply θ = ωt .
SET UP: For alignment, the earth must move through 60° more than Mars, in the same time t. ωe = 360° / yr . ωM = 360° /(1.9 yr) .
EXECUTE: θ e = θ M + 60° . ωet = ωMt + 60° .
60°
60°
60°
(1/[0.9yr /1.9 yr 2 ]) = 0.352 yr = 128 days .
=
=
360° 360° 360°
ωe − ω M
−
1 yr 1.9 yr
EVALUATE: Earth has a larger angular velocity than Mars, and completes one orbit in less time.
IDENTIFY and SET UP: Use Eqs.(9.3) and (9.5). As long as α z > 0, ω z increases. At the t when α z = 0, ω z is at
t= 9.66. its maximum positive value and then starts to decrease when α z becomes negative. θ (t ) = γ t 2 − β t 3 ; γ = 3.20 rad/s 2 , β = 0.500 rad/s3
dθ d (γ t 2 − β t 3 )
=
= 2γ t − 3β t 2
dt
dt
d ω z d (2γ t − 3β t 2 )
(b) α z (t ) =
=
= 2γ − 6 β t
dt
dt
(c) The maximum angular velocity occurs when α z = 0.
EXECUTE: (a) ω z (t ) = 2γ − 6 β t = 0 implies t = 9.67. 2γ
γ
3.20 rad/s 2
=
=
= 2.133 s
6 β 3β 3(0.500 rad/s3 ) At this t, ω z = 2γ t − 3β t 2 = 2(3.20 rad/s 2 )(2.133 s) − 3(0.500 rad/s3 )(2.133 s) 2 = 6.83 rad/s
The maximum positive angular velocity is 6.83 rad/s and it occurs at 2.13 s.
EVALUATE: For large t both ω z and α z are negative and ω z increases in magnitude. In fact, ω z → −∞ at
t → ∞. So the answer in (c) is not the largest angular speed, just the largest positive angular velocity.
IDENTIFY: The angular acceleration α of the disk is related to the linear acceleration a of the ball by a = Rα .
t t 0 0 Since the acceleration is not constant, use ω z − ω0 z = ∫ α z dt and θ − θ 0 = ∫ ω z dt to relate θ , ω z , α z and t for the
disk. ω0 z = 0 .
1 ∫ t dt = n + 1 t
n SET UP: n +1 . In a = Rα , α is in rad/s 2 . a 1.80 m/s 2
=
= 0.600 m/s3
t
3.00 s
a (0.600 m/s3 )t
(b) α = =
= (2.40 rad/s3 )t
R
0.250 m
EXECUTE: (a) A = t (c) ω z = ∫ (2.40 rad/s3 )tdt = (1.20 rad/s3 )t 2 . ω z = 15.0 rad/s for t =
0 t t 0 15.0 rad/s
= 3.54 s .
1.20 rad/s3 0 (d) θ − θ 0 = ∫ ω z dt = ∫ (1.20 rad/s3 )t 2 dt = (0.400 rad/s3 )t 3 . For t = 3.54 s , θ − θ 0 = 17.7 rad . 9.68. EVALUATE: If the disk had turned at a constant angular velocity of 15.0 rad/s for 3.54 s it would have turned
through an angle of 53.1 rad in 3.54 s. It actually turns through less than half this because the angular velocity is
increasing in time and is less than 15.0 rad/s at all but the end of the interval.
IDENTIFY and SET UP: The translational kinetic energy is K = 1 mv 2 and the kinetic energy of the rotating
2 flywheel is K = 1 I ω 2 . Use the scale speed to calculate the actual speed v. From that calculate K for the car and
2
then solve for ω that gives this K for the flywheel. Rotation of Rigid Bodies EXECUTE: (a) vtoy
vscale = 9-17 Ltoy
Lreal ⎛L ⎞
⎛ 0.150 m ⎞
vtoy = vscale ⎜ toy ⎟ = (700 km/h) ⎜
⎟ = 35.0 km/h
⎝ 3.0 m ⎠
⎝ Lreal ⎠
vtoy = (35.0 km/h)(1000 m/1 km)(1 h/3600 s) = 9.72 m/s
(b) K = 1 mv 2 = 1 (0.180 kg)(9.72 m/s)2 = 8.50 J
2
2
(c) K = 1 I ω 2 gives that ω =
2 9.69. 2K
2(8.50 J)
=
= 652 rad/s
I
4.00 ×10−5 kg ⋅ m 2 EVALUATE: K = 1 I ω 2 gives ω in rad/s. 652 rad/s = 6200 rev/min so the rotation rate of the flywheel is very
2
large.
!
!
IDENTIFY: atan = rα , arad = rω 2 . Apply the constant acceleration equations and ∑ F = ma .
!
2
2
SET UP: atan and arad are perpendicular components of a , so a = arad + atan . atan 3.00 m s 2
=
= 0.050 rad s 2
r
60.0 m
(b) α t = (0.05 rad s 2 )(6.00 s) = 0.300 rad s.
EXECUTE: (a) α = (c) arad = ω 2 r = (0.300 rad s) 2 (60.0 m) = 5.40 m s 2 .
(d) The sketch is given in Figure 9.69.
2 (e) a = a 2 rad + a 2 tan = (5.40 m s ) 2 + (3.00 m s 2 ) 2 = 6.18 m s 2 , and the magnitude of the force is F = ma = (1240 kg)(6.18 m s 2 ) = 7.66 kN.
⎛a ⎞
⎛ 5.40 ⎞
(f) arctan ⎜ rad ⎟ = arctan ⎜
⎟ = 60.9°.
atan ⎠
⎝ 3.00 ⎠
⎝
!
!
EVALUATE: atan is constant and arad increases as ω increases. At t = 0 , a is parallel to v . As t increases,
!
!
a moves toward the radial direction and the angle between a increases toward 90° . Figure 9.69
9.70. IDENTIFY: Apply conservation of energy to the system of drum plus falling mass, and compare the results for
earth and for Mars.
SET UP: K drum = 1 I ω 2 . K mass = 1 mv 2 . v = Rω so if K drum is the same, ω is the same and v is the same on both
2
2 planets. Therefore, K mass is the same. Let y = 0 at the initial height of the mass and take + y upward.
Configuration 1 is when the mass is at its initial position and 2 is when the mass has descended 5.00 m, so
y1 = 0 and y2 = − h , where h is the height the mass descends.
EXECUTE: (a) K1 + U1 = K 2 + U 2 gives 0 = K drum + K mass − mgh . K drum + K mass are the same on both planets, so ⎛g ⎞
⎛ 9.80 m/s 2 ⎞
mg E hE = mg M hM . hM = hE ⎜ E ⎟ = (5.00 m) ⎜
= 13.2 m .
2⎟
⎝ 3.71 m/s ⎠
⎝ gM ⎠ 9-18 Chapter 9 (b) mg M hM = K drum + K mass . v = 2 g M hM − 9.71. 1
2 mv 2 = mg M hM − K drum and 2 K drum
2(250.0 J)
= 2(3.71 m/s 2 )(13.2 m) −
= 8.04 m/s
m
15.0 kg EVALUATE: We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius
of the drum.
IDENTIFY and SET UP: All points on the belt move with the same speed. Since the belt doesn’t slip, the speed of
the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are
related to the angular speed of each circular object by v = rω .
EXECUTE: Figure 9.71
(a) v1 = r1ω1 ω1 = (60.0 rev/s)(2π rad/1 rev) = 377 rad/s
v1 = r1ω1 = (0.45 ×10−2 m)(377 rad/s) = 1.70 m/s
(b) v1 = v2 r1ω1 = r2ω2
ω2 = ( r1 / r2 )ω1 = (0.45 cm/2.00 cm)(377 rad/s) = 84.8 rad/s 9.72. EVALUATE: The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for
points on the rim.
IDENTIFY: The speed of all points on the belt is the same, so r1ω1 = r2ω2 applies to the two pulleys.
SET UP: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is
the angular velocity of the saw blade. π rad/s = 30 rev/min . ⎛ π rad s ⎞ ⎛ 0.208 m ⎞
(a) v2 = (2(3450 rev min)) ⎜
⎟⎜
⎟ = 75.1 m s.
2
⎠
⎝ 30 rev min ⎠ ⎝ EXECUTE: 2 ⎛
⎛π
rad s ⎞ ⎞ ⎛ 0.208 m ⎞
4
2
(b) arad = ω 2 r = ⎜ 2(3450 rev min) ⎜
⎜
⎟⎟ ⎜
⎟ = 5.43 ×10 m s ,
30 rev min ⎠ ⎟ ⎝
2
⎠
⎝
⎝
⎠
so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity.
EVALUATE: In v = rω and arad = rω 2 , ω must be in rad/s.
9.73. IDENTIFY and SET UP: Use Eq.(9.15) to relate arad to ω and then use a constant acceleration equation to
replace ω.
2
EXECUTE: (a) arad = rω 2 , arad,1 = rω12 , arad, 2 = rω2
2
Δarad = arad, 2 − arad,1 = r (ω2 − ω12 ) One of the constant acceleration equations can be written
2
2
ω2 z = ω12z + 2α (θ 2 − θ1 ), or ω2 z − ω12z = 2α z (θ 2 − θ1 )
Thus Δarad = r 2α z (θ 2 − θ1 ) = 2rα z (θ 2 − θ1 ), as was to be shown.
(b) α z = Δarad
85.0 m/s 2 − 25.0 m/s 2
=
= 8.00 rad/s 2
2r (θ 2 − θ1 ) 2(0.250 m)(15.0 rad) Then atan = rα = (0.250 m)(8.00 rad/s 2 ) = 2.00 m/s 2 ω 2 is proportional to α z and (θ − θ 0 ) so arad is also proportional to these quantities. arad increases
while r stays fixed, ω z increases, and α z is positive.
IDENTIFY and SET UP: Use Eq.(9.17) to relate K and ω and then use a constant acceleration equation to replace ω .
2
EXECUTE: (c) K = 1 I ω 2 ; K 2 = 1 I ω2 , K1 = 1 I ω12
2
2
2
EVALUATE: 2
ΔK = K 2 − K1 = 1 I (ω2 − ω12 ) = 1 I (2α z (θ 2 − θ1 )) = Iα z (θ 2 − θ1 ), as was to be shown.
2
2 (d) I = ΔK = 45.0 J − 20.0 J = 0.208 kg ⋅ m 2 α z (θ 2 − θ1 ) (8.00 rad/s 2 )(15.0 rad)
EVALUATE: α z is positive, ω increases, and K increases. Rotation of Rigid Bodies 9.74. IDENTIFY:
SET UP: 9-19 I = I wood + I lead . m = ρV , where ρ is the volume density and m = σ A , where σ is the area density.
2
2
4
For a solid sphere, I = 5 mR 2 . For the hollow sphere (foil), I = 3 mR 2 . For a sphere, V = 3 π R 3 and 4
A = 4π R 2 . mw = ρ wVw = ρ w π R 3 . mL = σ L AL = σ L 4π R 2 .
3
2
2
2⎛ 4
2
8
⎞
⎛ρ R
⎞
2
EXECUTE: I = mw R + mL R 2 = ⎜ ρ w π R 3 ⎟ R 2 + (σ L 4π R 2 )R 2 = π R 4 ⎜ w + σ L ⎟ .
5
3
5⎝ 3
3
3
⎠
⎝5
⎠
I= 9.75. 9.76. ⎡ (800 kg m3 )(0.20 m)
⎤
8π
(0.20 m) 4 ⎢
+ 20 kg m 2 ⎥ = 0.70 kg ⋅ m 2 .
3
5
⎣
⎦ EVALUATE: mW = 26.8 kg and I W = 0.429 kg ⋅ m 2 . mL = 10.1 kg and I L = 0.268 kg ⋅ m 2 . Even though the foil is
only 27% of the total mass its contribution to I is about 38% of the total.
IDENTIFY: Estimate the shape and dimensions of your body and apply the approximate expression from
Table 9.2.
SET UP: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and diameter 0.30 m
(radius 0.15 m)
1
1
EXECUTE: I = mR 2 = (80 kg) (0.15 m) 2 = 0.9 kg ⋅ m 2
2
2
EVALUATE: I depends on your mass and width but not on your height.
IDENTIFY: Treat the V like two thin 0.160 kg bars, each 25 cm long.
SET UP: For a slender bar with the axis at one end, I = 1 mL2 .
3 ⎛1
⎞
⎛1⎞
I = 2 ⎜ mL2 ⎟ = 2 ⎜ ⎟ (0.160 kg)(0.250 m) 2 = 6.67 ×10−3 kg ⋅ m 2
3
⎝
⎠
⎝3⎠
EVALUATE: The value of I is independent of the angle between the two sides of the V; the angle 70.0° didn't
enter into the calculation.
IDENTIFY: K = 1 I ω 2 . arad = rω 2 . m = ρV .
2
EXECUTE: 9.77. SET UP: For a disk with the axis at the center, I = 1 mR 2 . V = tπ R 2 , where t = 0.100 m is the thickness of the
2 flywheel. ρ = 7800 kg m 3 is the density of the iron.
EXECUTE: (a) ω = 90.0 rpm = 9.425 rad s . I = 2K ω2 = 2(10.0 × 106 J)
= 2.252 × 105 kg ⋅ m 2 .
(9.425 rad s) 2 1
1
m = ρV = ρπ R 2t . I = mR 2 = ρπ tR 4 . This gives R = (2 I ρπ t )1 4 = 3.68 m and the diameter is 7.36 m.
2
2
(b) arad = Rω 2 = 327 m s 2 9.78. EVALUATE: In K = 1 I ω 2 , ω must be in rad/s. arad is about 33g; the flywheel material must have large cohesive
2
strength to prevent the flywheel from flying apart.
IDENTIFY: K = 1 I ω 2 . To have the same K for any ω the two parts must have the same I. Use Table 9.2 for I.
2
SET UP: 9.79. 2
2
For a solid sphere, I solid = 5 M solid R 2 . For a hollow sphere, I hollow = 3 M hollow R 2 . 3
2
2
EXECUTE: I solid = I hollow gives 5 M solid R 2 = 3 M hollow R 2 and M hollow = 3 M solid = 5 M .
5
EVALUATE: The hollow sphere has less mass since all its mass is distributed farther from the rotation axis.
2π rad
IDENTIFY: K = 1 I ω 2 . ω =
, where T is the period of the motion. For the earth's orbital motion it can be
2
T
treated as a point mass and I = MR 2 .
SET UP: The earth's rotational period is 24 h = 86,164 s . Its orbital period is 1 yr = 3.156 ×107 s . M = 5.97 ×1024 kg . R = 6.38 ×106 m .
EXECUTE: (a) K =
2 2π 2 I 2π 2 (0.3308)(5.97 ×1024 kg)(6.38 × 106 m) 2
=
= 2.14 ×1029 J.
T2
(86,164 s) 2 1 ⎛ 2π R ⎞ 2π 2 (5.97 ×10 24 kg)(1.50 ×1011 m) 2
= 2.66 ×1033 J.
M⎜
⎟=
2 ⎝T⎠
(3.156 ×107 s) 2
(c) Since the Earth’s moment of inertia is less than that of a uniform sphere, more of the Earth’s mass must be
concentrated near its center.
EVALUATE: These kinetic energies are very large, because the mass of the earth is very large.
(b) 9-20 9.80. Chapter 9 IDENTIFY:
SET UP: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR.
1
1
The kinetic energy is K = I ω 2 + mv 2 , with I = 1 mR 2 for the disk. v = Rω .
2
2
2 EXECUTE:
EVALUATE: K= 4g
121
1
4g
.
I ω + m(ω R ) 2 = ( I + mR 2 ) . Using Ι = 1 mR 2 and solving for ω, ω 2 =
and ω =
2
3R
2
2
2
3R The small object has speed v = height h, it would attain a speed
9.81. 2
2 gR . If it was not attached to the disk and was dropped from a
3
2
.
3 2 gR . Being attached to the disk reduces its final speed by a factor of IDENTIFY: Use Eq.(9.20) to calculate I. Then use K = 1 I ω 2 to calculate K.
2
(a) SET UP: The object is sketched in Figure 9.81. Consider a small strip of width dy
and a distance y below the top of the
triangle.
The length of the strip is
x = ( y / h)b. Figure 9.81
EXECUTE: The strip has area x dy and the area of the sign is 1 bh, so the mass of the strip is
2 ⎛ x dy ⎞
⎛ yb ⎞⎛ 2 dy ⎞ ⎛ 2 M ⎞
dm = M ⎜ 1 ⎟ = M ⎜ ⎟⎜
⎟ = ⎜ 2 ⎟ y dy
⎝ h ⎠⎝ bh ⎠ ⎝ h ⎠
⎝ 2 bh ⎠
⎛ 2 Mb 2 ⎞ 3
dI = 1 (dm) x 2 = ⎜
y dy
3
4⎟
⎝ 3h ⎠
h
2 Mb 2 h 3
2 Mb 2 ⎛ 1 4 h ⎞ 1
2
I = ∫ dI =
y dy =
⎜ y 0 ⎟ = Mb
0
3h 4 ∫ 0
3h 4 ⎝ h
⎠6
(b) I = 1 Mb 2 = 2.304 kg ⋅ m 2
6
ω = 2.00 rev/s = 4.00π rad/s
K = 1 I ω 2 = 182 J
2 9.82. EVALUATE: From Table (9.2), if the sign were rectangular, with length b, then I = 1 Mb 2 . Our result is one-half
3
this, since mass is closer to the axis for the triangular than for the rectangular shape.
IDENTIFY: Apply conservation of energy to the system.
SET UP: For the falling mass K = 1 mv 2 . For the wheel K = 1 I ω 2 .
2
2
EXECUTE: ( 8.00 kg )( 5.00 m/s ) = 100 J.
mgh = ( 8.00 kg ) ( 9.8 m/s 2 ) ( 2.00 m ) = 156.8 J . The wheel must (a) The kinetic energy of the falling mass after 2.00 m is K = 1 mv 2 =
2 The change in its potential energy while falling is 1
2 2 have the “missing” 56.8 J in the form of rotational kinetic energy. Since its outer rim is moving at the same speed
v 5.00 m/s
1
as the falling mass, 5.00 m/s , v = rω gives ω = =
= 13.51 rad/s . K = I ω 2 ; therefore
r 0.370 m
2
2 ( 56.8 J )
2K
I= 2 =
= 0.622 kg ⋅ m 2 .
2
ω
(13.51 rad s )
(b) The wheel’s mass is (280 N) (9.8 m s 2 ) = 28.6 kg . The wheel with the largest possible moment of inertia
would have all this mass concentrated in its rim. Its moment of inertia would be
2
I = MR 2 = ( 28.6 kg )( 0.370 m ) = 3.92 kg ⋅ m 2 . The boss’s wheel is physically impossible. Rotation of Rigid Bodies 9.83. 9-21 EVALUATE: If the mass falls from rest in free-fall its speed after it has descended 2.00 m is
v = 2 g (2.00 m) = 6.26 m/s . Its actual speed is less because some of the energy of the system is in the form of
rotational kinetic energy of the wheel.
IDENTIFY: Use conservation of energy. The stick rotates about a fixed axis so K = 1 I ω 2 . Once we have ω use
2
v = rω to calculate v for the end of the stick.
SET UP: The object is sketched in Figure 9.83. Take the origin of coordinates at the lowest point reached by
the stick and take the positive y-direction to be upward. Figure 9.83
EXECUTE: (a) Use Eq.(9.18): U = Mgycm ΔU = U 2 − U1 = Mg ( ycm2 − ycm1 )
The center of mass of the meter stick is at its geometrical center, so
ycm1 = 1.00 m and ycm2 = 0.50 m
Then ΔU = (0.160 kg)(9.80 m/s 2 )(0.50 m − 1.00 m) = −0.784 J
(b) Use conservation of energy: K1 + U1 + Wother = K 2 + U 2 Gravity is the only force that does work on the meter stick, so Wother = 0.
K1 = 0.
Thus K 2 = U1 − U 2 = −ΔU , where ΔU was calculated in part (a).
2
K 2 = 1 I ω2 so
2 1
2 2
I ω2 = −ΔU and ω2 = 2( −ΔU ) / I For stick pivoted about one end, I = 1 ML2 where L = 1.00 m, so
3 ω2 = 6( −ΔU )
6(0.784 J)
=
= 5.42 rad/s
2
ML
(0.160 kg)(1.00 m) 2 (c) v = rω = (1.00 m)(5.42 rad/s) = 5.42 m/s
(d) For a particle in free-fall, with + y upward, v0 y = 0; y − y0 = −1.00 m; a y = −9.80 m/s 2 ; v y = ?
2
2
v y = v0 y + 2a y ( y − y0 ) v y = − 2a y ( y − y0 ) = − 2(−9.80 m/s 2 )( −1.00 m) = −4.43 m/s
EVALUATE: The magnitude of the answer in part (c) is larger. U1,grav is the same for the stick as for a particle 2
falling from a height of 1.00 m. For the stick K = 1 I ω2 =
2 9.84. 1
2 ( 1
3 ML2 ) (v / L) 2 = 1 Mv 2 . For the stick and for the
6 particle, K 2 is the same but the same K gives a larger v for the end of the stick than for the particle. The reason is
that all the other points along the stick are moving slower than the end opposite the axis.
IDENTIFY: Apply conservation of energy to the system of cylinder and rope.
SET UP: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the
rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance
π R below the axle, since the length of the rope is 2π R and half this distance is the position of the center of the
mass. Initially, every part of the rope is moving with speed ω0 R, and when the rope has unwound, and the cylinder
has angular speed ω , the speed of the rope is ω R (the upper end of the rope has the same tangential speed at the
edge of the cylinder). I = (1 2) MR 2 for a uniform cylinder, 9-22 Chapter 9 ⎛M m⎞
⎛M m⎞
2
K1 = K 2 + U 2 . ⎜ + ⎟ R 2ω0 = ⎜ + ⎟ R 2ω 2 − mgπ R. Solving for ω gives
⎝ 4 2⎠
⎝ 4 2⎠ EXECUTE: ω = ω02 + ( 4π mg R ) , and the speed of any part of the rope is
( M + 2m ) EVALUATE: 9.85. v = ω R. 2
When m → 0 , ω → ω0 . When m >> M , ω = ω0 + 2π g
2
and v = v0 + 2π gR . This is the final
R speed when an object with initial speed v0 descends a distance π R .
IDENTIFY: Apply conservation of energy to the system consisting of blocks A and B and the pulley.
SET UP: The system at points 1 and 2 of its motion is sketched in Figure 9.85. Figure 9.85 Use the work-energy relation K1 + U1 + Wother = K 2 + U 2 . Use coordinates where + y is upward and where the origin
is at the position of block B after it has descended. The tension in the rope does positive work on block A and
negative work of the same magnitude on block B, so the net work done by the tension in the rope is zero. Both
blocks have the same speed.
EXECUTE: Gravity does work on block B and kinetic friction does work on block A. Therefore
Wother = W f = − μ k mA gd .
K1 = 0 (system is released from rest)
U1 = mB gyB1 = mB gd ; U 2 = mB gyB 2 = 0
2
2
2
K 2 = 1 mAv2 + 1 mB v2 + 1 I ω2 .
2
2
2 But v(blocks) = Rω (pulley), so ω2 = v2 / R and
2
2
K 2 = 1 ( mA + mB )v2 + 1 I (v2 / R) 2 = 1 (m A + mB + I / R 2 )v2
2
2
2
Putting all this into the work-energy relation gives
2
mB gd − μ k mA gd = 1 (mA + mB + I / R 2 )v2
2
2
(mA + mB + I / R 2 )v2 = 2 gd (mB − μk mA ) v2 = 2 gd ( mB − μ k m A )
mA + mB + I / R 2 EVALUATE: 9.86. If mB >> mA and I / R 2 , then v2 = 2 gd ; block B falls freely. If I is very large, v2 is very small. Must have mB > μ k mA for motion, so the weight of B will be larger than the friction force on A. I / R 2 has units of
mass and is in a sense the “effective mass” of the pulley.
IDENTIFY: Apply conservation of energy to the system of two blocks and the pulley.
SET UP: Let the potential energy of each block be zero at its initial position. The kinetic energy of the system is
the sum of the kinetic energies of each object. v = Rω , where v is the common speed of the blocks and ω is the
angular velocity of the pulley.
EXECUTE: The amount of gravitational potential energy which has become kinetic energy is
K = ( 4.00 kg − 2.00 kg ) ( 9.80 m s 2 ) ( 5.00 m ) = 98.0 J. In terms of the common speed v of the blocks, the kinetic
2 1
1 ⎛v⎞
energy of the system is K = (m1 + m2 )v 2 + I ⎜ ⎟ .
2
2 ⎝R⎠
1⎛
(0.480 kg ⋅ m 2 ) ⎞ 2
98.0 J
K = v 2 ⎜ 4.00 kg + 2.00 kg +
= 2.81 m s.
⎟ = v (12.4 kg). Solving for v gives v =
2⎝
(0.160 m) 2 ⎠
12.4 kg Rotation of Rigid Bodies 9.87. 9-23 EVALUATE: If the pulley is massless, 98.0 J = 1 (4.00 kg + 2.00 kg)v 2 and v = 5.72 m/s . The moment of inertia
2
of the pulley reduces the final speed of the blocks.
IDENTIFY and SET UP: Apply conservation of energy to the motion of the hoop. Use Eq.(9.18) to calculate U grav . Use K = 1 I ω 2 for the kinetic energy of the hoop. Solve for ω . The center of mass of the hoop is at its geometrical
2
center. Take the origin to be at the original location
of the center of the hoop, before it is rotated
to one side, as shown in Figure 9.87. Figure 9.87 ycm1 = R − R cos β = R (1 − cos β )
ycm2 = 0 (at equilibrium position hoop is at original position)
K1 + U1 + Wother = K 2 + U 2 EXECUTE: Wother = 0 (only gravity does work)
2
K1 = 0 (released from rest), K 2 = 1 I ω2
2 For a hoop, I cm = MR 2 , so I = Md 2 + MR 2 with d = R and I = 2 MR 2 , for an axis at the edge. Thus
2
2
K 2 = 1 (2MR 2 )ω2 = MR 2ω2 .
2 U1 = Mgycm1 = MgR(1 − cos β ), U 2 = mgycm2 = 0
Thus K1 + U1 + Wother = K 2 + U 2 gives
2
MgR (1 − cos β ) = MR 2ω2 and ω = g (1 − cos β ) / R EVALUATE:
9.88. IDENTIFY:
SET UP:
EXECUTE: If β = 0, then ω2 = 0. As β increases, ω2 increases. energy
t
2
1
For a solid cylinder, I = 2 MR . 1 rev/min = (2π / 60) rad/s
K = 1 I ω 2 , with ω in rad/s. P =
2 (a) ω = 3000 rev/min = 314 rad/s . I = 1 (1000 kg)(0.900 m) 2 = 405 kg ⋅ m 2
2 K = 1 (405 kg ⋅ m 2 )(314 rad/s) 2 = 2.00 × 107 J .
2
K 2.00 ×107 J
=
= 1.08 ×103 s = 17.9 min .
P 1.86 ×104 W
EVALUATE: In K = 1 I ω 2 , we must use ω in rad/s.
2
(b) t = 9.89. IDENTIFY:
SET UP: I = I1 + I 2 . Apply conservation of energy to the system. The calculation is similar to Example 9.9. ω= v
v
for part (b) and ω =
for part (c).
R1
R2 1
1
1
2
(a) I = M 1R12 + M 2 R2 = ((0.80 kg)(2.50 ×10−2 m) 2 + (1.60 kg)(5.00 × 10−2 m) 2 )
2
2
2
−3
2
I = 2.25 ×10 kg ⋅ m . EXECUTE: (b) The method of Example 9.9 yields v = v= 2 gh
.
1 + ( I mR12 ) 2(9.80 m s 2 )(2.00 m)
= 3.40 m s.
(1 + ((2.25 ×10−3 kg ⋅ m 2 ) (1.50 kg)(0.025 m) 2 )) The same calculation, with R2 instead of R1 gives v = 4.95 m s.
EVALUATE: The final speed of the block is greater when the string is wrapped around the larger disk. v = Rω , so
when R = R2 the factor that relates v to ω is larger. For R = R2 a larger fraction of the total kinetic energy resides 9-24 9.90. Chapter 9 with the block. The total kinetic energy is the same in both cases (equal to mgh), so when R = R2 the kinetic energy
and speed of the block are greater.
IDENTIFY: Apply conservation of energy to the motion of the mass after it hits the ground.
2 gh
SET UP: From Example 9.9, the speed of the mass just before it hits the ground is v =
.
1 + M / 2m
(a) In the case that no energy is lost, the rebound height h′ is related to the speed v by h′ = EXECUTE: v2
, and
2g h
.
1 + M 2m
(b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy
of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not
conserved.
EVALUATE: If m >> M , h′ = h and the mass does rebound to its initial height.
IDENTIFY: Apply conservation of energy to relate the height of the mass to the kinetic energy of the cylinder.
SET UP: First use K (cylinder) = 250 J to find ω for the cylinder and v for the mass. with the form for v given in Example 9.9, h′ = 9.91. I = 1 MR 2 = 1 (10.0 kg)(0.150 m) 2 = 0.1125 kg ⋅ m 2
2
2 EXECUTE: K = 1 I ω 2 so ω = 2 K / I = 66.67 rad/s
2
v = Rω = 10.0 m/s
SET UP: Use conservation of energy K1 + U1 = K 2 + U 2 to solve for the distance the mass descends. Take y = 0
at lowest point of the mass, so y2 = 0 and y1 = h, the distance the mass descends.
K1 = U 2 = 0 so U1 = K 2 . EXECUTE: mgh = 1 mv + 1 I ω 2 , where m = 12.0 kg
2
2
2 For the cylinder, I = 1 MR 2 and ω = v / R, so
2 1
2 I ω 2 = 1 Mv 2 .
4 mgh = 1 mv 2 + 1 Mv 2
2
4
h= v2 ⎛
M⎞
⎜1 +
⎟ = 7.23 m
2 g ⎝ 2m ⎠ EVALUATE: For the cylinder K cyl = 1 I ω 2 = 1 ( 1 MR 2 ) (v / R ) 2 = 1 Mv 2 .
2
22
4 K mass = 1 mv 2 , so K mass = (2m / M ) K cyl = [2(12.0 kg)/10.0 kg](250 J) = 600 J. The mass has 600 J of kinetic energy
2
when the cylinder has 250 J of kinetic energy and at this point the system has total energy 850 J since U 2 = 0.
9.92. Initially the total energy of the system is U1 = mgy1 = mgh = 850 J, so the total energy is shown to be conserved.
IDENTIFY: Energy conservation: Loss of U of box equals gain in K of system. Both the cylinder and pulley have
1
1
1
2
2
2
kinetic energy of the form K = 1 I ω 2 . mbox gh = mbox vbox + I pulleyωpulley + I cylinderωcylinder .
2
2
2
2
v
v
SET UP: ωpulley = Box and ωcylinder = Box .
rp
rcylinder
EXECUTE: vB = 9.93. 1
1 ⎛1
⎞
2
mB gh = mBvB + ⎜ mP rp2 ⎟
2
2 ⎝2
⎠ 2 2 ⎛ vB ⎞ 1 ⎛ 1
1
1
1
2⎞ ⎛v ⎞
2
2
2
⎜ ⎟ + ⎜ mC rC ⎟ ⎜ B ⎟ . mB gh = mBvB + mP vB + mCvB and
⎜ rp ⎟ 2 ⎝ 2
2
4
4
⎠ ⎝ rC ⎠
⎝⎠ mB gh
(3.00 kg)(9.80 m s 2 )(1.50 m)
=
= 3.68 m s .
1
1
1
1.50 kg + 1 (7.00 kg)
2 mB + 4 m p + 4 mC
4 EVALUATE: If the box was disconnected from the rope and dropped from rest, after falling 1.50 m its speed
would be v = 2 g (1.50 m) = 5.42 m/s . Since in the problem some of the energy of the system goes into kinetic
energy of the cylinder and of the pulley, the final speed of the box is less than this.
IDENTIFY: I = I disk − I hole , where I hole is I for the piece punched from the disk. Apply the parallel-axis theorem to
calculate the required moments of inertia.
SET UP: For a uniform disk, I = 1 MR 2 .
2 Rotation of Rigid Bodies EXECUTE: (a) The initial moment of inertia is I 0 = 1 MR 2 . The piece punched has a mass of
2 9-25 M
and a moment
16 of inertia with respect to the axis of the original disk of
2
2
M ⎡1 ⎛ R ⎞ ⎛ R ⎞ ⎤
9
MR 2 .
⎢ ⎜ ⎟ +⎜ ⎟ ⎥ =
16 ⎢ 2 ⎝ 4 ⎠ ⎝ 2 ⎠ ⎥ 512
⎣
⎦ 9.94. 1
9
247
MR 2 =
MR 2 .
The moment of inertia of the remaining piece is then I = MR 2 −
2
512
512
383
(b) I = 1 MR 2 + M ( R / 2) 2 − 1 ( M /16)( R / 4) 2 = 512 MR 2 .
2
2
EVALUATE: For a solid disk and an axis at a distance R / 2 from the disk's center, the parallel-axis theorem gives
I = 1 MR 2 = 3 MR 2 = 384 MR 2 . For both choices of axes the presence of the hole reduces I, but the effect of the hole
2
4
512
is greater in part (a), when it is farther from the axis.
IDENTIFY: In part (a) use the parallel-axis theorem to relate the moment of inertia I cm for an axis through the
center of the sphere to I P , the moment of inertia for an axis at the pivot.
SET UP: I for a uniform solid sphere and the axis through its center is 2
5 MR 2 . I for a slender rod and an axis at one end is 1 mL2 , where m is the mass of the rod and L is its length.
3
EXECUTE: (a) From the parallel-axis theorem, the moment of inertia is I P = (2 5) MR 2 + ML2 , and ⎛ ⎛ 2 ⎞⎛ R ⎞ 2 ⎞
IP
= ⎜1 + ⎜ ⎟⎜ ⎟ ⎟ . If R = (0.05) L, the difference is (2 5)(0.05) 2 = 0.001 = 0.1%.
ML2 ⎜ ⎝ 5 ⎠⎝ L ⎠ ⎟
⎝
⎠
(b) ( I rod ML2 ) = (mrod 3M ), which is 0.33% when mrod = (0.01) M .
9.95. EVALUATE: In both these cases the correction to I = ML2 is very small.
IDENTIFY: Follow the instructions in the problem to derive the perpendicular-axis theorem. Then apply that
result in part (b).
SET UP: I = ∑ mi ri 2 . The moment of inertia for the washer and an axis perpendicular to the plane of the washer
i at its center is
1
12 1
2 2
M ( R12 + R2 ) . In part (b), I for an axis perpendicular to the plane of the square at its center is M ( L2 + L2 ) = 1 ML2 .
6 EXECUTE: (a) With respect to O, ri 2 = xi 2 + yi 2 , and so I O = ∑ mi ri 2 = ∑ mi ( xi 2 + yi 2 ) = ∑ mi xi 2 + ∑ mi yi 2 = I x + I y .
i i i i (b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about
those axes, and the perpendicular-axis theorem predicts that they will sum to the moment of inertia about the
washer axis, which is I = 1 M ( R12 + R2 2 ), and so I x = I y = 1 M ( R12 + R2 2 ).
2
4
(c) I 0 = 1 mL2 . Since I 0 = I x + I y , and I x = I y , both I x and I y must be
6 9.96. 1
12 mL2 . EVALUATE: The result in part (c) says that I is the same for an axis that bisects opposite sides of the square as for
an axis along the diagonal of the square, even though the distribution of mass relative to the two axes is quite
different in these two cases.
IDENTIFY: Apply the parallel-axis theorem to each side of the square.
SET UP: Each side has length a and mass M / 4, and the moment of inertia of each side about an axis
1
1
perpendicular to the side and through its center is 12 1 Ma 2 = 48 Ma 2 .
4
EXECUTE: The moment of inertia of each side about the axis through the center of the square is, from the () 2 2
2
perpendicular axis theorem, Ma + M a = Ma . The total moment of inertia is the sum of the contributions
48
42
12
2
2
from the four sides, or 4 × Ma = Ma .
12
3
EVALUATE: If all the mass of a side were at its center, a distance a / 2 from the axis, we would have
2 ⎛ M ⎞⎛ a ⎞ 1
I = 4 ⎜ ⎟⎜ ⎟ = Ma 2 . If all the mass was divided equally among the four corners of the square, a distance
⎝ 4 ⎠⎝ 2 ⎠ 4
2 ⎛ M ⎞⎛ a ⎞ 1
2
a / 2 from the axis, we would have I = 4 ⎜ ⎟⎜
⎟ = Ma . The actual I is between these two values.
⎝ 4 ⎠⎝ 2 ⎠ 2 9-26 9.97. Chapter 9 IDENTIFY: Use Eq.(9.20) to calculate I.
(a) SET UP: Let L be the length of the cylinder. Divide the cylinder into thin cylindrical shells of inner radius r
and outer radius r + dr. An end view is shown in Figure 9.97. ρ = αr
The mass of the thin cylindrical shell is
dm = ρ dV = ρ (2π r dr ) L = 2πα Lr 2 dr
Figure 9.97
EXECUTE: 2
I = ∫ r 2 dm = 2πα L ∫ r 4 dr = 2πα L ( 1 R 5 ) = 5 πα LR 5
5 R 0 Relate M to α : M = ∫ dm = 2πα L ∫ r 2 dr = 2πα L ( 1 R 3 ) = 2 πα LR 3 , so πα LR 3 = 3M / 2.
3
3
R 0 2
Using this in the above result for I gives I = 5 (3M / 2) R 2 = 3 MR 2 .
5 9.98. (b) EVALUATE: For a cylinder of uniform density I = 1 MR 2 . The answer in (a) is larger than this. Since the
2
density increases with distance from the axis the cylinder in (a) has more mass farther from the axis than for a
cylinder of uniform density.
IDENTIFY: Write K in terms of the period T and take derivatives of both sides of this equation to relate dK / dt to
dT / dt .
2π
SET UP: ω =
and K = 1 I ω 2 . The speed of light is c = 3.00 ×108 m/s .
2
T
2
2π I dK
4π 2 I dT
4π 2 I dT
EXECUTE: (a) K = 2 .
=− 3
. The rate of energy loss is
. Solving for the moment of
T
dt
T dt
T 3 dt
inertia I in terms of the power P, I=
(b) R =
(c) v = PT 3
1
(5 ×1031 W)(0.0331 s)3
1s
=
= 1.09 × 1038 kg ⋅ m 2
4π dT dt
4π 2
4.22 ×10−13 s 5I
5(1.08 ×1038 kg ⋅ m 2 )
=
= 9.9 × 103 m, about 10 km.
2M
2(1.4)(1.99 ×1030 kg) 2π R 2π (9.9 × 103 m)
=
= 1.9 × 106 m s = 6.3 × 10−3 c.
T
(0.0331 s) M
M
=
= 6.9 ×1017 kg m3 , which is much higher than the density of ordinary rock by 14 orders of
V (4π 3) R 3
magnitude, and is comparable to nuclear mass densities.
EVALUATE: I is huge because M is huge. A small rate of change in the period corresponds to a large release of
energy.
IDENTIFY: In part (a), do the calculations as specified in the hint. In part (b) calculate the mass of each shell of
inner radius R1 and outer radius R2 and sum to get the total mass. In part (c) use the expression in part (a) to
calculate I for each shell and sum to get the total I.
4
SET UP: m = ρV . For a solid sphere, V = 3 π R 3 .
EXECUTE: (a) Following the hint, the moment of inertia of a uniform sphere in terms of the mass density is
8
2
I = 5 MR 2 = 15 πρ R 5 , and so the difference in the moments of inertia of two spheres with the same density ρ but
(d) ρ = 9.99. 5
different radii R2 and R1 is I = ρ (8π 15)( R2 − R15 ).
(b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii
that bound the regions and multiplying by 4π 3, gives M = 5.97 ×1024 kg.
(c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii
that bound the regions and multiplying by 8π 15, gives I = 8.02 ×1022 kg ⋅ m 2 = 0.334 MR 2 . EVALUATE: The calculated value of I = 0.334 MR 2 agrees closely with the measured value of 0.3308MR 2 . This
simple model is fairly accurate. Rotation of Rigid Bodies 9.100. IDENTIFY:
SET UP: 9-27 Apply Eq.(9.20)
Let z be the coordinate along the vertical axis. r ( z ) = πρ R 4 R2 z 2
πρ R 4 4
zR
z dz .
. dm = πρ 2 and dI =
h
h
2 h4 πρ R 4 h
1
⎡ z 5 ⎤ = πρ R 4 h . The volume of a right circular cone is
⎣ ⎦ 0 10
2h
10 h
3 ⎛ πρ R 2 h ⎞ 2 3
2
V = 1 π R 2 h, the mass is 1 πρ R 2 h and so I = ⎜
⎟ R = MR .
3
3
10 ⎝ 3 ⎠
10 EXECUTE: 9.101. I = ∫ dI = ∫ h 0 z 4 dz = 4 EVALUATE: For a uniform cylinder of radius R and for an axis through its center, I = 1 MR 2 . I for the cone is
2
less, as expected, since the cone is constructed from a series of parallel discs whose radii decrease from R to zero
along the vertical axis of the cone.
IDENTIFY: Follow the steps outlined in the problem.
SET UP: ω z = dθ / dt . α z = d 2ω z / dt 2 .
EXECUTE: (a) ds = r dθ = r0 dθ + βθ dθ so s (θ ) = r0θ + (b) Setting s = vt = r0θ + θ (t ) = β
2 β
2 θ 2 . θ must be in radians. θ 2 gives a quadratic in θ . The positive solution is 1⎡ 2
r0 + 2 β vt − r0 ⎤ .
⎦
β⎣ (The negative solution would be going backwards, to values of r smaller than r0 .)
(c) Differentiating, ω z (t ) = dω z
β v2
dθ
v
=
, αz =
=−
. The angular acceleration α z is not
32
dt
dt
r02 + 2 β vt
( r02 + 2β vt ) constant.
(d) r0 = 25.0 mm. θ must be measured in radians, so β = (1.55 μ m rev ) (1 rev 2π rad ) = 0.247 μ m rad. Using θ (t ) from part (b), the total angle turned in 74.0 min = 4440 s is
θ= 2
1
⎛
⎞
−7
−3
−3
⎜ 2 ( 2.47 × 10 m/rad ) (1.25 m/s )( 4440 s ) + ( 25.0 × 10 m ) − 25.0 × 10 m ⎟
−7
2.47 × 10 m/rad ⎝
⎠ θ = 1.337 ×105 rad , which is 2.13 ×10 4 rev .
(e) The graphs are sketched in Figure 9.101.
EVALUATE: ω z must decrease as r increases, to keep v = rω constant. For ω z to decrease in time, α z must be
negative. Figure 9.101 10 DYNAMICS OF ROTATIONAL MOTION 10.1. IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in
Fig.(10.4) to calculate the torque direction.
(a) SET UP: Consider Figure 10.1a.
EXECUTE: τ = Fl
l = r sin φ = (4.00 m)sin 90°
l = 4.00 m
τ = (10.0 N)(4.00 m) = 40.0 N ⋅ m
Figure 10.1a !
This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is
directed out of the plane of the figure.
(b) SET UP: Consider Figure 10.1b.
EXECUTE: τ = Fl
l = r sin φ = (4.00 m)sin120°
l = 3.464 m
τ = (10.0 N)(3.464 m) = 34.6 N ⋅ m
Figure 10.1b !
This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is
directed out of the plane of the figure.
(c) SET UP: Consider Figure 10.1c.
EXECUTE: τ = Fl
l = r sin φ = (4.00 m)sin 30°
l = 2.00 m
τ = (10.0 N)(2.00 m) = 20.0 N ⋅ m
Figure 10.1c
!
This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector τ is
directed out of the plane of the figure.
(d) SET UP: Consider Figure 10.1d.
EXECUTE: τ = Fl
l = r sin φ = (2.00 m)sin 60° = 1.732 m
τ = (10.0 N)(1.732 m) = 17.3 N ⋅ m
Figure 10.1d !
This force tends to produce a clockwise rotation about the axis; by the right-hand rule the vector τ is directed into
the plane of the figure.
(e) SET UP: Consider Figure 10.1e.
EXECUTE: τ = Fl
r = 0 so l = 0 and τ = 0
Figure 10.1e 10-1 10-2 Chapter 10 (f) SET UP: Consider Figure 10.1f. EXECUTE: τ = Fl
l = r sin φ , φ = 180°,
so l = 0 and τ = 0 Figure 10.1f 10.2. EVALUATE: The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action of the force
passes through the axis.
IDENTIFY: τ = Fl with l = r sin φ . Add the two torques to calculate the net torque.
SET UP: Let counterclockwise torques be positive.
EXECUTE: τ 1 = − F1l1 = −(8.00 N)(5.00 m) = −40.0 N ⋅ m . τ 2 = + F2l2 = (12.0 N)(2.00 m)sin 30.0° = +12.0 N ⋅ m . ∑τ = τ
10.3. 1 + τ 2 = −28.0 N ⋅ m . The net torque is 28.0 N ⋅ m , clockwise. EVALUATE: Even though F1 < F2 , the magnitude of τ 1 is greater than the magnitude of τ 2 , because F1 has a
larger moment arm.
IDENTIFY and SET UP: Use Eq.(10.2) to calculate the magnitude of each torque and use the right-hand rule
(Fig.10.4) to determine the direction. Consider Figure 10.3 Figure 10.3 Let counterclockwise be the positive sense of rotation.
EXECUTE: r1 = r2 = r3 = (0.090 m) 2 + (0.090 m) 2 = 0.1273 m τ 1 = − F1l1
l1 = r1 sin φ1 = (0.1273 m)sin135° = 0.0900 m
τ 1 = −(18.0 N)(0.0900 m) = −1.62 N ⋅ m
!
τ 1 is directed into paper
τ 2 = + F2l2
l2 = r2 sin φ2 = (0.1273 m)sin135° = 0.0900 m
τ 2 = + (26.0 N)(0.0900 m) = +2.34 N ⋅ m
!
τ 2 is directed out of paper
τ 3 = + F3l3
l3 = r3 sin φ3 = (0.1273 m)sin 90° = 0.1273 m
τ 3 = +(14.0 N)(0.1273 m) = +1.78 N ⋅ m
!
τ 3 is directed out of paper ∑τ = τ
10.4. 1 + τ 2 + τ 3 = −1.62 N ⋅ m + 2.34 N ⋅ m + 1.78 N ⋅ m = 2.50 N ⋅ m EVALUATE: The net torque is positive, which means it tends to produce a counterclockwise rotation; the vector
torque is directed out of the plane of the paper. In summing the torques it is important to include + or − signs to
show direction.
IDENTIFY: Use τ = Fl = rF sin φ to calculate the magnitude of each torque and use the right-hand rule to
determine the direction of each torque. Add the torques to find the net torque. Dynamics of Rotational Motion SET UP: 10-3 Let counterclockwise torques be positive. For the 11.9 N force ( F1 ), r = 0 . For the 14.6 N force ( F2 ), r = 0.350 m and φ = 40.0° . For the 8.50 N force ( F3 ), r = 0.350 m and φ = 90.0°
EXECUTE: τ 1 = 0 . τ 2 = −(14.6 N)(0.350 m)sin 40.0° = −3.285 N ⋅ m . τ 3 = + (8.50 N)(0.350 m)sin 90.0° = +2.975 N ⋅ m . 10.5. ∑τ = −3.285 N ⋅ m + 2.975 N ⋅ m = −0.31 N ⋅ m .The net torque is 0.31 N ⋅ m and is clockwise.
!
!
EVALUATE: If we treat the torques as vectors, τ 2 is into the page and τ 3 is out of the page.
IDENTIFY and SET UP: Calculate the torque using Eq.(10.3) and also determine the direction of the torque using
the right-hand rule.
!
!
ˆ
ˆ
(a) r = (−0.450 m)i + (0.150 m) ˆ; F = (−5.00 N)i + (4.00 N) ˆ. The sketch is given in Figure 10.5.
j
j Figure 10.5 !
!
EXECUTE: (b) When the fingers of your right hand curl from the direction of r into the direction of F (through
!
the smaller of the two angles, angle φ ) your thumb points into the page (the direction of τ , the − z -direction).
!!!
ˆ
ˆ
(c) τ = r × F = ⎡( −0.450 m)i +(0.150 m) ˆ ⎤ × ⎡( −5.00 N)i + (4.00 N) ˆ ⎤
j⎦ ⎣
j⎦
⎣
!
τ = + (2.25 N ⋅ m)iˆ × iˆ − (1.80 N ⋅ m)iˆ × ˆ − (0.750 N ⋅ m) ˆ × iˆ + (0.600 N ⋅ m) ˆ × ˆ
j
j
jj
ˆˆjj
i ×i = ˆ× ˆ = 0 10.6. 10.7. ˆ
ˆj ˆjˆ
i × ˆ = k , ˆ × i = −k
!
ˆ
ˆ
ˆ
Thus τ = −(1.80 N ⋅ m)k − (0.750 N ⋅ m)( − k ) = (−1.05 N ⋅ m)k .
!
EVALUATE: The calculation gives that τ is in the − z -direction. This agrees with what we got from the righthand rule.
IDENTIFY: Use τ = Fl = rF sin φ for the magnitude of the torque and the right-hand rule for the direction.
SET UP: In part (a), r = 0.250 m and φ = 37°
EXECUTE: (a) τ = (17.0 N)(0.250 m)sin 37° = 2.56 N ⋅ m . The torque is counterclockwise.
(b) The torque is maximum when φ = 90° and the force is perpendicular to the wrench. This maximum torque is
(17.0 N)(0.250 m) = 4.25 N ⋅ m .
EVALUATE: If the force is directed along the handle then the torque is zero. The torque increases as the angle
between the force and the handle increases.
IDENTIFY: Apply ∑τ z = Iα z . ⎛ 2π rad/rev ⎞
⎟ = 41.9 rad/s
⎝ 60 s/min ⎠
ω − ω0 z
41.9 rad/s
= ( 2.50 kg ⋅ m 2 )
= 13.1 N ⋅ m.
EXECUTE: τ z = Iαz = I z
8.00 s
t
EVALUATE: In τ z = Iα z , α z must be in rad/s 2 .
SET UP: 10.8. IDENTIFY:
SET UP: ω0 z = 0 . ω z = (400 rev/min) ⎜ Use a constant acceleration equation to calculate α z and then apply ∑τ z = Iα z . I = MR + 2mR , where M = 8.40 kg, m = 2.00 kg , so I = 0.600 kg ⋅ m .
2
3 2 2 2 ω0 z = 75.0 rpm = 7.854 rad s; ωz = 50.0 rpm = 5.236 rad s; t = 30.0 s .
EXECUTE: 10.9. 2 ωz = ω0z + αzt gives αz = −0.08726 rad s . τ z = Iαz = −0.0524 N ⋅ m EVALUATE: The torque is negative because its direction is opposite to the direction of rotation, which must be
the case for the speed to decrease.
IDENTIFY: Use ∑τ z = Iα z to calculate α . Use a constant angular acceleration kinematic equation to relate α z , ω z and t.
2
SET UP: For a solid uniform sphere and an axis through its center, I = 5 MR 2 . Let the direction the sphere is
spinning be the positive sense of rotation. The moment arm for the friction force is l = 0.0150 m and the torque due
to this force is negative. 10-4 Chapter 10 (a) α z = EXECUTE: τz
I = −(0.0200 N)(0.0150 m)
= −14.8 rad/s 2
2
(0.225 kg)(0.0150 m) 2
5 (b) ω z − ω0 z = −22.5 rad/s . ω z = ω0 z + α z t gives t = ω z − ω0 z −22.5 rad/s
=
= 1.52 s .
αz
−14.8 rad/s 2 The fact that α z is negative means its direction is opposite to the direction of spin. The negative EVALUATE: α z causes ω z to decrease.
10.10. IDENTIFY: Apply ∑τ z = Iα z to the wheel. The acceleration a of a point on the cord and the angular acceleration α of the wheel are related by a = Rα .
SET UP: Let the direction of rotation of the wheel be positive. The wheel has the shape of a disk and I = 1 MR 2 .
2
The free-body diagram for the wheel is sketched in Figure 10.10a for a horizontal pull and in Figure 10.10b for a
vertical pull. P is the pull on the cord and F is the force exerted on the wheel by the axle.
τ
(40.0 N)(0.250 m)
EXECUTE: (a) α z = z = 1
= 34.8 rad/s 2 . a = Rα = (0.250 m)(34.8 rad/s 2 ) = 8.70 m/s 2 .
(9.20 kg)(0.250 m) 2
I
2
(b) Fx = − P , Fy = − Mg . F = P 2 + ( Mg ) 2 = (40.0 N) 2 + ([9.20 kg][9.80 m/s 2 ]) 2 = 98.6 N . tan φ = Fy
Fx = Mg (9.20 kg)(9.80 m/s 2 )
and φ = 66.1° . The force exerted by the axle has magnitude 98.6 N and
=
40.0 N
P is directed at 66.1° above the horizontal, away from the direction of the pull on the cord.
(c) The pull exerts the same torque as in part (a), so the answers to part (a) don’t change. In part (b),
F + P = Mg and F = Mg − P = (9.20 kg)(9.80 m/s 2 ) − 40.0 N = 50.2 N . The force exerted by the axle has
magnitude 50.2 N and is upward.
EVALUATE: The weight of the wheel and the force exerted by the axle produce no torque because they act at the
axle. Figure 10.10
10.11. IDENTIFY: Use a constant angular acceleration equation to calculate α z and then apply ∑τ z = Iα z to the motion of the cylinder. f k = μ k n .
SET UP: I = 1 mR 2 =
2 1
2 ( 8.25 kg )( 0.0750 m ) 2 = 0.02320 kg ⋅ m 2 . Let the direction the cylinder is rotating be positive. ω0 z = 220 rpm = 23.04 rad/s; ωz = 0; θ − θ0 = 5.25 rev = 33.0 rad .
EXECUTE: 2
ωz2 = ω0 z + 2αz ( θ − θ0 ) gives αz = −8.046 rad/s 2 . ∑ τ z = τ f = − f k R = − μk nR . Then ∑τ z Iαz
= 7.47 N .
μk R
EVALUATE: The friction torque is directed opposite to the direction of rotation and therefore produces an angular
acceleration that slows the rotation.
!
!
IDENTIFY: Apply ∑ F = ma to the stone and ∑τ z = Iα z to the pulley. Use a constant acceleration equation to
− μk nR = Iα z and n = 10.12. = Iα z gives find a for the stone.
SET UP: For the motion of the stone take + y to be downward. The pulley has I = 1 MR 2 . a = Rα .
2 Dynamics of Rotational Motion EXECUTE: ∑F (a) y − y0 = v0 yt + 1 a yt 2 gives 12.6 m = 1 a y ( 3.00 s ) and a y = 2.80 m s 2 . Then
2
2
2 to the stone gives mg − T = ma . ∑τ z y 10-5 = ma y applied = Iα z applied to the pulley gives TR = MR α = MR ( a / R ) . T = 1 Ma .
2
1
2 2 1
2 2 Combining these two equations to eliminate T gives M= 10.13. M
2 ⎞
⎛ a ⎞ ⎛ 10.0 kg ⎞ ⎛
2.80 m/s 2
⎟ = 2.00 kg .
⎜
⎟=⎜
⎟⎜
g − a ⎠ ⎝ 2 ⎠ ⎝ 9.80 m/s 2 − 2.80 m/s 2 ⎠
⎝ 1
1
(b) T = Ma = (10.0 kg ) ( 2.80 m/s 2 ) = 14.0 N
2
2
EVALUATE: The tension in the wire is less than the weight mg = 19.6 N of the stone, because the stone has a
downward acceleration.
IDENTIFY: Use the kinematic information to solve for the angular acceleration of the grindstone. Assume that the
grindstone is rotating counterclockwise and let that be the positive sense of rotation. Then apply Eq.(10.7) to
calculate the friction force and use f k = μ k n to calculate μ k . ω0 z = 850 rev/min(2π rad/1 rev)(1 min/60 s) = 89.0 rad/s
t = 7.50 s; ω z = 0 (comes to rest); α z = ?
EXECUTE: ω z = ω0 z + α z t
SET UP: 0 − 89.0 rad/s
= −11.9 rad/s 2
7.50 s
SET UP: Apply ∑τ z = Iα z to the grindstone. The free-body diagram is given in Figure 10.13. αz = Figure 10.13 The normal force has zero moment arm for rotation about an axis at the center of the grindstone, and therefore zero
torque. The only torque on the grindstone is that due to the friction force f k exerted by the ax; for this force the
moment arm is l = R and the torque is negative.
EXECUTE: ∑τ z = − f k R = − μ k nR
I = 1 MR 2 (solid disk, axis through center)
2
Thus ∑τ z = Iα z gives − μk nR = ( 1 MR 2 )α z
2 MRα z
(50.0 kg)(0.260 m)( − 11.9 rad/s 2 )
=−
= 0.483
2n
2(160 N)
EVALUATE: The friction torque is clockwise and slows down the counterclockwise rotation of the grindstone.
IDENTIFY: Apply ∑ Fy = ma y to the bucket, with + y downward. Apply ∑τ z = Iα z to the cylinder, with the μk = − 10.14. direction the cylinder rotates positive.
SET UP: The free-body diagram for the bucket is given in Fig.10.14a and the free-body diagram for the cylinder
is given in Fig.10.14b. I = 1 MR 2 . a (bucket) = Rα (cylinder)
2
EXECUTE: (a) For the bucket, mg − T = ma . For the cylinder, ∑τ z = Iα z gives TR = 1 MR 2α . α = a / R then
2 gives T = 1 Ma . Combining these two equations gives mg − 1 Ma = ma and
2
2 a= ⎛
⎞
mg
15.0 kg
2
2
=⎜
⎟ (9.80 m/s ) = 7.00 m/s .
m + M / 2 ⎝ 15.0 kg + 6.0 kg ⎠ T = m( g − a) = (15.0 kg)(9.80 m/s 2 − 7.00 m/s 2 ) = 42.0 N .
2
2
(b) v y = v0 y + 2a y ( y − y0 ) gives v y = 2(7.00 m/s 2 )(10.0 m) = 11.8 m/s . (c) a y = 7.00 m/s 2 , v0 y = 0 , y − y0 = 10.0 m . y − y0 = v0 y t + 1 α yt 2 gives t =
2 2( y − y0 )
2(10.0 m)
=
= 1.69 s
ay
7.00 m/s 2 10-6 Chapter 10 (d) ∑F y = ma y applied to the cylinder gives n − T − Mg = 0 and n = T + mg = 42.0 N + (12.0 kg)(9.80 m/s 2 ) = 160 N .
EVALUATE: The tension in the rope is less than the weight of the bucket, because the bucket has a downward
acceleration. If the rope were cut, so the bucket would be in free-fall, the bucket would strike the water in
2(10.0 m)
t=
= 1.43 s and would have a final speed of 14.0 m/s. The presence of the cylinder slows the fall of
9.80 m/s 2
the bucket. 10.15. Figure 10.14
!
!
IDENTIFY: Apply ∑ F = ma to each book and apply ∑τ z = Iα z to the pulley. Use a constant acceleration equation to find the common acceleration of the books.
SET UP: m1 = 2.00 kg , m2 = 3.00 kg . Let T1 be the tension in the part of the cord attached to m1 and T2 be the
tension in the part of the cord attached to m2 . Let the + x -direction be in the direction of the acceleration of each
book. a = Rα .
2( x − x0 ) 2(1.20 m)
=
= 3.75 m/s 2 . a1 = 3.75 m/s 2 so
EXECUTE: (a) x − x0 = v0 xt + 1 axt 2 gives ax =
2
t2
(0.800 s) 2
T1 = m1a1 = 7.50 N and T2 = m2 ( g − a1 ) = 18.2 N .
(b) The torque on the pulley is (T2 − T1 ) R = 0.803 N ⋅ m, and the angular acceleration is α = a1 R = 50 rad/s 2 , so I = τ α = 0.016 kg ⋅ m 2 . 10.16. EVALUATE: The tensions in the two parts of the cord must be different, so there will be a net torque on the
pulley.
!
!
IDENTIFY: Apply ∑ F = ma to each box and ∑τ z = Iα z to the pulley. The magnitude a of the acceleration of each box is related to the magnitude of the angular acceleration α of the pulley by a = Rα .
SET UP: The free-body diagrams for each object are shown in Figure 10.16a-c. For the pulley, R = 0.250 m and
I = 1 MR 2 . T1 and T2 are the tensions in the wire on either side of the pulley. m1 = 12.0 kg , m2 = 5.00 kg and
2
!
M = 2.00 kg . F is the force that the axle exerts on the pulley. For the pulley, let clockwise rotation be positive. ∑F
= m a . ∑τ EXECUTE: m2 g − T2 ∑F (a) x = max for the 12.0 kg box gives T1 = m1a . 2 z = Iα z for the pulley gives (T2 − T1 ) R = ( 1 MR 2 )α . a = Rα and T2 − T1 = 1 Ma . Adding these
2
2 y = ma y for the 5.00 kg weight gives three equations gives m2 g = (m1 + m2 + 1 M )a and
2
⎛
⎞
⎛
⎞
m2
5.00 kg
2
2
a =⎜
⎟g =⎜
⎟ (9.80 m/s ) = 2.72 m/s . Then
m1 + m2 + 1 M ⎠
12.0 kg + 5.00 kg + 1.00 kg ⎠
⎝
2
⎝
T1 = m1a = (12.0 kg)(2.72 m/s 2 ) = 32.6 N . m2 g − T2 = m2 a gives
T2 = m2 ( g − a ) = (5.00 kg)(9.80 m/s 2 − 2.72 m/s 2 ) = 35.4 N . The tension to the left of the pulley is 32.6 N and
below the pulley it is 35.4 N.
(b) a = 2.72 m/s 2
(c) For the pulley, ∑ Fx = max gives Fx = T1 = 32.6 N and ∑ Fy = ma y gives
Fy = Mg + T2 = (2.00 kg)(9.80 m/s 2 ) + 35.4 N = 55.0 N . Dynamics of Rotational Motion EVALUATE:
objects. 10.17. IDENTIFY: 10-7 The equation m2 g = (m1 + m2 + 1 M )a says that the external force m2 g must accelerate all three
2 Figure 10.16
!
!
!
Apply ∑τ z = Iα z to the post and ∑ F = ma to the hanging mass. The acceleration a of the mass has the same magnitude as the tangential acceleration atan = rα of the point on the post where the string is attached;
r = 1.75 m − 0.500 m = 1.25 m .
SET UP: The free-body diagrams for the post and mass are given in Figures 10.17a and b. The post has
I = 1 ML2 , with M = 15.0 kg and L = 1.75 m .
3
EXECUTE: (a) ∑τ z = Iα z for the post gives Tr = ( 1 ML2 )α . a = rα so α =
3 ⎛ ML2 ⎞
a
and T = ⎜ 2 ⎟ a .
r
⎝ 3r ⎠ ∑F y = ma y for the mass gives mg − T = ma . These two equations give mg = (m + ML2 /[3r 2 ])a and ⎛
⎞
⎛
⎞
m
5.00 kg
2
2
a =⎜
⎟ (9.80 m/s ) = 3.31 m/s .
⎟g = ⎜
5.00 kg + [15.0 kg][1.75 m]2 / 3[1.25 m]2 ⎠
m + ML2 /[3r 2 ] ⎠
⎝
⎝
a 3.31 m/s 2
=
= 2.65 rad/s 2 .
r
1.25 m
(b) No. As the post rotates and the point where the string is attached moves in an arc of a circle, the string is no
longer perpendicular to the post. The torque due to this tension changes and the acceleration due to this torque is
not constant.
(c) From part (a), a = 3.31 m/s 2 . The acceleration of the mass is not constant. It changes as α for the post changes.
EVALUATE: At the instant the cable breaks the tension in the string is less than the weight of the mass because
the mass accelerates downward and there is a net downward force on it. α= Figure 10.17
10.18. IDENTIFY:
SET UP:
EXECUTE: 10.19. Apply ∑τ z = Iα z to the rod. For the rod and axis at one end, I = 1 Ml 2 .
3 α= τ
I = 1
3 Fl
3F
.
=
Ml 2 Ml EVALUATE: Note that α decreases with the length of the rod, even though the torque increases.
IDENTIFY: Since there is rolling without slipping, vcm = Rω . The kinetic energy is given by Eq.(10.8). The
velocities of points on the rim of the hoop are as described in Figure 10.13 in chapter 10.
SET UP: ω = 3.00 rad/s and R = 0.600 m . For a hoop rotating about an axis at its center, I = MR 2 . 10-8 Chapter 10 (a) vcm = Rω = (0.600 m)(3.00 rad/s) = 1.80 m/s . EXECUTE: 2
2
2
(b) K = Mvcm + 1 I ω 2 = 1 Mvcm + 1 ( MR 2 )(vcm / R 2 ) = Mvcm = (2.20 kg)(1.80 m/s) 2 = 7.13 J
2
2
2
!
(c) (i) v = 2vcm = 3.60 m/s . v is to the right. (ii) v = 0
!
2
2
2
(iii) v = vcm + vtan = vcm + ( Rω ) 2 = 2vcm = 2.55 m/s . v at this point is at 45° below the horizontal.
1
2 10.20. (d) To someone moving to the right at v = vcm , the hoop appears to rotate about a stationary axis at its center.
(i) v = Rω = 1.80 m/s , to the right. (ii) v = 1.80 m/s , to the left. (iii) v = 1.80 m/s , downward.
EVALUATE: For the special case of a hoop, the total kinetic energy is equally divided between the motion of the
center of mass and the rotation about the axis through the center of mass. In the rest frame of the ground, different
points on the hoop have different speed.
IDENTIFY: Only gravity does work, so Wother = 0 and conservation of energy gives K i + U i = K f + U f .
2
K f = 1 Mvcm + 1 I cmω 2 .
2
2 SET UP: Let yf = 0 , so U f = 0 and yi = 0.750 m . The hoop is released from rest so K i = 0 . vcm = Rω . For a hoop with an axis at its center, I cm = MR 2 .
EXECUTE: (a) Conservation of energy gives U i = K f . K f = 1 MR 2ω 2 + 1 ( MR 2 )ω 2 = MR 2ω 2 , so MR 2ω 2 = Mgyi .
2
2 gyi
(9.80 m/s 2 )(0.750 m)
=
= 33.9 rad/s .
R
0.0800 m
(b) v = Rω = (0.0800 m)(33.9 rad/s) = 2.71 m/s
EVALUATE: An object released from rest and falling in free-fall for 0.750 m attains a speed of
2 g (0.750 m) = 3.83 m/s . The final speed of the hoop is less than this because some of its energy is in kinetic
energy of rotation. Or, equivalently, the upward tension causes the magnitude of the net force of the hoop to be less
than its weight.
IDENTIFY: Apply Eq.(10.8).
SET UP: For an object that is rolling without slipping, vcm = Rω .
EXECUTE: The fraction of the total kinetic energy that is rotational is ω= 10.21. 1
(1 2 ) I cmω 2
=
2
2
(1 2 ) Mvcm + (1 2 ) I cmω 2 1( M/I cm )vcm /ω 2 = 1
1 + ( MR 2 / I cm ) (a) I cm = (1 2) MR 2 , so the above ratio is 1 3.
(b) I cm = (2 5) MR 2 so the above ratio is 2 7 .
(c) I cm = (2 3) MR 2 so the ratio is 2 5 .
(d) I cm = (5 8)MR 2 so the ratio is 5 13. The moment of inertia of each object takes the form I = β MR 2 . The ratio of rotational kinetic
1
β
energy to total kinetic energy can be written as
=
. The ratio increases as β increases.
1 + 1/ β 1 + β
!
!
IDENTIFY: Apply ∑ F = ma to the translational motion of the center of mass and ∑τ z = Iα z to the rotation
EVALUATE: 10.22. about the center of mass.
SET UP: Let + x be down the incline and let the shell be turning in the positive direction. The free-body diagram
2
for the shell is given in Fig.10.22. From Table 9.2, I cm = 3 mR 2 .
EXECUTE: ∑F x = max gives mg sin β − f = macm . ∑τ z = Iα z gives fR = ( 2 mR 2 )α . With α = acm / R this
3 2
becomes f = 2 macm . Combining the equations gives mg sin β − 3 macm = macm and
3 3g sin β 3(9.80 m/s 2 )(sin 38.0°)
2
2
=
= 3.62 m/s 2 . f = 3 macm = 3 (2.00 kg)(3.62 m/s 2 ) = 4.83 N . The friction is
5
5
f
4.83 N
= 0.313 .
static since there is no slipping at the point of contact. n = mg cos β = 15.45 N . μs = =
n 15.45 N
(b) The acceleration is independent of m and doesn’t change. The friction force is proportional to m so will double;
f = 9.66 N . The normal force will also double, so the minimum μs required for no slipping wouldn’t change. acm = Dynamics of Rotational Motion 10-9 EVALUATE: If there is no friction and the object slides without rolling, the acceleration is g sin β . Friction and
rolling without slipping reduce a to 0.60 times this value. Figure 10.22
10.23. !
!
IDENTIFY: Apply ∑ Fext = macm and
(a) SET UP: ∑τ z = I cmα z to the motion of the ball. The free-body diagram is given in Figure 10.23a.
EXECUTE: ∑F y = ma y n = mg cosθ and f s = μs mg cosθ ∑F x = max mg sin θ − μs mg cosθ = ma
g (sin θ − μs cosθ ) = a (eq. 1) Figure 10.23a
SET UP: Consider Figure 10.23b. n and mg act at the
center of the ball and
provide no torque Figure 10.23b
EXECUTE: ∑τ z ∑τ = τ f 2
= μs mg cosθ R; I = 5 mR 2 2
= I cmα z gives μs mg cosθ R = 5 mR 2α 2
No slipping means α = a / R, so μs g cosθ = 5 a (eq.2) 5
We have two equations in the two unknowns a and μs . Solving gives a = 7 g sin θ and
2
2
μs = 7 tan θ = 7 tan 65.0° = 0.613
2
2
2
(b) Repeat the calculation of part (a), but now I = 3 mR 2 . a = 3 g sin θ and μs = 5 tan θ = 5 tan 65.0° = 0.858
5 10.24. The value of μs calculated in part (a) is not large enough to prevent slipping for the hollow ball.
(c) EVALUATE: There is no slipping at the point of contact. More friction is required for a hollow ball since for a
given m and R it has a larger I and more torque is needed to provide the same α . Note that the required μs is
independent of the mass or radius of the ball and only depends on how that mass is distributed.
IDENTIFY: Apply conservation of energy to the motion of the marble.
2
SET UP: K = 1 mv 2 + 1 I ω 2 , with I = 5 MR 2 . vcm = Rω for no slipping . Let y = 0 at the bottom of the bowl. The
2
2
marble at its initial and final locations is sketched in Figure 10.24. 10-10 Chapter 10 EXECUTE: 1
1
(a) Motion from the release point to the bottom of the bowl: mgh = mv 2 + I ω 2 .
2
2
2 1
1⎛ 2
10
⎞⎛ v ⎞
gh .
mgh = mv 2 + ⎜ mR 2 ⎟⎜ ⎟ and v =
2
2⎝ 5
7
⎠⎝ R ⎠
Motion along the smooth side: The rotational kinetic energy does not change, since there is no friction torque on
v 2 10 gh 5
1
=7
=h
the marble, mv 2 + K rot = mgh′ + K rot . h′ =
2g
2g
7
2
(b) mgh = mgh′ so h′ = h .
EVALUATE: (c) With friction on both halves, all the initial potential energy gets converted back to potential
energy. Without friction on the right half some of the energy is still in rotational kinetic energy when the marble is
at its maximum height. 10.25. Figure 10.24
IDENTIFY: Apply conservation of energy to the motion of the wheel.
SET UP: The wheel at points 1 and 2 of its motion is shown in Figure 10.25. Take y = 0 at the center
of the wheel when it is at
the bottom of the hill.
Figure 10.25
2
The wheel has both translational and rotational motion so its kinetic energy is K = 1 I cmω 2 + 1 Mvcm .
2
2 EXECUTE: K1 + U1 + Wother = K 2 + U 2 Wother = Wfric = −3500 J (the friction work is negative)
K1 = 1 I ω12 + 1 Mv12 ; v = Rω and I = 0.800 MR 2 so
2
2
K1 = 1 (0.800) MR 2ω12 + 1 MR 2ω12 = 0.900 MR 2ω12
2
2
K 2 = 0, U1 = 0, U 2 = Mgh
Thus 0.900MR 2ω12 + Wfric = Mgh
M = w / g = 392 N/(9.80 m/s 2 ) = 40.0 kg
h= 0.900MR 2ω12 + Wfric
Mg (0.900)(40.0 kg)(0.600 m) 2 (25.0 rad/s) 2 − 3500 J
= 11.7 m
(40.0 kg)(9.80 m/s 2 )
EVALUATE: Friction does negative work and reduces h.
!
!
IDENTIFY: Apply ∑τ z = Iα z and ∑ F = ma to the motion of the bowling ball.
h= 10.26. SET UP: acm = Rα . f s = μs n . Let + x be directed down the incline.
EXECUTE: (a) The free-body diagram is sketched in Figure 10.26.
The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill.
!
!
(b) The friction force results in an angular acceleration, given by Iα = fR. ∑ F = ma applied to the motion of the center of mass gives mg sin β − f = macm, and the acceleration and angular acceleration are related by acm = Rα .
I⎞
⎛
Combining, mg sin β = ma ⎜1 +
= ma ( 7 5) . acm = ( 5 7 ) g sin β .
2⎟
⎝ mR ⎠ Dynamics of Rotational Motion 10-11 2
2
(c) From either of the above relations between f and acm , f = macm = mg sin β ≤ μs n = μs mg cos β .
5
7
μs ≥ ( 2 7 ) tanβ .
EVALUATE: If μs = 0 , acm = mg sin β . acm is less when friction is present. The ball rolls farther uphill when
friction is present, because the friction removes the rotational kinetic energy and converts it to gravitational
potential energy. In the absence of friction the ball retains the rotational kinetic energy that is has initially. Figure 10.26
10.27. (a) IDENTIFY: Use Eq.(10.7) to find α z and then use a constant angular acceleration equation to find ω z .
SET UP: The free-body diagram is given in Figure 10.27.
EXECUTE: Apply ∑τ z = Iα z to find the angular acceleration:
FR = Iα z αz = FR (18.0 N)(2.40 m)
=
= 0.02057 rad/s 2
I
2100 kg ⋅ m 2 Figure 10.27
SET UP: Use the constant α z kinematic equations to find ω z . ω z = ?; ω0 z (initially at rest); α z = 0.02057 rad/s 2 ; t = 15.0 s
EXECUTE: ω z = ω0 z + α z t = 0 + (0.02057 rad/s 2 )(15.0 s) = 0.309 rad/s
(b) IDENTIFY and SET UP: Calculate the work from Eq.(10.21), using a constant angular acceleration equation to
calculate θ − θ 0 , or use the work-energy theorem. We will do it both ways.
EXECUTE: (1) W = τ z Δθ (Eq.(10.21)) Δθ = θ − θ 0 = ω0 zt + 1 α zt 2 = 0 + 1 (0.02057 rad/s 2 )(15.0 s) 2 = 2.314 rad
2
2
t z = FR = (18.0 N)(2.40 m) = 43.2 N ⋅ m
Then W = τ z Δθ = (43.2 N ⋅ m)(2.314 rad) = 100 J.
or
(2) Wtot = K 2 − K1 (the work-energy relation from chapter 6)
Wtot = W , the work done by the child
K1 = 0; K 2 = 1 I ω 2 = 1 (2100 kg ⋅ m 2 )(0.309 rad/s)2 = 100 J
2
2
Thus W = 100 J, the same as before.
EVALUATE: Either method yields the same result for W.
(c) IDENTIFY and SET UP: Use Eq.(6.15) to calculate Pav
ΔW 100 J
=
= 6.67 W
Δt 15.0 s
EVALUATE: Work is in joules, power is in watts.
IDENTIFY: Apply P = τω and W = τΔθ .
SET UP: P must be in watts, Δθ must be in radians, and ω must be in rad/s. 1 rev = 2π rad . 1 hp = 746 W .
π rad/s = 30 rev/min .
P
(175 hp ) ( 746 W / hp ) = 519 N ⋅ m.
EXECUTE: (a) τ = =
π rad/s ⎞
ω
( 2400 rev/min ) ⎛
⎜
⎟
⎝ 30 rev/min ⎠
(b) W = τΔθ = ( 519 N ⋅ m )( 2π rad ) = 3260 J
EXECUTE:
10.28. Pav = 10-12 Chapter 10 10.29. EVALUATE: ω = 40 rev/s , so the time for one revolution is 0.025 s. P = 1.306 × 105 W , so in one revolution,
W = Pt = 3260 J , which agrees with our previous result.
IDENTIFY: Apply ∑τ z = Iα z and constant angular acceleration equations to the motion of the wheel. 1 rev = 2π rad . π rad/s = 30 rev/min .
ω − ω0 z
EXECUTE: (a) τ z = Iα z = I z
.
t
SET UP: π rad s
((1 2)(1.50 kg )( 0.100 m ) ) (1200 rev min ) ⎛ 30 rev min ⎞
⎜
⎟
⎝
⎠
2 τz =
(b) ωav Δt = 2.5 s = 0.377 N ⋅ m ( 600 rev/min )( 2.5 s ) = 25.0 rev = 157 rad. 60 s/min
(c) W = τΔθ = (0.377 N ⋅ m)(157 rad) = 59.2 J . 2 (d) K = 10.30. 121
⎛
⎛ π rad/s ⎞ ⎞
I ω = ( (1/ 2)(1.5 kg)(0.100 m) 2 ) ⎜ (1200 rev/min) ⎜
⎟ ⎟ = 59.2 J .
2
2
⎝ 30 rev/min ⎠ ⎠
⎝ the same as in part (c).
EVALUATE: The agreement between the results of parts (c) and (d) illustrates the work-energy theorem
IDENTIFY: The power output of the motor is related to the torque it produces and to its angular velocity by
P = τ zω z , where ω z must be in rad/s.
The work output of the motor in 60.0 s is SET UP: 2
6.00 kJ
(9.00 kJ) = 6.00 kJ , so P =
= 100 W .
3
60.0 s ω z = 2500 rev/min = 262 rad/s .
100 W
= 0.382 N ⋅ m
262 rad/s
For a constant power output, the torque developed decreases and the rotation speed of the motor τz = EXECUTE: 10.31. 10.32. P ωz = EVALUATE:
increases.
IDENTIFY: Apply τ = FR and P = τω .
SET UP: 1 hp = 746 W . π rad/s = 30 rev/min
EXECUTE: (a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction),
and the bearing radius is small compared to the blade radius, so any frictional torque can be neglected.
τ P /ω
(1.9 hp)(746 W/hp)
(b) F = =
=
= 65.6 N.
⎛ π rad/s ⎞
R
R
(2400 rev/min) ⎜
⎟ (0.086 m)
⎝ 30 rev/min ⎠
EVALUATE: In P = I ω , τ must be in watts and ω must be in rad/s.
IDENTIFY: Apply ∑τ z = Iα z to the motion of the propeller and then use constant acceleration equations to analyze the motion. W = τΔθ .
SET UP: I = 1 mL2 = 1 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 .
2
2
(a) α = EXECUTE: τ
I = 1950 N ⋅ m
= 46.2 rad/s 2 .
42.2 kg ⋅ m 2 (b) ω = ω + 2α z (θ − θ 0 ) gives ω = 2αθ = 2(46.2 rad/s 2 )(5.0 rev)(2π rad/rev) = 53.9 rad/s.
2
z 2
0z (c) W = τθ = (1950 N ⋅ m)(5.00 rev)(2π rad/rev) = 6.13 × 104 J. ω z − ω0 z
53.9 rad/s
W 6.13 × 104 J
=
= 1.17 s . Pav =
=
= 52.5 kW .
2
αz
46.2 rad/s
Δt
1.17 s
EVALUATE: P = τω . τ is constant and ω is linear in t, so Pav is half the instantaneous power at the end of the
(d) t = 5.00 revolutions. We could also calculate W from W = ΔK = 1 I ω 2 = 1 (42.2 kg ⋅ m 2 )(53.9 rad/s) 2 = 6.13 × 104 J .
2
2
10.33. (a) IDENTIFY and SET UP: Use Eq.(10.23) and solve for τ z . P = τ zω z , where ω z must be in rad/s
EXECUTE: τz = P ωz = ω z = ( 4000 rev/min)(2π rad/1 rev)(1 min/60 s) = 418.9 rad/s 1.50 × 105 W
= 358 N ⋅ m
418.9 rad/s Dynamics of Rotational Motion 10-13 !
!
(b) IDENTIFY and SET UP: Apply ∑ F = ma to the drum. Find the tension T in the rope using τ z from part (a).
The system is sketched in Figure 10.33.
EXECUTE: v constant implies a = 0 and T = w
τ z = TR implies T = τ z / R = 358 N ⋅ m/0.200 m = 1790 N
Thus a weight w = 1790 N can be lifted.
Figure 10.33 10.34. (c) IDENTIFY and SET UP: Use v = Rω.
EXECUTE: The drum has ω = 418.9 rad/s, so v = (0.200 m)(418.9 rad/s) = 83.8 m/s
EVALUATE: The rate at which T is doing work on the drum is P = Tv = (1790 N)(83.8 m/s) = 150 kW. This
agrees with the work output of the motor.
IDENTIFY: L = I ω and I = I disk + I woman .
SET UP: 10.35. ω = 0.50 rev/s = 3.14 rad/s . I disk = 1 mdisk R 2 and I woman = mwoman R 2 .
2 EXECUTE: I = (55 kg + 50.0 kg)(4.0 m) 2 = 1680 kg ⋅ m 2 . L = (1680 kg ⋅ m 2 )(3.14 rad/s) = 5.28 × 103 kg ⋅ m 2 /s
EVALUATE: The disk and the woman have similar values of I, even though the disk has twice the mass.
(a) IDENTIFY: Use L = mvr sin φ (Eq.(10.25)):
SET UP: Consider Figure 10.35.
EXECUTE: L = mvr sin φ =
(2.00 kg)(12.0 m/s)(8.00 m)sin143.1° L = 115 kg ⋅ m 2 / s 10.36. Figure 10.35
!
!
!
To find the direction of L apply the right-hand rule by turning r into the direction of v by pushing on it with the
!
fingers of your right hand. Your thumb points into the page, in the direction of L.
(b) IDENTIFY and SET UP: By Eq.(10.26) the rate of change of the angular momentum of the rock equals the
torque of the net force acting on it.
EXECUTE: τ = mg (8.00 m)cos36.9° = 125 kg ⋅ m 2 / s 2
!
!
!
To find the direction of τ and hence of dL / dt , apply the right-hand rule by turning r into the direction of the
gravity force by pushing on it with the fingers of your right hand. Your thumb points out of the page, in the
!
direction of dL / dt.
!
!
EVALUATE: L and dL / dt are in opposite directions, so L is decreasing. The gravity force is accelerating the
rock downward, toward the axis. Its horizontal velocity is constant but the distance l is decreasing and hence L is
decreasing.
IDENTIFY: Lz = I ω z
SET UP: For a particle of mass m moving in a circular path at a distance r from the axis, I = mr 2 and v = rω . For
2
a uniform sphere of mass M and radius R and an axis through its center, I = 5 MR 2 . The earth has mass mE = 5.97 × 1024 kg , radius RE = 6.38 × 106 m and orbit radius r = 1.50 × 1011 m . The earth completes one rotation
on its axis in 24 h = 86,400 s and one orbit in 1 y = 3.156 × 107 s .
⎛ 2π rad ⎞
40
2
(a) Lz = I ω z = mr 2ω z = (5.97 × 1024 kg)(1.50 × 1011 m) 2 ⎜
⎟ = 2.67 × 10 kg ⋅ m /s .
7
⎝ 3.156 × 10 s ⎠
The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this
calculation.
⎛ 2π rad ⎞
33
2
2
2
(b) Lz = I ω z = ( 5 MR 2 ) ω = 5 (5.97 × 1024 kg)(6.38 × 106 m) 2 ⎜
⎟ = 7.07 × 10 kg ⋅ m /s
86,400 s ⎠
⎝
EVALUATE: The angular momentum associated with each of these motions is very large.
EXECUTE: 10-14 Chapter 10 10.37. IDENTIFY and SET UP: Use L = I ω
EXECUTE: The second hand makes 1 revolution in 1 minute, so
ω = (1.00 rev/min)(2π rad/1 rev)(1 min/60 s) = 0.1047 rad/s
For a slender rod, with the axis about one end,
I = 1 ML2 = 1 (6.00 × 10−3 kg)(0.150 m)2 = 4.50 × 10−5 kg ⋅ m 2
3
3 10.38. Then L = I ω = (4.50 × 10−5 kg ⋅ m 2 )(0.1047 rad/s) = 4.71 × 10−6 kg ⋅ m 2 / s.
!
EVALUATE: L is clockwise.
IDENTIFY: ω z = dθ / dt . Lz = I ω z and τ z = dLz dt .
2
For a hollow, thin-walled sphere rolling about an axis through its center, I = 3 MR 2 . R = 0.240 m . SET UP:
EXECUTE: (a) A = 1.50 rad/s 2 and B = 1.10 rad/s 4 , so that θ (t ) will have units of radians. dθ
= 2 At + 4 Bt 3 . At t = 3.00 s , ω z = 2(1.50 rad/s 2 )(3.00 s) + 4(1.10 rad/s 4 )(3.00 s)3 = 128 rad/s .
dt
2
2
Lz = ( 3 MR 2 )ω z = 3 (12.0 kg)(0.240 m) 2 (128 rad/s) = 59.0 kg ⋅ m 2 /s . (b) (i) ω z = dLz
dω z
=I
= I (2 A + 12 Bt 2 ) and
dt
dt
2
τ z = 3 (12.0 kg)(0.240 m) 2 (2[1.50 rad/s 2 ] + 12[1.10 rad/s 4 ][3.00 s]2 ) = 56.1 N ⋅ m . (ii) τ z = EVALUATE: 10.39. The angular speed of rotation is increasing. This increase is due to an acceleration α z that is produced by the torque on the sphere. When I is constant, as it is here, τ z = dLz dt = Idω z / dt = Iα z and
Equations (10.29) and (10.7) are identical.
IDENTIFY: Apply conservation of angular momentum.
2
SET UP: For a uniform sphere and an axis through its center, I = 5 MR 2 .
EXECUTE: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be
proportional to the inverse of the square of the radius, and the final angular velocity is
2 2 ⎛ R1 ⎞ ⎛
⎞ ⎛ 7.0 × 105 km ⎞
2π rad
3
⎟ =⎜
⎟ = 4.6 × 10 rad s.
⎟⎜
⎝ R2 ⎠ ⎝ (30 d)(86,400 s d) ⎠⎝ 16 km ⎠ ω2 = ω1 ⎜ 10.40. EVALUATE: K = 1 I ω 2 = 1 Lω . L is constant and ω increases by a large factor, so there is a large increase in the
2
2
rotational kinetic energy of the star. This energy comes from potential energy associated with the gravity force
within the star.
!
IDENTIFY and SET UP: L is conserved if there is no net external torque.
Use conservation of angular momentum to find ω at the new radius and use K = 1 I ω 2 to find the change in
2
kinetic energy, which is equal to the work done on the block.
EXECUTE: (a) Yes, angular momentum is conserved. The moment arm for the tension in the cord is zero so this
force exerts no torque and there is no net torque on the block.
(b) L1 = L2 so I1ω1 = I 2ω2. Block treated as a point mass, so I = mr 2, where r is the distance of the block from the
hole.
mr12ω1 = mr22ω2
2 2 ⎛r ⎞
⎛ 0.300 m ⎞
ω2 = ⎜ 1 ⎟ ω1 = ⎜
⎟ (1.75 rad/s) = 7.00 rad/s
r2 ⎠
⎝ 0.150 m ⎠
⎝
(c) K1 = 1 I1ω12 = 1 mr12ω12 = 1 mv12
2
2
2
v1 = r1ω1 = (0.300 m)(1.75 rad/s) = 0.525 m/s
K1 = 1 mv12 = 1 (0.0250 kg)(0.525 m/s)2 = 0.00345 J
2
2
2
K 2 = 1 mv2
2 v2 = r2ω2 = (0.150 m)(7.00 rad/s) = 1.05 m/s
2
K 2 = 1 mv2 = 1 (0.0250 kg)(1.05 m/s) 2 = 0.01378 J
2
2 ΔK = K 2 − K1 = 0.01378 J − 0.00345 J = 0.0103 J Dynamics of Rotational Motion 10-15 (d) Wtot = ΔK 10.41. But Wtot = W , the work done by the tension in the cord, so W = 0.0103 J
EVALUATE: Smaller r means smaller I. L = I ω is constant so ω increases and K increases. The work done by
the tension is positive since it is directed inward and the block moves inward, toward the hole.
IDENTIFY: Apply conservation of angular momentum to the motion of the skater.
SET UP: For a thin-walled hollow cylinder I = mR 2 . For a slender rod rotating about an axis through its center,
1
I = 12 ml 2 .
EXECUTE: Li = Lf so I iωi = I f ωf . 1
I i = 0.40 kg ⋅ m 2 + 12 (8.0 kg)(1.8 m)2 = 2.56 kg ⋅ m 2 . I f = 0.40 kg ⋅ m 2 + (8.0 kg)(0.25 m) 2 = 0.90 kg ⋅ m 2 . ⎛ Ii ⎞
⎛ 2.56 kg ⋅ m 2 ⎞
(0.40 rev/s)=1.14 rev/s .
⎟ ωi = ⎜
2⎟
⎝ 0.90 kg ⋅ m ⎠
⎝ If ⎠
EVALUATE: K = 1 I ω 2 = 1 Lω . ω increases and L is constant, so K increases. The increase in kinetic energy
2
2
comes from the work done by the skater when he pulls in his hands.
IDENTIFY: Apply conservation of angular momentum to the diver.
SET UP: The number of revolutions she makes in a certain time is proportional to her angular velocity. The ratio
of her untucked to tucked angular velocity is (3.6 kg ⋅ m 2 ) /(18 kg ⋅ m 2 ) . ωf = ⎜ 10.42. EXECUTE: 10.43. If she had tucked, she would have made (2 rev)(3.6 kg ⋅ m 2 ) (18 kg ⋅ m 2 ) = 0.40 rev in the last 1.0 s, so she would have made (0.40 rev)(1.5 1.0) = 0.60 rev in the total 1.5 s.
EVALUATE: Untucked she rotates slower and completes fewer revolutions.
IDENTIFY and SET UP: There is no net external torque about the rotation axis so the angular momentum L = I ω
is conserved.
EXECUTE: (a) L1 = L2 gives I1ω1 = I 2ω2 , so ω2 = ( I1 / I 2 )ω1
I1 = I tt = 1 MR 2 = 1 (120 kg)(2.00 m) 2 = 240 kg ⋅ m 2
2
2
I 2 = I tt + I p = 240 kg ⋅ m 2 + mR 2 = 240 kg ⋅ m 2 + (70 kg)(2.00 m) 2 = 520 kg ⋅ m 2 ω2 = ( I1 / I 2 )ω1 = (240 kg ⋅ m 2 / 520 kg ⋅ m 2 )(3.00 rad/s) = 1.38 rad/s
(b) K1 = 1 I1ω12 = 1 (240 kg ⋅ m 2 )(3.00 rad/s) 2 = 1080 J
2
2 10.44. 2
K 2 = 1 I 2ω2 = 1 (520 kg ⋅ m 2 )(1.38 rad/s) 2 = 495 J
2
2
EVALUATE: The kinetic energy decreases because of the negative work done on the turntable and the parachutist
by the friction force between these two objects.
The angular speed decreases because I increases when the parachutist is added to the system.
IDENTIFY: Apply conservation of angular momentum to the collision.
SET UP: Let the width of the door be l. The initial angular momentum of the mud is mv (l / 2) , since it strikes the door at its center. For the axis at the hinge, I door = 1 Ml 2 and I mud = m(l / 2) 2 .
3
EXECUTE: ω= 10.45. ω= mv ( l 2 )
L
=
.
I (1 3) Ml 2 + m ( l 2 )2 ( 0.500 kg ) (12.0 m s )( 0.500 m )
2
2
(1 3)( 40.0 kg )(1.00 m ) + ( 0.500 kg )( 0.500 m ) = 0.223 rad s. Ignoring the mass of the mud in the denominator of the above expression gives ω = 0.225 rad s, so the mass of
the mud in the moment of inertia does affect the third significant figure.
EVALUATE: Angular momentum is conserved but there is a large decrease in the kinetic energy of the system.
!
(a) IDENTIFY and SET UP: Apply conservation of angular momentum L, with the axis at the nail. Let object A
be the bug and object B be the bar. Initially, all objects are at rest and L1 = 0. Just after the bug jumps, it has
angular momentum in one direction of rotation and the bar is rotating with angular velocity ωB in the opposite
direction.
EXECUTE: L2 = mAv Ar − I BωB where r = 1.00 m and I B = 1 mB r 2
3
L1 = L2 gives mAv A r = 1 mB r 2ωB
3 ωB = 3mAv A
= 0.120 rad/s
mB r 10-16 Chapter 10
2
2
(b) K1 = 0; K 2 = 1 mAv A + 1 I BωB =
2
2
1
2 10.46. (0.0100 kg)(0.200 m/s)2 + 1 ( 1 [0.0500 kg][1.00 m]2 ) (0.120 rad/s) 2 = 3.2 × 10−4 J.
23 The increase in kinetic energy comes from work done by the bug when it pushes against the bar in order to jump.
EVALUATE: There is no external torque applied to the system and the total angular momentum of the system is
constant. There are internal forces, forces the bug and bar exert on each other. The forces exert torques and change
the angular momentum of the bug and the bar, but these changes are equal in magnitude and opposite in direction.
These internal forces do positive work on the two objects and the kinetic energy of each object and of the system
increases.
IDENTIFY: Apply conservation of angular momentum to the system of earth plus asteroid.
SET UP: Take the axis to be the earth’s rotation axis. The asteroid may be treated as a point mass and it has zero
angular momentum before the collision, since it is headed toward the center of the earth. For the earth,
2π rad
2
Lz = I ω z and I = 5 MR 2 ,where M is the mass of the earth and R is its radius. The length of a day is T =
, ω where ω is the earth’s angular rotation rate.
EXECUTE: Conservation of angular momentum applied to the collision between the earth and asteroid gives
⎛ ω − ω2 ⎞
1 1.250
2
2
2
MR 2ω1 = (mR 2 + 5 MR 2 )ω2 and m = 5 M ⎜ 1
=
and ω1 = 1.250ω2 .
⎟ . T2 = 1.250T1 gives
5
ω2
ω1
ω2 ⎠
⎝
ω1 − ω2
2
= 0.250 . m = 5 (0.250) M = 0.100 M . ω2 10.47. EVALUATE: If the asteroid hit the surface of the earth tangentially it could have some angular momentum with
respect to the earth’s rotation axis, and could either speed up or slow down the earth’s rotation rate.
IDENTIFY: Apply conservation of angular momentum to the collision.
SET UP: The system before and after the collision is sketched in Figure 10.47. Let counterclockwise rotation be
positive. The bar has I = 1 m2 L2 .
3
EXECUTE: (a) Conservation of angular momentum: m1v0 d = − m1vd + 1 m2 L2ω .
3 1 ⎛ 90.0 N ⎞
2
(3.00 kg)(10.0 m s)(1.50 m) = −(3.00 kg)(6.00 m s)(1.50 m) + ⎜
⎟ (2.00 m) ω
3 ⎝ 9.80 m s 2 ⎠ ω = 5.88 rad s .
(b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an
unbalanced horizontal external force on the system, so the linear momentum is not conserved.
EVALUATE: Kinetic energy is not conserved in the collision. 10.48. Figure 10.47
!
!!
!
IDENTIFY: dL = τ dt , so dL is in the direction of τ .
!
SET UP: The direction of ω is given by the right-hand rule, as described in Figure 10.26 in the textbook.
EXECUTE: The sketches are given in Figures 10.48a–d. Dynamics of Rotational Motion 10-17 EVALUATE: In figures (a) and (c) the precession is counterclockwise and in figures (b) and (d) it is clockwise.
!
!
When the direction of either ω or τ reverses, the direction of precession reverses. Figure 10.48
10.49. IDENTIFY: The precession angular velocity is Ω = wr
, where ω is in rad/s. Also apply
Iω ! ! ∑ F = ma to the gyroscope.
SET UP: The total mass of the gyroscope is mr + mf = 0.140 kg + 0.0250 kg = 0.165 kg .
2π rad 2π rad
=
= 2.856 rad/s .
T
2.20 s
EXECUTE: (a) Fp = wtot = (0.165 kg)(9.80 m/s 2 ) = 1.62 N
Ω= wr (0.165 kg)(9.80 m/s 2 )(0.0400 m)
=
= 189 rad/s = 1.80 × 103 rev/min
I Ω (1.20 × 10−4 kg ⋅ m 2 )(2.856 rad/s)
!
!
(c) If the figure in the problem is viewed from above, τ is in the direction of the precession and L is along the
axis of the rotor, away from the pivot.
EVALUATE: There is no vertical component of acceleration associated with the motion, so the force from the
pivot equals the weight of the gyroscope. The larger ω is, the slower the rate of precession.
IDENTIFY: The precession angular speed is related to the acceleration due to gravity by Eq.(10.33), with w = mg .
(b) ω = 10.50. SET UP: Ω E = 0.50 rad/s , g E = g and g M = 0.165 g . For the gyroscope, m, r, I, and ω are the same on the moon
as on the earth.
Ω
Ω
mgr Ω mr
EXECUTE: Ω =
.
=
= constant , so E = M .
Iω
g Iω
gE
gM 10.51. 10.52. ⎛g ⎞
Ω M = Ω E ⎜ M ⎟ = 0.165Ω E = (0.165)(0.50 rad/s) = 0.0825 rad/s .
⎝ gE ⎠
EVALUATE: In the limit that g → 0 the precession rate → 0 .
IDENTIFY and SET UP: Apply Eq.(10.33).
EXECUTE: (a) halved
(b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed)
(c) halved (assuming that w and r are not changed)
(d) doubled
(e) unchanged.
EVALUATE: Ω is directly proportional to w and r and is inversely proportional to I and ω .
IDENTIFY: Apply Eq.(10.33), where τ = wr .
SET UP: 1 day = 86,400 s . 1 yr = 3.156 × 107 s . The earth has mass M = 5.97 × 1024 kg and radius
2
R = 6.38 × 106 m . For a uniform sphere and an axis through its center, I = 5 MR 2 . 2π rad
(a) τ = I ωΩ = (2 / 5) MR 2ωΩ. Using ω = 2π rad and Ω =
, and the mass
86,400 s
(26,000 y)(3.156 × 107 s/y)
and radius of the earth from Appendix F, τ = 5.4 N ⋅ m .
EVALUATE: If the torque is applied by the sun, r = 1.5 × 1011 m and F⊥ = 3.6 × 1011 N . EXECUTE: 10-18 Chapter 10 10.53. IDENTIFY: Apply ∑τ z = Iα z and constant acceleration equations to the motion of the grindstone. Let the direction of rotation of the grindstone be positive. The friction force is f = μ k n and produces SET UP: ⎛ 2π rad ⎞⎛ 1 min ⎞
2
2
1
torque fR . ω = ⎜
⎟⎜
⎟ = 4π rad . I = 2 MR = 1.69 kg ⋅ m .
⎝ 1 rev ⎠⎝ 60 s ⎠
EXECUTE: (a) The net torque must be
4π rad/s
= (1.69 kg ⋅ m 2 )
= 2.36 N ⋅ m.
t
9.00 s
This torque must be the sum of the applied force FR and the opposing frictional torques
1
τ f at the axle and fR = μ k nR due to the knife. F = (τ + τ f + μ k nR ) .
R
1
F=
( (2.36 N ⋅ m) + (6.50 N ⋅ m) + (0.60)(160 N)(0.260 m) ) = 67.6 N.
0.500 m
(b) To maintain a constant angular velocity, the net torque τ is zero, and the force F ′ is
1
F′ =
(6.50 N ⋅ m + 24.96 N ⋅ m) = 62.9 N.
0.500 m
(c) The time t needed to come to a stop is found by taking the magnitudes in Eq.(10.27), with τ = τ f constant; τ = Iα = I ω z − ω0 z 2
ω I (4π rad/s) (1.69 kg ⋅ m )
=
= 3.27 s.
τf τf
6.50 N ⋅ m
EVALUATE: The time for a given change in ω is proportional to α , which is in turn proportional to the net torque, t= 10.54. L = so the time in part (c) can also be found as t = ( 9.00 s ) 2.36 N ⋅ m .
6.50 N ⋅ m
IDENTIFY: Apply ∑τ z = Iα z and use the constant acceleration equations to relate α to the motion.
Let the direction the wheel is rotating be positive. 100 rev/min = 10.47 rad/s
ω − ω0 z 10.47 rad/s − 0
=
= 5.23 rad/s 2 .
EXECUTE: (a) ω z = ω0 z + α z t gives α z = z
t
2.00 s
SET UP: I= ∑τ
αz z = 5.00 N ⋅ m
= 0.956 kg ⋅ m 2
5.23 rad/s 2 (b) ω0 z = 10.47 rad/s , ω z = 0 , t = 125 s . ω z = ω0 z + α z t gives α z = ∑τ 10.55. z ω z − ω0 z
t = 0 − 10.47 rad/s
= −0.0838 rad/s 2
125 s = Iα z = (0.956 kg ⋅ m 2 )(−0.0838 rad/s 2 ) = −0.0801 N ⋅ m ⎛ ω + ω z ⎞ ⎛ 10.47 rad/s + 0 ⎞
(c) θ = ⎜ 0 z
⎟t = ⎜
⎟ (125 s) = 654 rad = 104 rev
2
2
⎝
⎠⎝
⎠
EVALUATE: The applied net torque ( 5.00 N ⋅ m ) is much larger than the magnitude of the friction torque
( 0.0801 N ⋅ m ), so the time of 2.00 s that it takes the wheel to reach an angular speed of 100 rev/min is much less
than the 125 s it takes the wheel to be brought to rest by friction.
IDENTIFY and SET UP: Apply v = rω. v is the tangential speed of a point on the rim of the wheel and equals the
linear speed of the car.
EXECUTE: (a) v = 60 mph = 26.82 m/s
r = 12 in. = 0.3048 m
v
ω = = 88.0 rad/s = 14.0 rev/s = 840 rpm
r
(b) Same ω as in part (a) since speedometer reads same.
r = 15 in. = 0.381 m
v = rω = (0.381 m)(88.0 rad/s) = 33.5 m/s = 75 mph
(c) v = 50 mph = 22.35 m/s
r = 10 in. = 0.254 m
v
ω = = 88.0 rad/s. This is the same as for 60 mph with correct tires, so speedometer read 60 mph.
r
EVALUATE: For a given ω , v increases when r increases. Dynamics of Rotational Motion 10.56. 10-19 2
IDENTIFY: The kinetic energy of the disk is K = 1 Mvcm + 1 I ω 2 . As it falls its gravitational potential energy
2
2
decreases and its kinetic energy increases. The only work done on the disk is the work done by gravity, so
K1 + U1 = K 2 + U 2 . SET UP: 2
I cm = 1 M ( R2 + R12 ) , where R1 = 0.300 m and R2 = 0.500 m . vcm = R2ω . Take y1 = 0 , so
2 y2 = −1.20 m .
K1 + U1 = K 2 + U 2 . K1 = 0 , U1 = 0 . K 2 = −U 2 . EXECUTE:
1
2 1
2 2
Mvcm + 1 I cmω 2 = − Mgy2 .
2 2
2
2
I cmω 2 = 1 M (1 + [ R1 / R2 ]2 ) vcm = 0.340 Mvcm . Then 0.840 Mvcm = − Mgy2 and
4 − gy2
−(9.80 m/s 2 )( −1.20 m)
=
= 3.74 m/s
0.840
0.840
EVALUATE: A point mass in free-fall acquires a speed of 4.85 m/s after falling 1.20 m. The disk has a value of
vcm that is less than this, because some of the original gravitational potential energy has been converted to
rotational kinetic energy.
IDENTIFY: Use ∑τ z = Iα z to find the angular acceleration just after the ball falls off and use conservation of
vcm = 10.57. energy to find the angular velocity of the bar as it swings through the vertical position.
1
SET UP: The axis of rotation is at the axle. For this axis the bar has I = 12 mbar L2 , where mbar = 3.80 kg and L = 0.800 m . Energy conservation gives K1 + U1 = K 2 + U 2 . The gravitational potential energy of the bar doesn’t
change. Let y1 = 0 , so y2 = − L / 2 .
1
(a) τ z = mball g ( L / 2) and I = I ball + I bar = 12 mbar L2 + mball ( L / 2) 2 . EXECUTE: ∑τ z = Iα z gives ⎞
⎞
mball g ( L / 2)
2g ⎛
mball
2(9.80 m/s ) ⎛
2.50 kg
2
=
⎜
⎟ and α z =
⎜
⎟ = 16.3 rad/s .
2
2
mbar L + mball ( L / 2)
L ⎝ mball + mbar / 3 ⎠
0.800 m ⎝ 2.50 kg + [3.80 kg]/ 3 ⎠
(b) As the bar rotates, the moment arm for the weight of the ball decreases and the angular acceleration of the bar
decreases.
(c) K1 + U1 = K 2 + U 2 . 0 = K 2 + U 2 . 1 ( I bar + I ball )ω 2 = − mball g ( − L / 2) .
2 αz = 1
12 mball gL
=
mball L / 4 + mbar L2 /12 ω= 10.58. 2 2 ⎞
⎞
g⎛
4mball
9.80 m/s 2 ⎛
4[2.50 kg]
⎜
⎟=
⎜
⎟
L ⎝ mball + mbar / 3 ⎠
0.800 m ⎝ 2.50 kg + [3.80 kg]/ 3 ⎠ ω = 5.70 rad/s .
EVALUATE: As the bar swings through the vertical, the linear speed of the ball that is still attached to the bar is
v = (0.400 m)(5.70 rad/s) = 2.28 m/s . A point mass in free-fall acquires a speed of 2.80 m/s after falling 0.400 m;
the ball on the bar acquires a speed less than this.
IDENTIFY: Use ∑τ z = Iα z to find α z , and then use the constant α z kinematic equations to solve for t.
SET UP: The door is sketched in Figure 10.58.
EXECUTE: ∑τ z = Fl = (220 N)(1.25 m) = 275 N ⋅ m From Table 9.2(d), I = 1 Ml 2
3
I = 1 (750 N/9.80 m/s 2 )(1.25 m) 2 = 39.9 kg ⋅ m 2
3
Figure 10.58 ∑τ z = Iα z so α z = SET UP: ∑τ
I z = 275 N ⋅ m
= 6.89 rad/s 2
39.9 kg ⋅ m 2 α z = 6.89 rad/s 2 ; θ − θ 0 = 90°(π rad/180°) = π /2 rad; ω0 z = 0 (door initially at rest); t = ? θ − θ 0 = ω0 z t + 1 α z t 2
2
EXECUTE:
10.59. t= 2(θ − θ 0 ) αz − 2(π / 2 rad)
= 0.675 s
6.89 rad/s 2 EVALUATE: The forces and the motion are connected through the angular acceleration.
IDENTIFY: τ = rF sin φ
SET UP: Let x be the distance from the left end of the rod where the string is attached. For the value of x
where f ( x) is a maximum, df / dx = 0 . 10-20 Chapter 10 ""
!
!
EXECUTE: (a) From geometric consideration, the lever arm and the sine of the angle between F and r are both
maximum if the string is attached at the end of the rod. (b) In terms of the distance x where the string is attached, the magnitude of the torque is Fxh x 2 + h 2 . This
function attains its maximum at the boundary, where x = h, so the string should be attached at the right end of
the rod.
xh
(c) As a function of x, l and h, the torque has magnitude τ = F
. Differentiating τ with respect to x
( x − l 2) 2 + h 2 10.60. and setting equal to zero gives xmax = (l 2)(1 + (2 h l ) 2 ). This will be the point at which to attach the string unless
2h > l , in which case the string should be attached at the furthest point to the right, x = l .
EVALUATE: In part (a) the maximum torque is independent of h. In part (b) the maximum torque is independent
of l. In part (c) the maximum torque depends on both h and l.
IDENTIFY: Apply ∑τ z = Iα z , where τ z is due to the gravity force on the object.
SET UP: I = I rod + I clay . I rod = 1 ML2 . In part (b), I clay = ML2 . In part (c), I clay = 0 .
3
(a) A distance L 4 from the end with the clay. EXECUTE: (b) In this case I = (4 3) ML2 and the gravitational torque is (3L 4)(2Mg )sin θ = (3Mg L 2)sin θ , so α = (9 g 8L)sin θ . 10.61. (c) In this case I = (1 3) ML2 and the gravitational torque is ( L 4)(2Mg )sin θ = ( Mg L 2)sin θ , so α = (3 g 2 L)sin θ .
This is greater than in part (b).
(d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome
deviations from the vertical.
EVALUATE: In part (b), I is 4 times larger than in part (c) and τ is 3 times larger. α = τ / I , so the net effect is
that α is smaller in (b) than in (c).
IDENTIFY: Calculate W using the procedure specified in the problem. In part (c) apply the work-energy theorem.
In part (d), atan = Rα and ∑τ z = Iα z . arad = Rω 2 .
SET UP: Let θ be the angle the disk has turned through. The moment arm for F is R cosθ . EXECUTE: (a) The torque is τ = FR cosθ . W = ∫ π2
0 FR cosθ dθ = FR . (b) In Eq.(6.14), dl is the horizontal distance the point moves, and so W = F ∫ dl = FR , the same as part (a).
(c) From K 2 = W = (MR 2 4)ω 2 , ω = 4 F MR .
(d) The torque, and hence the angular acceleration, is greatest when θ = 0, at which point α = (τ I ) = 2 F MR , and so the maximum tangential acceleration is 2 F M .
(e) Using the value for ω found in part (c), arad = ω 2 R = 4 F M .
EVALUATE: 10.62. atan = ω 2 R is maximum initially, when the moment arm for F is a maximum, and it is zero after the disk has rotated one-quarter of a revolution. arad is zero initially and is a maximum at the end of the motion, after
the disk has rotated one-quarter of a revolution.
!
!
IDENTIFY: Apply ∑ F = ma to the crate and ∑τ z = Iα z to the cylinder. The motions are connected by
a (crate) = Rα (cylinder).
SET UP: The force diagram for the crate is given in Figure 10.62a.
EXECUTE: ∑F y = ma y T − mg = ma
T = m( g + a ) = 50 kg(9.80 m/s 2 + 0.80 m/s 2 ) = 530 N
Figure 10.62a Dynamics of Rotational Motion SET UP: 10-21 The force diagram for the cylinder is given in Figure 10.62b. EXECUTE: ∑τ z = Iα z Fl − TR = Iα z , where l = 0.12 m and R = 0.25 m
a = Rα so α z = a / R
Fl = TR + Ia / R Figure 10.62b
2
2
⎛ R ⎞ Ia
⎛ 0.25 m ⎞ (2.9 kg ⋅ m )(0.80 m/s )
= 530 N ⎜
+
= 1200 N
F =T⎜ ⎟+
⎟
(0.25 m)(0.12 m)
⎝ l ⎠ Rl
⎝ 0.12 m ⎠ 10.63. EVALUATE: The tension in the rope is greater than the weight of the crate since the crate accelerates upward. If F
were applied to the rim of the cylinder (l = 0.25 m), it would have the value F = 567 N. This is greater than T
because it must accelerate the cylinder as well as the crate. And F is larger than this because it is applied closer to
the axis than R so has a smaller moment arm and must be larger to give the same torque.
!
!
IDENTIFY: Apply ∑ Fext = macm and ∑τ z = I cmα z to the roll.
SET UP: At the point of contact, the wall exerts a friction force f directed downward and a normal force n
directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero.
EXECUTE: (a) Balancing vertical forces, Frod cosθ = f + w + F , and balancing horizontal forces Frod sin θ = n. With f = μk n, these equations become Frod cosθ = μk n + F + w, Frod sin θ = n. Eliminating n and solving for Frod gives
Frod = w+ F
(16.0 kg) (9.80 m/s 2 ) + (40.0 N)
=
= 266 N.
cosθ − μ k sin θ
cos 30° − (0.25)sin 30° (b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net
torque is ( F − f ) R, and f = μk n may be found by insertion of the value found for Frod into either of the above relations; i.e., f = μk Frod sin θ = 33.2 N. Then, α= 10.64. τ
I = (40.0 N − 31.54 N)(18.0 × 10−2 m)
= 4.71 rad/s 2 .
(0.260 kg ⋅ m 2 ) EVALUATE: If the applied force F is increased, Frod increases and this causes n and f to increase. The angle
φ changes as the amount of paper unrolls and this affects α for a given F.
!
!
IDENTIFY: Apply ∑τ z = Iα z to the flywheel and ∑ F = ma to the block. The target variables are the tension in the string and the acceleration of the block.
(a) SET UP: Apply ∑τ z = Iα z to the rotation of the flywheel about the axis. The free-body diagram for the
flywheel is given in Figure 10.64a.
EXECUTE: The forces n and Mg act
at the axis so have zero torque.
∑τ z = TR TR = Iα z
Figure 10.64a 10-22 Chapter 10 Apply SET UP: ! ! ∑ F = ma to the translational motion of the block. The free-body diagram for the block is given in Figure 10.64b. EXECUTE: ∑F y = ma y n − mg cos36.9° = 0
n = mg cos36.9°
f k = μ k n = μ k mg cos36.9° Figure 10.64b ∑F x = max mg sin 36.9° − T − μ k mg cos36.9° = ma
mg (sin 36.9° − μ k cos36.9°) − T = ma But we also know that ablock = Rα wheel , so α = a / R. Using this in the
T = ( I / R ) a. Use this to replace T in the
2 ∑F x ∑τ z = Iα z equation gives TR = Ia / R and = max equation: mg (sin 36.9° − μ k cos36.9°) − ( I / R )a = ma
2 mg (sin 36.9° − μ k cos36.9°)
m + I / R2
(5.00 kg)(9.80 m/s 2 )(sin 36.9° − (0.25)cos36.9°)
a=
= 1.12 m/s 2
5.00 kg + 0.500 kg ⋅ m 2 /(0.200 m) 2 a= 0.500 kg ⋅ m 2
(1.12 m/s 2 ) = 14.0 N
(0.200 m)2
EVALUATE: If the string is cut the block will slide down the incline with
a = g sin 36.9° − μ k g cos36.9° = 3.92 m/s 2 . The actual acceleration is less than this because mg sin 36.9° must also
(b) T = 10.65. accelerate the flywheel. mg sin 36.9° − f k = 19.6 N. T is less than this; there must be more force on the block
directed down the incline than up then incline since the block accelerates down the incline.
!
!
IDENTIFY: Apply ∑ F = ma to the block and ∑τ z = Iα z to the combined disks.
SET UP: For a disk, I disk = 1 MR 2 , so I for the disk combination is I = 2.25 × 10−3 kg ⋅ m 2 .
2 For a tension T in the string, mg − T = ma and TR = Iα = I a . Eliminating T and solving for a gives
R
m
g
a=g
=
, where m is the mass of the hanging block and R is the radius of the disk to which the
m + I / R 2 1 + I / mR 2
string is attached.
(a) With m = 1.50 kg and R = 2.50 × 10−2 m, a = 2.88 m/s 2 .
EXECUTE: 10.66. (b) With m = 1.50 kg and R = 5.00 × 10−2 m, a = 6.13 m/s 2 .
The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable
of applying a larger torque.
EVALUATE: ω = v / R , where v is the speed of the block and ω is the angular speed of the disks. When R is
larger, in part (b), a smaller fraction of the kinetic energy resides with the disks. The block gains more speed as it
falls a certain distance and therefore has a larger acceleration.
!
!
IDENTIFY: Apply both ∑ F = ma and ∑τ z = Iα z to the motion of the roller. Rolling without slipping means acm = Rα . Target variables are acm and f. Dynamics of Rotational Motion SET UP: 10-23 The free-body diagram for the roller is given in Figure 10.66. EXECUTE: Apply ! ! ∑ F = ma to the translational motion of the center of mass:
∑ Fx = max
F − f = Macm Figure 10.66 Apply ∑τ z ∑τ z = Iα z to the rotation about the center of mass: = fR thin-walled hollow cylinder: I = MR 2
Then ∑τ z = Iα z implies fR = MR 2α .
But α cm = Rα , so f = Macm .
Using this in the 10.67. ∑F x = max equation gives F − Macm = Macm acm = F / 2 M , and then f = Macm = M ( F / 2M ) = F / 2.
EVALUATE: If the surface were frictionless the object would slide without rolling and the acceleration would be
acm = F / M . The acceleration is less when the object rolls.
!
!
IDENTIFY: Apply ∑ F = ma to each object and apply ∑τ z = Iα z to the pulley.
SET UP: Call the 75.0 N weight A and the 125 N weight B. Let TA and TB be the tensions in the cord to the left and to the right of the pulley. For the pulley, I = 1 MR 2 , where Mg = 50.0 N and R = 0.300 m . The 125 N weight
2
accelerates downward with acceleration a, the 75.0 N weight accelerates upward with acceleration a and the pulley
rotates clockwise with angular acceleration α , where a = Rα .
!
!
!
!
EXECUTE: ∑ F = ma applied to the 75.0 N weight gives TA − wA = mAa . ∑ F = ma applied to the 125.0 N
weight gives wB − TB = mB a . ∑τ z = Iα z applied to the pulley gives (TB − TA ) R = ( 1 MR 2 )α z and TB − TA = 1 M .
2
2 Combining these three equations gives wB − wA = (m A + mB + M / 2)a and
⎛
⎞
wB − wA
125 N − 75.0 N
⎛
⎞
a =⎜
g =⎜
⎟ g = 0.222 g . TA = wA (1 + a / g ) = 1.222 wA = 91.65 N .
⎜ wA + wB + wpulley / 2 ⎟
⎟
⎝ 75.0 N + 125 N + 25.0 N ⎠
⎝
⎠
!
!
TB = wB (1 − a / g ) = 0.778wB = 97.25 N . ∑ F = ma applied to the pulley gives that the force F applied by the hook
to the pulley is F = TA + TB + wpulley = 239 N . The force the ceiling applies to the hook is 239 N. 10.68. EVALUATE: The force the hook exerts on the pulley is less than the total weight of the system, since the net
effect of the motion of the system is a downward acceleration of mass.
!
!
IDENTIFY: This problem can be done either with conservation of energy or with ∑ Fext = ma. We will do it both ways.
(a) SET UP: (1) Conservation of energy: K1 + U1 + Wother = K 2 + U 2 .
Take position 1 to be the location of the disk
at the base of the ramp and 2 to be where the
disk momentarily stops before rolling back
down, as shown in Figure 10.68a.
Figure 10.68a Take the origin of coordinates at the center of the disk at position 1 and take + y to be upward. Then y1 = 0 and
y2 = d sin 30°, where d is the distance that the disk rolls up the ramp. “Rolls without slipping” and neglect rolling friction says W f = 0; only gravity does work on the disk, so Wother = 0 10-24 Chapter 10 EXECUTE: U1 = Mgy1 = 0 K1 = Mv + 1 I cmω12 (Eq.10.11). But ω1 = v1 / R and I cm = 1 MR 2 , so
2
2
1
2 2
1 1
2 I cmω12 = 1 ( 1 MR 2 ) (v1 / R) 2 = 1 Mv12 . Thus
22
4 K1 = 1 Mv12 + 1 Mv12 = 3 Mv12 .
2
4
4
U 2 = Mgy2 = Mgd sin 30°
K 2 = 0 (disk is at rest at point 2). Thus 3
4 Mv12 = Mgd sin 30° 3v12
3(2.50 m/s) 2
=
= 0.957 m
4 g sin 30° 4(9.80 m/s 2 )sin 30°
SET UP: (2) force and acceleration The free-body diagram is given in Figure 10.68b.
d= Apply EXECUTE: ∑F x = max to the translational motion of the center of mass:
Mg sin θ − f = Macm
Apply ∑τ z = Iα z to the rotation about the center of mass:
f R = ( 1 MR 2 )α z
2
f = 1 MRα z
2
Figure 10.68b But acm = Rα in this equation gives f = 1 Macm . Use this in the
2 ∑F x = max equation to eliminate f. Mg sin θ − 1 Macm = Macm
2
2
M divides out and 3 acm = g sin θ . acm = 2 g sin θ = 3 (9.80 m/s 2 )sin 30° = 3.267 m/s 2
3
2
SET UP: Apply the constant acceleration equations to the motion of the center of mass. Note that in our
coordinates the positive x-direction is down the incline.
v0 x = −2.50 m/s (directed up the incline); ax = +3.267 m/s 2 ; vx = 0 (momentarily comes to rest); x − x0 = ?
2
2
vx = v0 x + 2ax ( x − x0 )
2
v0 x
( −2.50 m/s) 2
=−
= −0.957 m
2a x
2(3.267 m/s 2 )
(b) EVALUATE: The results from the two methods agree; the disk rolls 0.957 m up the ramp before it stops.
The mass M enters both in the linear inertia and in the gravity force so divides out. The mass M and radius R enter
in both the rotational inertia and the gravitational torque so divide out.
!
!
IDENTIFY: Apply ∑ Fext = macm to the motion of the center of mass and apply ∑τ z = I cmα z to the rotation about EXECUTE: 10.69. x − x0 = − the center of mass.
SET UP: I = 2 ( 1 MR 2 ) = MR 2 . The moment arm for T is b.
2
EXECUTE: The tension is related to the acceleration of the yo-yo by (2m) g − T = (2m)a, and to the angular acceleration by Tb = Iα = I a . Dividing the second equation by b and adding to the first to eliminate T yields
b
2m
2
2
=g
. The tension is found by substitution into either of the two
, α=g
a=g
(2m + I b 2 )
2 + ( R b) 2
2b + R 2 b
equations:
⎛
⎞
2
( R b) 2
2mg
=
.
T = (2m)( g − a ) = (2mg ) ⎜1 −
⎟ = 2mg
2
2
2 + ( R b)
(2(b R) 2 + 1)
⎝ 2 + ( R b) ⎠
EVALUATE: a → 0 when b → 0 . As b → R , a → 2 g / 3 . Dynamics of Rotational Motion 10.70. 10-25 IDENTIFY: Apply conservation of energy to the motion of the shell, to find its linear speed v at points A and B.
!
!
Apply ∑ F = ma to the circular motion of the shell in the circular part of the track to find the normal force exerted by the track at each point. Since r << R the shell can be treated as a point mass moving in a circle of radius R when
!
!
applying ∑ F = ma . But as the shell rolls along the track, it has both translational and rotational kinetic energy.
SET UP: K1 + U1 = K 2 + U 2 . Let 1 be at the starting point and take y = 0 to be at the bottom of the track, so y1 = h0 . K = 1 mv 2 + 1 I ω 2 . I = 2 mr 2 and ω = v / r , so K = 5 mv 2 . During the circular motion, arad = v 2 / R .
2
2
3
6
!
!
v2
EXECUTE: (a) ∑ F = ma at point A gives n + mg = m . The minimum speed for the shell not to fall off the
R
2
track is when n → 0 and v 2 = gR . Let point 2 be A, so y2 = 2 R and v2 = mR . Then K1 + U1 = K 2 + U 2 gives
mgh0 = 2mgR + 5 m( gR) . h0 = (2 + 5 ) R = 17 R .
6
6
6 10.71. 2
(b) Let point 2 be B, so y2 = R . Then K1 + U1 = K 2 + U 2 gives mgh0 = mgR + 5 mv2 . With h = 167 R this gives
6
!
!
v2
v 2 = 11 gR . Then ∑ F = ma at B gives n = m = 11 mg .
5
R5
(c) Now K = 1 mv 2 instead of 5 mv 2 . The shell would be moving faster at A than with friction and would still make
2
6
the complete loop.
!
!
v2
(d) In part (c): mgh0 = mg (2 R ) + 1 mv 2 . h0 = 167 R gives v 2 = 5 gR . ∑ F = ma at point A gives mg + n = m and
2
3
R
2
⎛v
⎞2
n = m ⎜ − g ⎟ = 3 mg . In part (a), n = 0 , since at this point gravity alone supplies the net downward force that is
⎝R
⎠
required for the circular motion.
EVALUATE: The normal force at A is greater when friction is absent because the speed of the shell at A is greater
when friction is absent than when there is rolling without slipping.
IDENTIFY: Consider the direction of the net force and the sense of the net torque in each case.
SET UP: The free-body diagram in each case is shown in Figure 10.71.
→ EXECUTE: In the first case, F and the friction force act in opposite directions, and the friction force causes a
larger torque to tend to rotate the yo-yo to the right. The net force to the right is the difference F − f , so the net
force is to the right while the net torque causes a clockwise rotation. For the second case, both the torque and the
friction force tend to turn the yo-yo clockwise, and the yo-yo moves to the right. In the third case, friction tends to
move the yo-yo to the right, and since the applied force is vertical, the yo-yo moves to the right.
EVALUATE: In the first case the torque due to friction must be larger than the torque due to F, so the net torque is
clockwise. In the third case the torque due to F must be larger than the torque due to f, so the net torque will be clockwise. 10.72. IDENTIFY: Figure 10.71
!
!
Apply ∑ Fext = macm to the motion of the center of mass and ∑τ z = I cmα z to the rotation about the center of mass.
SET UP: For a hoop, I = MR 2 . For a solid disk, I = 1 MR 2 .
2
EXECUTE: (a) Because there is no vertical motion, the tension is just the weight of the hoop:
T = Mg = ( 0.180 kg ) ( 9.8 N kg ) = 1.76 N .
(b) Use τ = Iα to find α . The torque is RT , so α = RT / I = RT MR 2 = T / MR = Mg MR , so α = g/R = (9.8 m/s 2 ) (0.08 m) = 122.5 rad/s 2 .
(c) a = Rα = 9.8 m s 2
(d) T would be unchanged because the mass M is the same, α and a would be twice as great because I is now
EVALUATE: atan for a point on the rim of the hoop or disk equals a for the free end of the string. Since I is
smaller for the disk, the same value of T produces a greater angular acceleration. 1
2 MR 2 . 10-26 Chapter 10 10.73. IDENTIFY: Apply ∑τ z = Iα z to the cylinder or hoop. Find a for the free end of the cable and apply constant acceleration equations.
SET UP: atan for a point on the rim equals a for the free end of the cable, and atan = Rα .
EXECUTE: (a) ∑τ z 1
1
⎛a
= Iα z and atan = Rα gives FR = MR 2α = MR 2 ⎜ tan
2
2
⎝R Distance the cable moves: x − x0 = v0 xt + 1 axt 2 gives 50 m =
2
vx = v0 x + axt = 0 + ( 50 m/s 2 ) (1.41 s ) = 70.5 m s . 2F
200 N
⎞
=
= 50 m/s 2 .
⎟ . atan =
M 4.00 kg
⎠ 1
( 50 m/s2 ) t 2 and t = 1.41 s .
2 (b) For a hoop, I = MR 2 , which is twice as large as before, so α and atan would be half as large. Therefore the time would be longer by a factor of 10.74. 2
2
2 . For the speed, vx = v0 x + 2ax x, in which x is the same, so vx would be half as large since ax is smaller.
EVALUATE: The acceleration a that is produced depends on the mass of the object but is independent of its
radius. But a depends on how the mass is distributed and is different for a hoop versus a cylinder.
IDENTIFY: Use projectile motion to find the speed v the marble needs at the edge of the pit to make it to the level
ground on the other side. Apply conservation of energy to the motion down the hill in order to relate the initial
height to the speed v at the edge of the pit. Wother = 0 so conservation of energy gives K i + U i = K f + U f .
SET UP: In the projectile motion the marble must travel 36 m horizontally while falling vertically 20 m. Let + y
be downward. For the motion down the hill, let yf = 0 so U f = 0 and yi = h . K i = 0 . Rolling without slipping
7
2
means v = Rω . K = 1 I cmω 2 + 1 mv 2 = 1 ( 5 mR 2 )ω 2 + 1 mv 2 = 10 mv 2 .
2
2
2
2 EXECUTE: t= (a) Projectile motion: v0 y = 0 . a y = 9.80 m/s 2 . y − y0 = 20 m . y − y0 = v0 yt + 1 a yt 2 gives
2 2( y − y0 )
x − x0 36 m
=
= 17.8 m/s .
= 2.02 s . Then x − x0 = v0 xt gives v = v0 x =
t
2.02 s
ay 7
Motion down the hill: U i = K f . mgh = 10 mv 2 . h = 7v 2
7(17.8 m/s) 2
=
= 22.6 m .
10 g 10(9.80 m/s 2 ) (b) 1 I ω 2 = 1 mv 2 , independent of R. I is proportional to R 2 but ω 2 is proportional to 1/ R 2 for a given
2
5
translational speed v.
(c) The object still needs v = 17.8 m/s at the bottom of the hill in order to clear the pit. But now K f = 1 mv 2 and
2 v2
= 16.6 m .
2g
EVALUATE: The answer to part (a) also does not depend on the mass of the marble. But, it does depend on how
the mass is distributed within the object. The answer would be different if the object were a hollow spherical shell.
In part (c) less height is needed to give the object the same translational speed because in (c) none of the energy
goes into rotational motion.
IDENTIFY: Apply conservation of energy to the motion of the boulder.
2
SET UP: K = 1 mv 2 + 1 I ω 2 and v = Rω when there is rolling without slipping. I = 5 mR 2 .
2
2
EXECUTE: Break into 2 parts, the rough and smooth sections.
2
1
1⎛ 2
10
⎞⎛ v ⎞
Rough: mgh1 = 1 mv 2 + 1 I ω 2 . mgh1 = mv 2 + ⎜ mR 2 ⎟⎜ ⎟ . v 2 = gh1 .
2
2
2
2⎝ 5
7
R⎠
⎠⎝
h= 10.75. 1 ⎛ 10
1
12
⎞ 12
Smooth: Rotational kinetic energy does not change. mgh2 + mv 2 + K rot = mvBottom + K rot . gh2 + ⎜ gh1 ⎟ = vB .
2⎝ 7
2
2
⎠2
vB = 10
10
gh1 + 2 gh2 =
(9.80 m/s 2 )(25 m) + 2(9.80 m/s 2 )(25 m) = 29.0 m/s .
7
7 10
g (50 m) = 26.5 m/s . A
7
smaller fraction of the initial gravitational potential energy goes into translational kinetic energy of the center of
mass than if part of the hill is smooth. If the entire hill is smooth and the boulder slides without slipping,
vB = 2 g (50 m) = 31.3 m/s . In this case all the initial gravitational potential energy goes into the kinetic energy of
the translational motion.
EVALUATE: If all the hill was rough enough to cause rolling without slipping, vB = Dynamics of Rotational Motion 10.76. 10-27 IDENTIFY: Apply conservation of energy to the motion of the ball as it rolls up the hill. After the ball leaves the
edge of the cliff it moves in projectile motion and constant acceleration equations can be used.
(a) SET UP: Use conservation of energy to find the speed v2 of the ball just before it leaves the top of the cliff.
Let point 1 be at the bottom of the hill and point 2 be at the top of the hill. Take y = 0 at the bottom of the hill, so y1 = 0 and y2 = 28.0 m.
K1 = U1 = K 2 + U 2 EXECUTE:
1
2 2
2
mv + I ω = mgy2 + 1 mv2 + 1 I ω2
2
2
2
1 1
2 2
1 Rolling without slipping means ω = v / r and
7
10 1
2 2
I ω 2 = 1 ( 5 mr 2 ) (v / r ) 2 = 1 mv 2
2
5 2
7
mv12 = mgy2 + 10 mv2 v2 = v12 − 10 gy2 = 15.26 m/s
7
SET UP: Consider the projectile motion of the ball, from just after it leaves the top of the cliff until just before it
lands. Take + y to be downward. Use the vertical motion to find the time in the air: v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 28.0 m, t = ?
EXECUTE: y − y0 = v0 yt + 1 a yt 2 gives t = 2.39 s
2 During this time the ball travels horizontally
x − x0 = v0 xt = (15.26 m/s)(2.39 s) = 36.5 m.
Just before it lands, v y = v0 y + a yt = 23.4 m/s and vx = v0 x = 15.3 m/s
2
2
v = vx + v y = 28.0 m/s 10.77. (b) EVALUATE: At the bottom of the hill, ω = v / r = (25.0 m/s) / r. The rotation rate doesn’t change while the ball
is in the air, after it leaves the top of the cliff, so just before it lands ω = (15.3 m/s) / r . The total kinetic energy is
the same at the bottom of the hill and just before it lands, but just before it lands less of this energy is rotational
kinetic energy, so the translational kinetic energy is greater.
IDENTIFY: Apply conservation of energy to the motion of the wheel. K = 1 mv 2 + 1 I ω 2 .
2
2
SET UP: No slipping means that ω = v R. Uniform density means mr = λ 2π R and ms = λ R , where mr is the mass of the rim and ms is the mass of each spoke. For the wheel, I = I rim + I spokes . For each spoke, I = 1 ms R 2 .
3
EXECUTE: 1
1
⎛1
⎞
(a) mgh = mv 2 + I ω 2 . I = I rim + I spokes = mr R 2 + 6 ⎜ ms R 2 ⎟
2
2
3
⎝
⎠ Also, m = mr + ms = 2π Rλ + 6 Rλ = 2 Rλ (π + 3) . Substituting into the conservation of energy equation gives
2 Rλ (π + 3) gh = ω= 10.78. 1
1⎡
1
⎤
2
( 2 Rλ )(π + 3)( Rω ) + ⎢ 2π Rλ R 2 + 6 ⎛ π RR 2 ⎞ ⎥ ω 2 .
⎜
⎟
2
2⎣
⎝3
⎠⎦ (π + 3) gh = (π + 3) ( 9.80 m s 2 ) ( 58.0 m ) = 124
2
R 2 (π + 2 )
( 0.210 m ) (π + 2 ) rad s and v = Rω = 26.0 m s (b) Doubling the density would have no effect because it does not appear in the answer. ω is inversely proportional
to R so doubling the diameter would double the radius which would reduce ω by half, but v = Rω would be
unchanged.
EVALUATE: Changing the masses of the rim and spokes by different amounts would alter the speed v at the
bottom of the hill.
IDENTIFY: Apply v = Rω .
SET UP: For the antique bike, v is the same for points on the rim of each wheel and equals the linear speed of the
bike. 1 rev = 2π rad .
EXECUTE: (a) The front wheel is turning at ω = 1.00 rev s = 2π rad s. v = rω = (0.330 m)(2π rad s) = 2.07 s .
(b) ω = v r = ( 2.07 m s) (0.655 m) = 3.16 rad s = 0.503 rev s
(c) ω = v r = ( 2.07 m s) (0.220 m) = 9.41 rad s = 1.50 rev s
EVALUATE: Since the front wheel has a larger radius for the antique bike, that wheel doesn't have to rotate at as
many rev/s to achieve the same linear speed of the bike. 10-28 Chapter 10 10.79. IDENTIFY: Apply conservation of energy to the motion of the ball. Once the ball leaves the track the ball moves
in projectile motion.
2
SET UP: The ball has I = 5 mR 2 ; the silver dollar has I = 1 mR 2 . For the projectile motion take + y downward,
2 so ax = 0 and a y = + g .
EXECUTE: (a) The kinetic energy of the ball when it leaves the track (when it is still rolling without slipping) is
(7 10) mv 2 and this must be the work done by gravity, W = mgh , so v = 10 gh 7. The ball is in the air for a time 10.80. t = 2 y g , so x = vt = 20hy 7.
(b) The answer does not depend on g, so the result should be the same on the moon.
(c) The presence of rolling friction would decrease the distance.
(e) For the dollar coin, modeled as a uniform disc, K = (3 4)mv 2 , and so x = 8hy 3.
EVALUATE: The sphere travels a little farther horizontally, because its moment of inertia is a smaller fraction of
MR 2 than for the disk. The result is independent of the mass and radius of the object but it does depend on how
that mass is distributed within the object.
IDENTIFY and SET UP: Apply conservation of energy to the motion of the ball. The ball ends up with both
translational and rotational kinetic energy. Use Fig.(10.13) in the textbook to relate the speed of different points on
the ball to vcm .
(a) U el = 1 kx 2 = 1 (400 N ⋅ m)(0.15 m)2 = 4.50 J and K1 = 0.800U el = 3.60 J
2
2 EXECUTE: 2
K1 = 1 mvcm + 1 I cmω 2 rolling without slipping says ω = vcm / R
2
2
2
I cm = 5 mR 2 2
2
2
7
2
Thus K1 = 1 mvcm + 1 ( 5 mR 2 ) (vcm / R) 2 = mvcm ( 1 + 1 ) = 10 mvcm
2
2
2
5 10 K1
10(3.60 J)
=
= 9.34 m/s.
7m
7(0.0590 kg) and vcm = (b) Consider Figure 10.80a. From Fig.(10.13) in the textbook,
at the top of the ball
v = 2vcm = 18.7 m/s
Figure 10.80a
(c) From Fig.(10.13) in the textbook,
v = 0 at the bottom of the ball.
Figure 10.80b
(d) The problem says that U 2 = 0.900 K1 = 3.24 J. Thus U 2 = mgh = 3.24 J and h= 10.81. 3.24 J
3.24 J
=
= 5.60 m
mg
(0.0590 kg)(9.80 m/s 2 ) EVALUATE: Not all the potential energy stored in the spring goes into kinetic energy at the base of the ramp or
into gravitational potential energy at the top of the ramp because of loss of mechanical energy due to negative
2
work done by friction. If the ball slides without rolling, then K1 = 1 mvcm and vcm = 11.0 m/s. vcm is less than this
2
when the ball rolls and some of its total kinetic energy is rotational.
IDENTIFY: vx = dx / dt , v y = dy / dt . ax = dvx / dt , a y = dv y / dt .
SET UP: d cos(ωt ) / dt = −ω sin(ωt ) . d sin(ωt ) / dt = ω cos(ωt ) .
EXECUTE: (a) The sketch is shown in Figure 10.81.
(b) R is the radius of the wheel (y varies from 0 to 2R) and T is the period of the wheel’s rotation.
2
2π R ⎡
2π R ⎛ 2π t ⎞
⎛ 2π ⎞
⎛ 2π t ⎞
⎛ 2π t ⎞ ⎤
(c) Differentiating, vx =
1 − cos ⎜
, ax = ⎜
sin ⎜
⎟
⎟,
⎟ R sin ⎜
⎟ and v y =
T⎢
T ⎠⎥
T⎠
T⎠
T
⎝T⎠
⎝
⎝
⎝
⎣
⎦
2 ⎛ 2π ⎞
⎛ 2π t ⎞
ay = ⎜
⎟ R cos ⎜
⎟.
⎝T ⎠
⎝T⎠ Dynamics of Rotational Motion 10-29 ⎛ 2π t ⎞
(d) vx = v y = 0 when ⎜
⎟ = 2π or any multiple of 2π , so the times are integer multiples of the period T. The
⎝T⎠
acceleration components at these times are ax = 0, a y = 4π 2 R
.
T2 2 4π 2 R
⎛ 2π ⎞
2
2
2 ⎛ 2π t ⎞
2 ⎛ 2π t ⎞
(e) a = ax + a y = ⎜
, independent of time. This is the magnitude of the
⎟ R cos ⎜
⎟ + sin ⎜
⎟=
T2
⎝T ⎠
⎝T⎠
⎝T⎠
radial acceleration for a point moving on a circle of radius R with constant angular velocity 2π / T . For motion that
!
consists of this circular motion superimposed on motion with constant velocity ( a = 0 ) , the acceleration due to the
circular motion will be the total acceleration.
EVALUATE: a is independent of time, but v does depend on time. Figure 10.81
10.82. IDENTIFY:
SET UP:
EXECUTE: Apply the work-energy theorem to the motion of the basketball. K = 1 mv 2 + 1 I ω 2 and v = Rω .
2
2
2
For a thin-walled, hollow sphere I = 3 mR 2 . For rolling without slipping, the kinetic energy is (1 2 ) ( m + I R 2 ) v 2 = ( 5 6 ) mv 2 ; initially, this is 32.0 J and at the return to the bottom it is 8.0 J. Friction has done −24.0 J of work, −12.0 J each going up and
down. The potential energy at the highest point was 20.0 J, so the height above the ground was
20.0 J
= 3.40 m.
( 0.600 kg ) ( 9.80 m s2 ) 10.83. EVALUATE: All of the kinetic energy of the basketball, translational and rotational, has been removed at the point
where the basketball is at its maximum height up the ramp.
IDENTIFY: Use conservation of energy to relate the speed of the block to the distance it has descended. Then use
a constant acceleration equation to relate these quantities to the acceleration.
SET UP: For the cylinder, I = 1 M (2 R) 2 , and for the pulley, I = 1 MR 2 .
2
2
EXECUTE: Doing this problem using kinematics involves four unknowns (six, counting the two angular
accelerations), while using energy considerations simplifies the calculations greatly. If the block and the cylinder
both have speed v, the pulley has angular velocity v/R and the cylinder has angular velocity v/2R, the total kinetic
energy is ⎤3
1⎡
M (2 R) 2
MR 2
K = ⎢ Mv 2 +
(v 2 R ) 2 +
(v R ) 2 + Mv 2 ⎥ = Mv 2 .
2⎣
2
2
⎦2
This kinetic energy must be the work done by gravity; if the hanging mass descends a distance y, K = Mgy, or 10.84. v 2 = (2 3) gy. For constant acceleration, v 2 = 2ay, and comparison of the two expressions gives a = g 3.
EVALUATE: If the pulley were massless and the cylinder slid without rolling, Mg = 2Ma and a = g / 2 . The
rotation of the objects reduces the acceleration of the block.
IDENTIFY: Apply ∑τ z = Iα z to the drawbridge and calculate α z . For part (c) use conservation of energy.
SET UP:
EXECUTE: The free-body diagram for the drawbridge is given in Fig.10.84. For an axis at the lower end, I = 1 ml 2 .
3
(a) ∑τ z = Iα z gives mg (4.00 m)(cos60.0°) = 1 ml 2α z and α z =
3 3 g (4.00 m)(cos60.0°)
= 0.919 rad/s 2 .
(8.00 m) 2 (b) α z depends on the angle the bridge makes with the horizontal. α z is not constant during the motion and ω z = ω0 z + α z t cannot be used.
(c) Use conservation of energy. Take y = 0 at the lower end of the drawbridge, so yi = (4.00 m)(sin 60.0°) and yf = 0 . K f + U f = K i + U i + Wother gives U i = K f , mgyi = 1 I ω 2 . mgyi = 1 ( 1 ml 2 )ω 2 and
2
23 ω= 6 gyi
6(9.80 m/s 2 )(4.00 m)(sin 60.0°)
=
= 1.78 rad/s .
l
8.00 m 10-30 Chapter 10 EVALUATE: If we incorrectly assume that α z is constant and has the value calculated in part (a), then ω = ω + 2α z (θ − θ 0 ) gives ω = 139 rad/s . The angular acceleration increases as the bridge rotates and the actual
angular velocity is larger than this.
2
z 2
0z Figure 10.84
10.85. IDENTIFY: Apply conservation of energy to the motion of the first ball before the collision and to the motion of
the second ball after the collision. Apply conservation of angular momentum to the collision between the first ball
and the bar.
SET UP: The speed of the ball just before it hits the bar is v = 2 gy = 15.34 m/s. Use conservation of angular
momentum to find the angular velocity ω of the bar just after the collision. Take the axis at the center of the bar.
EXECUTE: L1 = mvr = (5.00 kg)(15.34 m/s)(2.00 m) = 153.4 kg ⋅ m 2
Immediately after the collision the bar and both balls are rotating together.
L2 = I totω
1
1
I tot = 12 Ml 2 + 2mr 2 = 12 (8.00 kg)(4.00 m) 2 + 2(5.00 kg)(2.00 m) 2 = 50.67 kg ⋅ m 2 L2 = L1 = 153.4 kg ⋅ m 2 ω = L2 / I tot = 3.027 rad/s
Just after the collision the second ball has linear speed v = rω = (2.00 m)(3.027 rad/s) = 6.055 m/s and is moving 10.86. 10.87. upward. 1 mv 2 = mgy gives y = 1.87 m for the height the second ball goes.
2
EVALUATE: Mechanical energy is lost in the inelastic collision and some of the final energy is in the rotation of the bar
with the first ball stuck to it. As a result, the second ball does not reach the height from which the first ball was dropped.
IDENTIFY: The rings and the rod exert forces on each other, but there is no net force or torque on the system, and
so the angular momentum will be constant.
1
SET UP: For the rod, I = 12 ML2 . For each ring, I = mr 2 , where r is their distance from the axis.
EXECUTE: (a) As the rings slide toward the ends, the moment of inertia changes, and the final angular velocity is
⎡ 1 ML2 + 2mr12 ⎤
5.00 × 10−4 kg ⋅ m 2 ω1
I
given by ω2 = ω1 1 = ω1 ⎢ 12
= ω1
= , so ω2 = 7.5 rev min.
2
2⎥
1
2.00 × 10−3 kg ⋅ m 2 4
I2
⎣ 12 ML + 2mr2 ⎦
(b) The forces and torques that the rings and the rod exert on each other will vanish, but the common angular
velocity will be the same, 7.5 rev/min.
EVALUATE: Note that conversion from rev/min to rad/s was not necessary. The angular velocity of the rod
decreases as the rings move away from the rotation axis.
IDENTIFY: Apply conservation of angular momentum to the collision. Linear momentum is not conserved
because of the force applied to the rod at the axis. But since this external force acts at the axis, it produces no
torque and angular momentum is conserved.
SET UP: The system before and after the collision is sketched in Figure 10.87.
EXECUTE: (a) mb = 1 mrod
4
EXECUTE: L1 = mbvr = 1 mrod v ( L / 2)
4 L1 = 1 mrod vL
8
L2 = ( I rod + I b )ω
I rod = 1 mrod L2
3
I b = mb r 2 = 1 mrod ( L / 2) 2
4
1
I b = 16 mrod L2 Figure 10.87 Dynamics of Rotational Motion 10-31 1
Thus L1 = L2 gives 1 mrod vL = ( 1 mrod L2 + 16 mrod L2 ) ω
8
3
1
8 v = 19 Lω
48 6
ω = 19 v / L (b) K1 = 1 mv 2 = 1 mrod v 2
2
8 K 2 = 1 I ω 2 = 1 ( I rod + I b )ω 2 =
2
2 1
2 ( 1
3 1
mrod L2 + 16 mrod L2 ) (6v /19 L) 2 6
3
K 2 = 1 ( 19 ) ( 19 ) mrod v 2 = 152 mrod v 2
2 48
2 Then 10.88. K2
=
K1 EVALUATE: The collision is inelastic and K 2 < K1.
IDENTIFY: Apply Eq.(10.29).
SET UP: The door has I = 1 ml 2 . The torque applied by the force is rFav , where r = l / 2 .
3
EXECUTE: ω= 10.89. mrod v 2
= 3/19.
mrod v 2 3
152
1
8 Στ av = rFav, and ΔL = rFav Δt = rJ . The angular velocity ω is then ΔL rFav Δt ( l 2 ) Fav Δt 3 Fav Δt
=
=12=
, where l is the width of the door. Substitution of the given numeral
2 ml
I
I
ml
3 values gives ω = 0.514rad s.
EVALUATE: The final angular velocity of the door is proportional to both the magnitude of the average force and
also to the time it acts.
(a) IDENTIFY: Apply conservation of angular momentum to the collision between the bullet and the board:
SET UP: The system before and after the collision is sketched in Figure 10.89a. Figure 10.89a
EXECUTE: L1 = L2 L1 = mvr sin φ = mvl = (1.90 × 10−3 kg)(360 m/s)(0.125 m) = 0.0855 kg ⋅ m 2 / s
L2 = I 2ω2
I 2 = I board + I bullet = 1 ML2 + mr 2
3
I 2 = 1 (0.750 kg)(0.250 m) 2 + (1.90 × 10−3 kg)(0.125 m) 2 = 0.01565 kg ⋅ m 2
3
L1 0.0855 kg ⋅ m 2 /s
=
= 5.46 rad/s
I2
0.1565 kg ⋅ m 2
(b) IDENTIFY: Apply conservation of energy to the motion of the board after the collision.
SET UP: The position of the board at points 1 and 2 in its motion is shown in Figure 10.89b. Take the origin of
coordinates at the center of the board and + y to be upward, so ycm,1 = 0 and ycm, 2 = h, the height being asked for.
Then L1 = L2 gives that ω2 = K1 + U1 + Wother = K 2 + U 2
EXECUTE: K1 = I ω
1
2 Only gravity does work, so Wother = 0. 2 U1 = mgycm,1 = 0
K2 = 0
U 2 = mgycm, 2 = mgh
Figure 10.89b 10-32 Chapter 10 Thus 1
2 I ω 2 = mgh. Iω 2
(0.01565 kg ⋅ m 2 )(5.46 rad/s) 2
=
= 0.0317 m = 3.17 cm
2mg 2(0.750 kg + 1.90 × 10 −3 kg)(9.80 m/s 2 )
(c) IDENTIFY and SET UP: The position of the board at points 1 and 2 in its motion is shown in Figure 10.89c.
h= Apply conservation of energy as in part (b),
except now we want ycm,2 = h = 0.250 m.
Solve for the ω after the collision that is
required for this to happen. Figure 10.89c
EXECUTE: ω= 1
2 I ω 2 = mgh 2mgh
2(0.750 kg + 1.90 × 10−3 kg)(9.80 m/s2 )(0.250 m)
=
I
0.01565 kg ⋅ m 2 ω = 15.34 rad/s
Now go back to the equation that results from applying conservation of angular momentum to the collision and
solve for the initial speed of the bullet. L1 = L2 implies mbullet vl = I 2ω2
I 2ω2
(0.01565 kg ⋅ m 2 )(15.34 rad/s)
=
= 1010 m/s
mbulletl
(1.90 × 10−3 kg)(0.125 m)
EVALUATE: We have divided the motion into two separate events: the collision and the motion after the
collision. Angular momentum is conserved in the collision because the collision happens quickly. The board
doesn’t move much until after the collision is over, so there is no gravity torque about the axis. The collision is
inelastic and mechanical energy is lost in the collision. Angular momentum of the system is not conserved during
this motion, due to the external gravity torque. Our answer to parts (b) and (c) say that a bullet speed of 360 m/s
causes the board to swing up only a little and a speed of 1010 m/s causes it to swing all the way over.
IDENTIFY: Angular momentum is conserved, so I 0ω0 = I 2ω2 .
SET UP: For constant mass the moment of inertia is proportional to the square of the radius.
2
2
EXECUTE: R02ω0 = R2 ω 2 , or R02ω0 = ( R0 + ΔR ) (ω0 + Δω ) = R02ω0 + 2 R0 ΔRω0 + R02 Δω , where the terms in
v= 10.90. ΔRΔω and (Δω ) 2 have been omitted. Canceling the R02ω0 term gives
ΔR = − 10.91. R0 Δω
= −1.1 cm.
2 ω0 EVALUATE: ΔR / R0 and Δω / ω0 are each very small so the neglect of terms containing ΔRΔω or (Δω ) 2 is an
accurate simplifying approximation.
IDENTIFY: Apply conservation of angular momentum to the collision between the bird and the bar and apply
conservation of energy to the motion of the bar after the collision.
SET UP: For conservation of angular momentum take the axis at the hinge. For this axis the initial angular
momentum of the bird is mbird (0.500 m)v , where mbird = 0.500 kg and v = 2.25 m/s . For this axis the moment of inertia is I = 1 mbar L2 = 1 (1.50 kg)(0.750 m) 2 = 0.281 kg ⋅ m 2 . For conservation of energy, the gravitational
3
3
potential energy of the bar is U = mbar gycm , where ycm is the height of the center of the bar. Take ycm,1 = 0 , so
ycm,2 = −0.375 m .
EXECUTE: ω= (a) L1 = L2 gives mbird (0.500 m)v = ( 1 mbar L2 )ω .
3 3mbird (0.500 m)v 3(0.500 kg)(0.500 m)(2.25 m/s)
=
= 2.00 rad/s .
mbar L2
(1.50 kg)(0.750 m) 2 Dynamics of Rotational Motion (b) U1 + K1 = U 2 + K 2 applied to the motion of the bar after the collision gives 1
2 10-33 2
I ω12 = mbar g ( −0.375 m) + 1 I ω2 .
2 2
I 2
(1.50 kg)(9.80 m/s 2 )(0.375 m) = 6.58 rad/s
0.281 kg ⋅ m 2
EVALUATE: Mechanical energy is not conserved in the collision. The kinetic energy of the bar just after the
collision is less than the kinetic energy of the bird just before the collision.
IDENTIFY: Angular momentum is conserved, since the tension in the string is in the radial direction and therefore
!
!
produces no torque. Apply ∑ F = ma to the block, with a = arad = v 2 / r . ω2 = ω12 + mbar g (0.375 m) . ω2 = (2.00 rad/s)2 + 10.92. The block’s angular momentum with respect to the hole is L = mvr . SET UP: The tension is related to the block’s mass and speed, and the radius of the circle, by T = m EXECUTE: v2
.
r L2
1 m 2v 2 r 2 ( mvr )
=
=
= 3 . The radius at which the string breaks is
r
m r3
mr 3
mr
2 T = mv 2 L2
( mv1r1 ) = ( ( 0.250 kg ) ( 4.00 m s )( 0.800 m ) ) , from which r = 0.440 m.
=
mTmax
mTmax
( 0.250 kg )( 30.0 N )
2 2 r3 = ⎛ 0.800 m ⎞
Just before the string breaks the speed of the rock is (4.00 m/s) ⎜
⎟ = 7.27 m/s . We can
⎝ 0.440 m ⎠
verify that v = 7.27 m/s and r = 0.440 m do give T = 30.0 N .
IDENTIFY and SET UP: Apply conservation of angular momentum to the system consisting of the disk and train.
SET UP: L1 = L2 , counterclockwise positive. The motion is sketched in Figure 10.93.
EVALUATE: 10.93. L1 = 0 (before you switch on the train’s engine;
both the train and the platform are at rest)
L2 = Ltrain + Ldisk
Figure 10.93 The train is EXECUTE: 1
2 (0.95 m) = 0.475 m from the axis of rotation, so for it I t = mt R = (1.20 kg)(0.475 m) 2 = 0.2708 kg ⋅ m 2
2
t ωrel = vrel / Rt = (0.600 m/s)/0.475 s = 1.263 rad/s
This is the angular velocity of the train relative to the disk. Relative to the earth ωt = ωrel + ωd .
Thus Ltrain = I tωt = I t (ωrel + ωd ).
L2 = L1 says Ldisk = − Ltrain
Ldisk = I dωd , where I d = 1 md Rd2
2
1
2 md Rd2ωd = − I t (ωrel + ωd )
I tωrel
(0.2708 kg ⋅ m 2 )(1.263 rad/s)
=−1
= −0.30 rad/s.
2
md Rd + I t
(7.00 kg)(0.500 m) 2 + 0.2708 kg ⋅ m 2
2
2 ωd = − 1 10.94. EVALUATE: The minus sign tells us that the disk is rotating clockwise relative to the earth. The disk and train
rotate in opposite directions, since the total angular momentum of the system must remain zero. Note that we
applied L1 = L2 in an inertial frame attached to the earth.
IDENTIFY: I for the wheel is the sum of the values of I for each of its parts, the rim and each spoke. The total
length of wire is constant. The motion is related to the friction torque by ∑τ z = Iα z .
SET UP: 4 R + 2π R = L0 , where R is the radius of the wheel and therefore the length of each of the four spokes.
The mass of a piece is proportional to the length of that piece.
2π R
L0
⎛ 2π ⎞
EXECUTE: (a) R =
. I rim = mrim R 2 . mrim =
M0 = ⎜
⎟M0 .
4 + 2π
L0
⎝ 4 + 2π ⎠ I rim = M 0 L2
0 R
M0
2π
= (5.778 × 10−3 ) M 0 L2 . I spoke = 1 mspoke R 2 . mspoke = M 0 =
and
0
3
3
L0
2π + 4
(2π + 4) I spoke = M 0 L2
0 1
= (3.065 × 10−4 ) M 0 L2 . I = I rim + 4 I spoke = (7.00 × 10−3 ) M 0 L2 .
0
0
3(2π + 4)3 10-34 Chapter 10 (b) ω z = ω0 z + α z t gives α z = − 10.95. ω0 . Then ∑τ z = Iα z gives τ f = (7.00 × 10−3 ) M 0 L2
0 ω0 T
T
EVALUATE: If the wire were bent into a circle, without spokes, the moment of inertia would be
M 0 L2
0
M 0R2 =
= (9.46 × 10−3 ) M 0 L2 . The actual value of I for the wheel is less than this because the mass in the
0
(4 + 2π ) 2
spokes is closer to the axis than the rim.
IDENTIFY and SET UP: Use the methods stipulated in the problem.
EXECUTE: (a) The initial angular momentum with respect to the pivot is mvr , and the final total moment of
inertia is I + mr 2 , so the final angular velocity is ω = mvr ( mr 2 + I ) .
(b) The kinetic energy after the collision is converted to gravitational potential energy, so 2 ( M + m ) gh
12
.
ω ( mr 2 + I ) = ( M + m ) gh , or ω =
2
( mr 2 + I )
⎛m⎞
(c) Substitution of Ι = Μr 2 into the result of part (a) gives ω = ⎜
⎟ ( v r ) , and into the result of part (b),
⎝m+M ⎠ ω = 2 gh (1 r ), which are consistent with the forms for v. 10.96. EVALUATE: I = Mr 2 applies approximately when the pendulum consists of a heavy catcher mounted on a light
arm. In the actual apparatus some of the mass is distributed closer to the axis and I < Mr 2 .
IDENTIFY: Apply conservation of momentum to the system of the runner and turntable
SET UP: Let the positive sense of rotation be the direction the turntable is rotating initially.
EXECUTE: The initial angular momentum is I ω1 − mRv1 , with the minus sign indicating that runner’s motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is ω2 ( I + mR 2 ), so ω2 = I ω1 − mRv1
.
I + mR 2 (80 kg ⋅ m 2 )(0.200 rad s) − (55.0 kg)(3.00 m)(2.8 m s)
= −0.776 rad s .
(80 kg ⋅ m 2 ) + (55.0 kg)(3.00 m) 2
EVALUATE: The minus sign indicates that the turntable has reversed its direction of motion. This happened
because the man had the larger magnitude of angular momentum initially.
IDENTIFY: Treat the moon as a point mass, so L = I ω = mr 2ω , where r is the distance of the moon from the
center of the earth. Conservation of angular momentum says dL / dt = 0 .
SET UP: dr / dt = 3.0 cm/y = 3.0 × 10−2 m/y . The period of the moon’s orbital motion is 27.3 d = 2.36 × 106 s . ω2 = 10.97. r = 3.84 × 108 m .
d
dr
dω
dω
2ω dr
( mr 2ω ) = mω (2r ) + mr 2
= 0 , so
=−
.
dt
dt
dt
dt
r dt
dω
2(2.66 × 10−6 rad/s)
2π rad
2π rad
=−
(3.0 × 10−2 m/y) = −4.2 × 10 −16 rad/s per year .
=
= 2.66 × 10−6 rad/s .
ω=
6
dt
3.84 × 108 m
T
2.36 × 10 s
dω
is negative, so the angular velocity is decreasing.
dt
EVALUATE: L = mr 2ω . If L is constant, then ω decreases when r increases. The fractional changes in r and
ω are very, very small.
IDENTIFY: Follow the method outlined in the hint.
SET UP: J = mΔvcm . ΔL = J ( x − xcm ) .
EXECUTE: 10.98. dL / dt = EXECUTE: The velocity of the center of mass will change by Δvcm = J / m and the angular velocity will change by J ( x − xcm )
J J (x − xcm ) xcm
−
⋅
. The change is velocity of the end of the bat will then be Δvend = Δvcm − Δω xcm =
I
m
I
Setting Δvend = 0 allows cancellation of J and gives I = ( x − xcm ) xcm m, which when solved for x is
Δω = x= I (5.30 × 10−2 kg ⋅ m 2 )
+ (0.600 m) = 0.710 m.
(0.600 m)(0.800 kg)
The center of percussion is farther from the handle than the center of mass. + xcm = xcm m
EVALUATE: Dynamics of Rotational Motion 10-35 IDENTIFY and SET UP: Follow the analysis that led to Eq.(10.33).
"
!
!
EXECUTE: In Figure 10.33a in the textbook, if the vector r and hence the vector L are not horizontal but make
an angle β with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical
weight). The magnitude of the torque will be ω r cos β , and this torque will change the direction of the horizontal
component of the angular momentum, which has magnitude L cos β . Thus, the situation of Figure 10.35 in the
"
!
"
!
textbook is reproduced, but with Lhoriz instead of L . Then, the expression found in Eq. (10.33) becomes
"
!"
!
τ
dφ d L Lhoriz
mgr cosβ wr
Ω=
=
="
=
=
⋅
!
dt
dt
L cos β
Iω
Lhoriz
!
EVALUATE: The torque and the horizontal component of L both depend on β by the same factor, cos β .
10.100. IDENTIFY: Apply conservation of energy to the motion of the ball.
2
2
SET UP: In relating 1 mvcm and 1 I ω 2 , instead of vcm = Rω use the relation derived in part (a). I = 5 mR 2 .
2
2
EXECUTE: (a) Consider the sketch in Figure 10.100.
The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact
10.99. with the rails is R 2 − ( d 2 ) , so vcm = ω R 2 − d 2 4 . When d = 0, this reduces to vcm = ω R, the same as rolling
2 on a flat surface. When d = 2 R, the rolling radius approaches zero, and vcm → 0 for any ω.
2
⎡
⎛
⎞⎤
2
1 2 1 2 1⎢ 2
vcm
2⎜
⎟ ⎥ = mvcm
(b) K = mv + I ω = ⎢ mvcm + ( 2 5 ) mR
⎥
⎜ R2 − d 2 4 ⎟
2
2
2
⎜
( ) ⎟ ⎥ 10
⎢
⎝
⎠⎦
⎣
Setting this equal to mgh and solving for vcm gives the desired result. ⎡
⎤
2
⎢5 +
⎥
2
2
⎢ (1 − d 4 R ) ⎥
⎣
⎦ (c) The denominator in the square root in the expression for vcm is larger than for the case d = 0, so vcm is smaller.
For a given speed, ω is larger than in the d = 0 case, so a larger fraction of the kinetic energy is rotational, and the translational kinetic energy, and hence vcm , is smaller.
(d) Setting the expression in part (b) equal to 0.95 of that of the d = 0 case and solving for the ratio d R gives d R = 1.05. Setting the ratio equal to 0.995 gives d R = 0.37.
EVALUATE: 10 gh
, the same as for a sphere rolling down a
7 If we set d = 0 in the expression in part (b), vcm = ramp. When d → 2 R , the expression gives vcm = 0 , as it should. Figure 10.100 !
!
10.101. IDENTIFY: Apply ∑ Fext = macm and ∑τ z = I cmα z to the motion of the cylinder. Use constant acceleration equations to relate ax to the distance the object travels. Use the work-energy theorem to find the work done by friction.
SET UP: The cylinder has I cm = 1 MR 2 .
2
EXECUTE: (a) The free-body diagram is sketched in Figure 10.101. The friction force is
μ MgR
2μ g
fR
f = μ k n = μ k Mg , so a = μ k g. The magnitude of the angular acceleration is
=k
= k.
2
I
R
(1 2 ) MR
(b) Setting v = at = ω R = (ω0 − ωt ) R and solving for t gives t =
2 2
⎛ Rω0 ⎞
1
1
R 2ω0
and d = at 2 = ( μ k g ) ⎜
.
⎟=
2
2
⎝ 3μ k g ⎠ 18μ k g Rω0
Rω0
Rω0
=
=
,
a + Rα μ k g + 2 μ k g 3μ k g 10-36 Chapter 10 (c) The final kinetic energy is ( 3 4 ) Mv 2 = ( 3 4 ) M ( at ) , so the change in kinetic energy is
2 2 3⎛
Rω0 ⎞ 1
1
22
22
ΔK = M ⎜ μ k g
⎟ − MR ω0 = − MR ω0 .
4⎝
3μ k g ⎠ 4
6
EVALUATE: The fraction of the initial kinetic energy that is removed by friction work is 1
6
1
4 2
MRω0 2
= . This
2
MRω0 3 fraction is independent of the initial angular speed ω0 . Figure 10.101
10.102. IDENTIFY: The vertical forces must sum to zero. Apply Eq.(10.33).
SET UP: Denote the upward forces that the hands exert as FL and FR . τ = ( FL − FR )r , where r = 0.200 m . Iω
, where the second
r
equation is τ = ΩL, divided by r. These two equations can be solved for the forces by first adding and then subtracting, EXECUTE: The conditions that FL and FR must satisfy are FL + FR = w and FL − FR = Ω 1⎛
Iω ⎞
1⎛
Iω ⎞
2
yielding FL = ⎜ w + Ω
⎟ and FR = ⎜ w − Ω
⎟ . Using the values w = mg = (8.00 kg)(9.80 m s ) = 78.4 N and
2⎝
r⎠
2⎝
r⎠
I ω (8.00 kg)(0.325 m) 2 (5.00 rev s × 2π rad rev)
=
= 132.7 kg ⋅ m s gives
r
(0.200 m)
FL = 39.2 N + Ω(66.4 N ⋅ s), FR = 39.2 N − Ω(66.4 N ⋅ s).
(a) Ω = 0, FL = FR = 39.2 N .
(b) Ω = 0.05 rev s = 0.314 rad s, FL = 60.0 N, FR = 18.4 N.
(c) Ω = 0.3 rev s = 1.89 rad s, FL = 165 N, FR = −86.2 N , with the minus sign indicating a downward force. 39.2 N
= 0.575 rad s, which is 0.0916 rev s.
66.4 N ⋅ s
EVALUATE: The larger the precession rate Ω , the greater the torque on the wheel and the greater the difference
between the forces exerted by the two hands.
10.103. IDENTIFY: The answer to part (a) can be taken from the solution to Problem 10.92. The work-energy theorem
says W = ΔK .
SET UP: Problem 10.92 uses conservation of angular momentum to show that r1v1 = r2v2 .
(e) FR = 0 gives Ω = EXECUTE: (a) T = mv12 r12 r 3 .
!
!!
!
(b) T and dr are always antiparallel. T ⋅ dr = −Tdr . dr mv12 2 ⎡ 1 1 ⎤
=
r1 ⎢ 2 − 2 ⎥ .
r1
r2 r 3
2
⎣ r2 r1 ⎦
1
mv 2
2
(c) v2 = v1 (r1 r2 ), so ΔK = m(v2 − v12 ) = 1 ⎡(r1 / r2 ) 2 − 1⎤ , which is the same as the work found in part (b).
⎦
2
2⎣
!
EVALUATE: The work done by T is positive, since T is toward the hole in the surface and the block moves
toward the hole. Positive work means the kinetic energy of the object increases.
r2 W = − ∫ T dr = mv12 r12 ∫ r1 EQUILIBRIUM AND ELASTICITY 11.1. 11 IDENTIFY: Use Eq.(11.3) to calculate xcm . The center of gravity of the bar is at its center and it can be treated as
a point mass at that point.
SET UP: Use coordinates with the origin at the left end of the bar and the + x axis along the bar. m1 = 2.40 kg, m2 = 1.10 kg, m3 = 2.20 kg.
m1 x1 + m2 x2 + m3 x3 (2.40 kg)(0.250 m) + 0 + (2.20 kg)(0.500 m)
=
= 0.298 m . The fulcrum
m1 + m2 + m3
2.40 kg + 1.10 kg + 2.20 kg
should be placed 29.8 cm to the right of the left-hand end.
EVALUATE: The mass at the right-hand end is greater than the mass at the left-hand end. So the center of gravity
is to the right of the center of the bar.
IDENTIFY: Use Eq.(11.3) to calculate xcm of the composite object.
SET UP: Use coordinates where the origin is at the original center of gravity of the object and + x is to the right.
With the 1.50 g mass added, xcm = −2.20 cm , m1 = 5.00 g and m2 = 1.50 g . x1 = 0 .
EXECUTE: 11.2. ⎛ m + m2 ⎞
⎛ 5.00 g + 1.50 g ⎞
m2 x2
. x2 = ⎜ 1
⎟ xcm = ⎜
⎟ (−2.20 cm) = −9.53 cm .
m1 + m2
1.50 g
m2 ⎠
⎝
⎠
⎝
The additional mass should be attached 9.53 cm to the left of the original center of gravity.
EVALUATE: The new center of gravity is somewhere between the added mass and the original center of gravity.
IDENTIFY: The center of gravity of the combined object must be at the fulcrum. Use Eq.(11.3) to calculate xcm
SET UP: The center of gravity of the sand is at the middle of the box. Use coordinates with the origin at the
fulcrum and + x to the right. Let m1 = 25.0 kg , so x1 = 0.500 m . Let m2 = msand , so x2 = −0.625 m . xcm = 0 .
EXECUTE: 11.3. 11.5. xcm = m1 x1 + m2 x2
x
⎛ 0.500 m ⎞
= 0 and m2 = − m1 1 = −(25.0 kg) ⎜
⎟ = 20.0 kg .
x2
m1 + m2
⎝ −0.625 m ⎠
EVALUATE: The mass of sand required is less than the mass of the plank since the center of the box is farther
from the fulcrum than the center of gravity of the plank is.
IDENTIFY: Apply the first and second conditions for equilibrium to the trap door.
SET UP: For ∑τ z = 0 take the axis at the hinge. Then the torque due to the applied force must balance the
EXECUTE: 11.4. xcm = xcm = torque due to the weight of the door.
EXECUTE: (a) The force is applied at the center of gravity, so the applied force must have the same magnitude as
the weight of the door, or 300 N. In this case the hinge exerts no force.
(b) With respect to the hinges, the moment arm of the applied force is twice the distance to the center of mass, so
the force has half the magnitude of the weight, or 150 N . The hinges supply an upward force of
300 N − 150 N = 150 N.
EVALUATE: Less force must be applied when it is applied farther from the hinges.
IDENTIFY: Apply ∑τ z = 0 to the ladder.
SET UP: Take the axis to be at point A. The free-body diagram for the ladder is given in Figure 11.5. The torque
due to F must balance the torque due to the weight of the ladder.
EXECUTE: F (8.0 m)sin 40° = (2800 N)(10.0 m), so F = 5.45 kN . 11-1 11-2 Chapter 11 EVALUATE: The force required is greater than the weight of the ladder, because the moment arm for F is less
than the moment arm for w. Figure 11.5
11.6. IDENTIFY: Apply the first and second conditions of equilibrium to the board.
SET UP: The free-body diagram for the board is given in Figure 11.6. Since the board is uniform its center of
gravity is 1.50 m from each end. Apply ∑ Fy = 0 , with + y upward. Apply ∑τ = 0 with the axis at the end where the first person applies a force and with counterclockwise torques positive.
EXECUTE: ∑ Fy = 0 gives F1 + F2 − w = 0 and F2 = w − F1 = 160 N − 60 N = 100 N . ∑τ = 0 gives ⎛w⎞
⎛ 160 N ⎞
F2 x − w(1.50 m) = 0 and x = ⎜ ⎟ (1.50 m) = ⎜
⎟ (1.50 m) = 2.40 m . The other person lifts with a force of
F2 ⎠
⎝ 100 N ⎠
⎝
100 N at a point 2.40 m from the end where the other person lifts.
EVALUATE: By considering the axis at the center of gravity we can see that a larger force is applied by the
person who pushes closer to the center of gravity. Figure 11.6
11.7. IDENTIFY: Apply ∑F y = 0 and ∑τ z = 0 to the board. SET UP: Let + y be upward. Let x be the distance of the center of gravity of the motor from the end of the board
where the 400 N force is applied.
EXECUTE: (a) If the board is taken to be massless, the weight of the motor is the sum of the applied forces,
(2.00 m)(600 N)
1000 N. The motor is a distance
= 1.20 m from the end where the 400 N force is applied, and so
(1000 N)
is 0.800 m from the end where the 600 N force is applied.
(b) The weight of the motor is 400 N + 600 N − 200 N = 800 N. Applying ∑τ z = 0 with the axis at the end of the 11.8. board where the 400 N acts gives (600 N)(2.00 m) = (200 N)(1.00 m) + (800 N) x and x = 1.25 m . The center of
gravity of the motor is 0.75 m from the end of the board where the 600 N force is applied.
EVALUATE: The motor is closest to the end of the board where the larger force is applied.
IDENTIFY: Apply the first and second conditions of equilibrium to the shelf.
SET UP: The free-body diagram for the shelf is given in Figure 11.8. Take the axis at the left-hand end of the
shelf and let counterclockwise torque be positive. The center of gravity of the uniform shelf is at its center.
EXECUTE: (a) ∑τ z = 0 gives − wt (0.200 m) − ws (0.300 m) + T (0.400 m) = 0 .
(25.0 N)(0.200 m) + (50.0 N)(0.300 m)
= 50.0 N
0.400 m
∑ Fy = 0 gives T1 + T2 − wt − ws = 0 and T1 = 25.0 N . The tension in the left-hand wire is 25.0 N and the tension in T= the right-hand wire is 50.0 N. Equilibrium and Elasticity EVALUATE: We can verify that ∑τ z 11-3 = 0 is zero for any axis, for example for an axis at the right-hand end of the shelf. Figure 11.8
11.9. IDENTIFY: Apply the conditions for equilibrium to the bar. Set each tension equal to its maximum value.
SET UP: Let cable A be at the left-hand end. Take the axis to be at the left-hand end of the bar and x be the
distance of the weight w from this end. The free-body diagram for the bar is given in Figure 11.9.
EXECUTE: (a) ∑ Fy = 0 gives TA + TB − w − wbar = 0 and w = TA + TB − wbar = 500.0 N + 400.0 N − 350.0 N = 550 N .
(b) ∑τ z = 0 gives TB (1.50 m) − wx − wbar (0.750 m) = 0 . TB (1.50 m) − wbar (0.750 m) (400.0 N)(1.50 m) − (350 N)(0.750 m)
=
= 0.614 m . The weight should be placed
w
550 N
0.614 m from the left-hand end of the bar.
EVALUATE: If the weight is moved to the left, TA exceeds 500.0 N and if it is moved to the right TB exceeds
400.0 N.
x= Figure 11.9
11.10. IDENTIFY: Apply the first and second conditions for equilibrium to the ladder.
SET UP: Let n2 be the upward normal force exerted by the ground and let n1 be the horizontal normal force exerted by the wall. The maximum possible static friction force that can be exerted by the ground is μs n2 .
EXECUTE: (a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder,
and the normal force n2 . For the vertical forces to balance, n2 = w1 + wm = 160 N + 740 N = 900 N, and the
maximum frictional force is μs n2 = (0.40)(900N) = 360N .
(b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground. Balancing torques
about the point of contact with the ground, (4.0 m)n1 = (1.5 m)(160 N) + (1.0 m)(3 5)(740 N) = 684 N ⋅ m, so
n1 = 171.0 N . This horizontal force about must be balanced by the friction force, which must then be 170 N to two
figures.
(c) Setting the friction force, and hence n1 , equal to the maximum of 360 N and solving for the distance x along the
ladder, (4.0 m)(360 N) = (1.50 m)(160 N) + x(3 5)(740 N), so x = 2.7 m. 11-4 Chapter 11 11.11. EVALUATE: The normal force exerted by the ground doesn’t change as the man climbs up the ladder. But the
normal force exerted by the wall and the friction force exerted by the ground both increase as he moves up the ladder.
IDENTIFY: The system of the person and diving board is at rest so the two conditions of equilibrium apply.
(a) SET UP: The free-body diagram for the diving board is given in Figure 11.11. Take the origin of coordinates
at the left-hand end of the board (point A). !
F1 is the force applied at
!
the support point and F2 is
the force at the end that is
held down. Figure 11.11
EXECUTE: ∑τ A = 0 gives + F1 (1.0 m) − (500 N)(3.00 m) − (280 N)(1.50 m) = 0 (500 N)(3.00 m) + (280 N)(1.50 m)
= 1920 N
1.00 m
(b) ∑ Fy = ma y
F1 = F1 − F2 − 280 N − 500 N = 0
F2 = F1 − 280 N − 500 N = 1920 N − 280 N − 500 N = 1140 N 11.12. EVALUATE: We can check our answers by calculating the net torque about some point and checking that τ z = 0
for that point also. Net torque about the right-hand of the board:
(1140 N)(3.00 m)+(280 N)(1.50 m) − (1920 N)(2.00 m) = 3420 N ⋅ m + 420 N ⋅ m − 3840 N ⋅ m = 0, which checks.
IDENTIFY: Apply the first and second conditions of equilibrium to the beam.
SET UP: The boy exerts a downward force on the beam that is equal to his weight.
EXECUTE: (a) The graphs are given in Figure 11.12.
(b) x = 6.25 m when FA = 0, which is 1.25 m beyond point B.
(c) Take torques about the right end. When the beam is just balanced, FA = 0, so FB = 900 N. The distance that (300 N)(4.50 m)
= 1.50 m.
(900 N)
EVALUATE: When the beam is on the verge of tipping it starts to lift off the support A and the normal force
FA exerted by the support goes to zero. point B must be from the right end is then Figure 11.12 Equilibrium and Elasticity 11.13. 11-5 IDENTIFY: Apply the first and second conditions of equilibrium to the strut.
(a) SET UP: The free-body diagram for the strut is given in Figure 11.13a. Take the origin of coordinates at the
!
hinge (point A) and + y upward. Let Fh and Fv be the horizontal and vertical components of the force F exerted
on the strut by the pivot. The tension in the vertical cable is the weight w of the suspended object. The weight w of
the strut can be taken to act at the center of the strut. Let L be the length of the strut. EXECUTE:
∑ Fy = ma y Fv − w − w = 0
Fv = 2w Figure 11.13a Sum torques about point A. The pivot force has zero moment arm for this axis and so doesn’t enter into the torque
equation.
τA = 0
TL sin 30.0° − w ( ( L / 2)cos30.0° ) − w( L cos30.0°) = 0 T sin 30.0° − (3w / 2)cos30.0° = 0
3w cos30.0°
= 2.60 w
2sin 30.0°
Then ∑ Fx = max implies T − Fh = 0 and Fh = 2.60w.
!
We now have the components of F so can find its magnitude and direction (Figure 11.13b)
T= F = Fh2 + Fv2
F = (2.60w) 2 + (2.00w) 2
F = 3.28w
F 2.00w
tan θ = v =
Fh 2.60w
θ = 37.6°
Figure 11.13b
(b) SET UP: The free-body diagram for the strut is given in Figure 11.13c. Figure 11.13c The tension T has been replaced by its x and y components. The torque due to T equals the sum of the torques of its
components, and the latter are easier to calculate. 11-6 Chapter 11 EXECUTE: ∑τ A = 0 + (T cos30.0°)( L sin 45.0°) − (T sin 30.0°)( L cos 45.0°) − w(( L / 2)cos 45.0°) − w( L cos 45.0°) = 0
The length L divides out of the equation. The equation can also be simplified by noting that sin 45.0° = cos 45.0°.
Then T (cos30.0° − sin 30.0°) = 3w / 2.
T= 3w
= 4.10 w
2(cos30.0° − sin 30.0°) ∑F x = max Fh − T cos30.0° = 0
Fh = T cos30.0° = (4.10w)(cos30.0°) = 3.55w ∑F y = ma y Fv − w − w − T sin 30.0° = 0
Fv = 2 w + (4.10 w)sin 30.0° = 4.05w From Figure 11.13d,
F = Fh2 + Fv2
F = (3.55w) 2 + (4.05w) 2 = 5.39 w
Fv 4.05w
=
Fh 3.55w
θ = 48.8° tan θ = Figure 11.13d 11.14. EVALUATE: In each case the force exerted by the pivot does not act along the strut. Consider the net torque about
the upper end of the strut. If the pivot force acted along the strut, it would have zero torque about this point. The
two forces acting at this point also have zero torque and there would be one nonzero torque, due to the weight of
the strut. The net torque about this point would then not be zero, violating the second condition of equilibrium.
IDENTIFY: Apply the first and second conditions of equilibrium to the beam.
SET UP: The free-body diagram for the beam is given in Figure 11.14. H v and H h are the vertical and
horizontal components of the force exerted on the beam at the wall (by the hinge). Since the beam is uniform, its
center of gravity is 2.00 m from each end. The angle θ has cosθ = 0.800 and sin θ = 0.600 . The tension T has
been replaced by its x and y components.
EXECUTE: (a) H v , H h and Tx = T cosθ all produce zero torque. ∑τ = 0 gives − w(2.00 m) − wload (4.00 m) + T sin θ (4.00 m) = 0 and T =
(b) ∑F x (150 N)(2.00 m) + (300 N)(4.00 m)
= 625 N .
(4.00 m)(0.600) = 0 gives H h = T cosθ = 0 and H h = (625 N)(0.800) = 500 N . ∑F y = 0 gives H v − w − wload + T sin θ = 0 and H v = w + wload − T sin θ = 150 N + 300 N − (625 N)(0.600) = 75 N .
EVALUATE: For an axis at the right-hand end of the beam, only w and H v produce torque. The torque due to w is counterclockwise so the torque due to H v must be clockwise. To produce a counterclockwise torque, H v must be
upward, in agreement with our result from ∑F y = 0. Figure 11.14 Equilibrium and Elasticity 11.15. 11-7 IDENTIFY: Apply the first and second conditions of equilibrium to the door.
!
!
SET UP: The free-body diagram for the door is given in Figure 11.15. Let H1 and H 2 be the forces exerted by
the upper and lower hinges. Take the origin of coordinates at the bottom hinge (point A) and + y upward. EXECUTE:
We are given that
H1v = H 2v = w / 2 = 140 N. ∑F x = max H 2 h − H1h = 0
H1h = H 2h
The horizontal components of the
hinge forces are equal in magnitude
and opposite in direction. Figure 11.15 Sum torques about point A. H1v , H 2 v , and H 2 h all have zero moment arm and hence zero torque about an axis at
this point. Thus 11.16. ∑τ A = 0 gives H1h (1.00 m) − w(0.50 m) = 0 ⎛ 0.50 m ⎞ 1
H1h = w ⎜
⎟ − 2 (280 N) = 140 N.
⎝ 1.00 m ⎠
The horizontal component of each hinge force is 140 N.
EVALUATE: The horizontal components of the force exerted by each hinge are the only horizontal forces so must
be equal in magnitude and opposite in direction. With an axis at A, the torque due to the horizontal force exerted by
the upper hinge must be counterclockwise to oppose the clockwise torque exerted by the weight of the door. So,
the horizontal force exerted by the upper hinge must be to the left. You can also verify that the net torque is also
zero if the axis is at the upper hinge.
IDENTIFY: Apply the conditions of equilibrium to the wheelbarrow plus its contents. The upward force applied
by the person is 650 N.
SET UP: The free-body diagram for the wheelbarrow is given in Figure 11.16. F = 650 N , wwb = 80.0 N and w is
the weight of the load placed in the wheelbarrow.
EXECUTE: (a) ∑τ z = 0 with the axis at the center of gravity gives n (0.50 m) − F (0.90 m) = 0 and
⎛ 0.90 m ⎞
n = F⎜
⎟ = 1170 N . ∑ Fy = 0 gives F + n − wwb − w = 0 and
⎝ 0.50 m ⎠
w = F + n − wwb = 650 N + 1170 N − 80.0 N = 1740 N .
(b) The extra force is applied by the ground pushing up on the wheel.
EVALUATE: You can verify that ∑τ z = 0 for any axis, for example for an axis where the wheel contacts the ground. Figure 11.16 11-8 Chapter 11 11.17. IDENTIFY: Apply the first and second conditions of equilibrium to Clea.
SET UP: Consider the forces on Clea. The free-body diagram is given in Figure 11.17
EXECUTE:
nr = 89 N, nf = 157 N nr + nf = w so w = 246 N
Figure 11.17 ∑τ 11.18. z = 0, axis at rear feet Let x be the distance from the rear feet to the center of gravity.
nf (0.95 m) − xw = 0
x = 0.606 m from rear feet so 0.34 m from front feet.
EVALUATE: The normal force at her front feet is greater than at her rear feet, so her center of gravity is closer to
her front feet.
IDENTIFY: Apply the conditions for equilibrium to the crane.
SET UP: The free-body diagram for the crane is sketched in Figure 11.18. Fh and Fv are the components of the
!
!
force exerted by the axle. T pulls to the left so Fh is to the right. T also pulls downward and the two weights are
downward, so Fv is upward.
EXECUTE: (a) ∑τ z = 0 gives T ([13 m]sin 25° − wc ([7.0 m]cos55°) − wb ([16.0 m]cos55° = 0 . T= (11,000 N)([16.0 m]cos55°) + (15,000 N)([7.0 m]cos55°)
= 2.93 × 104 N .
(13.0 m)sin 25° (b) ∑F x ∑F y = 0 gives Fh − T cos30° = 0 and Fh = 2.54 × 104 N . = 0 gives Fv − T sin 30° − wc − wb = 0 and Fv = 4.06 × 104 N . EVALUATE: tan θ = Fv 4.06 × 104 N
=
and θ = 58° . The force exerted by the axle is not directed along the crane.
Fh 2.54 × 104 N Figure 11.18
11.19. IDENTIFY: Apply the first and second conditions of equilibrium to the rod.
SET UP: The force diagram for the rod is given in Figure 11.19. Figure 11.19 Equilibrium and Elasticity EXECUTE: ∑τ z 11-9 = 0, axis at right end of rod, counterclockwise torque is positive (240 N)(1.50 m) + (90 N)(0.50 m) − (T1 sin 30.0°)(3.00 m) = 0
360 N ⋅ m + 45 N ⋅ m
= 270 N
1.50 m
∑ Fx = max T1 = T2 cosθ − T1 cos30° = 0 and T2 cosθ = 234 N ∑F y = ma y T1 sin 30° + T2 sin θ − 240 N − 90 N = 0
T2 sin θ = 330 N − (270 N)sin 30° = 195 N Then T2 sin θ 195 N
=
gives tan θ = 0.8333 and θ = 40°
T2 cosθ 234 N 195 N
= 303 N.
sin 40°
EVALUATE: The monkey is closer to the right rope than to the left one, so the tension is larger in the right rope.
The horizontal components of the tensions must be equal in magnitude and opposite in direction. Since T2 > T1 , the
rope on the right must be at a greater angle above the horizontal to have the same horizontal component as the
tension in the other rope.
IDENTIFY: Apply the first and second conditions for equilibrium to the beam.
SET UP: The free-body diagram for the beam is given in Figure 11.20.
EXECUTE: The cable is given as perpendicular to the beam, so the tension is found by taking torques about the
pivot point; T (3.00 m) = (1.00 kN)(2.00 m)cos 25.0° + (5.00 kN)(4.50 m)cos 25.0° , and T = 7.40 kN . The vertical
component of the force exerted on the beam by the pivot is the net weight minus the upward component of T,
6.00 kN − T cos 25.0° = 0.17 kN. The horizontal force is T sin 25.0° = 3.13 kN.
EVALUATE: The vertical component of the tension is nearly the same magnitude as the total weight of the object
and the vertical component of the force exerted by the pivot is much less than its horizontal component.
And T2 = 11.20. Figure 11.20
11.21. (a) IDENTIFY and SET UP: Use Eq.(10.3) to calculate the torque (magnitude and direction) for each force and add
the torques as vectors. See Figure 11.21a.
EXECUTE:
τ 1 = F1l1 = +(8.00 N)(3.00 m) τ 1 = +24.0 N ⋅ m
τ 2 = − F2l2 = −(8.00 N)(l + 3.00 m)
τ 2 = −24.0 N ⋅ m − (8.00 N)l
Figure 11.21a ∑τ z = τ 1 + τ 2 = +24.0 N ⋅ m − 24.0 N ⋅ m − (8.00 N)l = −(8.00 N)l Want l that makes ∑τ z = −6.40 N ⋅ m (net torque must be clockwise) −(8.00 N)l = −6.40 N ⋅ m
l = (6.40 N ⋅ m)/8.00 N = 0.800 m 11-10 Chapter 11 (b) τ 2 > τ 1 since F2 has a larger moment arm; the net torque is clockwise.
(c) See Figure 11.21b. τ 1 = − F1l1 = −(8.00 N)l
! τ 2 = 0 since F2 is at the axis
Figure 11.21b ∑τ
11.22. z = −6.40 N ⋅ m gives −(8.00 N)l = −6.40 N ⋅ m l = 0.800 m, same as in part (a).
EVALUATE: The force couple gives the same magnitude of torque for the pivot at any point.
lF
IDENTIFY: Y = 0 ⊥
AΔl
SET UP: A = 50.0 cm 2 = 50.0 × 10−4 m 2 .
(0.200 m)(25.0 N)
EXECUTE: relaxed: Y =
= 3.33 × 104 Pa
(50.0 × 10−4 m 2 )(3.0 × 10−2 m)
(0.200 m)(500 N)
= 6.67 × 105 Pa
(50.0 × 10−4 m 2 )(3.0 × 10−2 m)
EVALUATE: The muscle tissue is much more difficult to stretch when it is under maximum tension.
IDENTIFY and SET UP: Apply Eq.(11.10) and solve for A and then use A = π r 2 to get the radius and d = 2r to
calculate the diameter.
lF
lF
EXECUTE: Y = 0 ⊥ so A = 0 ⊥ (A is the cross-section area of the wire)
A Δl
Y Δl
For steel, Y = 2.0 × 1011 Pa (Table 11.1)
(2.00 m)(400 N)
= 1.6 × 10−6 m 2 .
Thus A =
(2.0 × 1011 Pa)(0.25 × 10−2 m)
maximum tension: Y = 11.23. A = π r 2 , so r = A / π = 1.6 × 10−6 m 2 / π = 7.1 × 10−4 m 11.24. d = 2r = 1.4 × 10−3 m = 1.4 mm
EVALUATE: Steel wire of this diameter doesn’t stretch much; Δl / l0 = 0.12%.
IDENTIFY: Apply Eq.(11.10).
SET UP: From Table 11.1, for steel, Y = 2.0 × 1011 Pa and for copper, Y = 1.1 × 1011 Pa .
A = π (d 2 / 4) = 1.77 × 10−4 m 2 . F⊥ = 4000 N for each rod.
EXECUTE: 11.25. 11.26. Δl
(4000N)
=
= 1.1 × 10−4. Similarly, the
(a) The strain is Δl = F . For steel
11
l0 YA
l0 (2.0 × 10 Pa)(1.77 × 10−4 m 2 ) strain for copper is 2.1 × 10−4.
(b) Steel: (1.1 × 10−4 )(0.750 m) = 8.3 × 10−5 m . Copper: (2.1 × 10−4 )(0.750 m) = 1.6 × 10−4 m .
EVALUATE: Copper has a smaller Y and therefore a greater elongation.
lF
IDENTIFY: Y = 0 ⊥
AΔl
SET UP: A = 0.50 cm 2 = 0.50 × 10−4 m 2
(4.00 m)(5000 N)
EXECUTE: Y =
= 2.0 × 1011 Pa
(0.50 × 10−4 m 2 )(0.20 × 10−2 m)
EVALUATE: Our result is the same as that given for steel in Table 11.1.
lF
IDENTIFY: Y = 0 ⊥
AΔl
SET UP: A = π r 2 = π (3.5 × 10−3 m)2 = 3.85 × 10−5 m 2 . The force applied to the end of the rope is the weight of the
climber: F⊥ = (65.0 kg)(9.80 m/s 2 ) = 637 N .
(45.0 m)(637 N)
= 6.77 × 108 Pa
(3.85 × 10−5 m 2 )(1.10 m)
EVALUATE: Our result is a lot smaller than the values given in Table 11.1. An object made of rope material is
much easier to stretch than if the object were made of metal.
EXECUTE: Y= Equilibrium and Elasticity 11.27. 11-11 IDENTIFY: Use the first condition of equilibrium to calculate the tensions T1 and T2 in the wires (Figure 11.27a).
Then use Eq.(11.10) to calculate the strain and elongation of each wire. Figure 11.27a
SET UP: The free-body diagram for m2 is given in Figure 11.27b.
EXECUTE:
∑ Fy = ma y T2 − m2 g = 0
T2 = 98.0 N
Figure 11.27b
SET UP: The free-body-diagram for m1 is given in Figure 11.27c
EXECUTE:
∑ Fy = ma y T1 − T2 − m1 g = 0
T1 = T2 + m1 g
T1 = 98.0 N + 58.8 N = 157 N
Figure 11.27c stress
stress F⊥
so strain =
=
strain
Y
AY
T1
157 N
=
= 3.1 × 10−3
upper wire: strain =
AY (2.5 × 10−7 m 2 )(2.0 × 1011 Pa) (a) Y = lower wire: strain = T2
98 N
=
= 2.0 × 10−3
AY (2.5 × 10−7 m 2 )(2.0 × 1011 Pa) (b) strain = Δl / l0 so Δl = l0 (strain) upper wire: Δl = (0.50 m)(3.1 × 10−3 ) = 1.6 × 10−3 m = 1.6 mm 11.28. lower wire: Δl = (0.50 m)(2.0 × 10−3 ) = 1.0 × 10−3 m = 1.0 mm
EVALUATE: The tension is greater in the upper wire because it must support both objects. The wires have the
same length and diameter, so the one with the greater tension has the greater strain and elongation.
IDENTIFY: Apply Eqs.(11.8), (11.9) and (11.10).
SET UP: The cross-sectional area of the post is A = π r 2 = π (0.125 m) 2 = 0.0491 m 2 . The force applied to the end
of the post is F⊥ = (8000 kg)(9.80 m/s 2 ) = 7.84 × 104 N . The Young’s modulus of steel is Y = 2.0 × 1011 Pa . F⊥ 7.84 × 104 N
=
= 1.60 × 106 Pa
A
0.0491 m 2
stress
1.60 × 106 Pa
=−
= −8.0 × 10−6 . The minus sign indicates that the length decreases.
(b) strain =
Y
2.0 × 1011 Pa
(c) Δl = l0 (strain) = (2.50 m)( −8.0 × 10−6 ) = −2.0 × 10−5 m
EVALUATE: The fractional change in length of the post is very small.
IDENTIFY: F⊥ = pA , so Fnet = (Δp ) A .
EXECUTE: 11.29. (a) stress = SET UP: 1 atm = 1.013 × 105 Pa .
EXECUTE: (2.8 atm − 1.0 atm)(1.013 × 105 Pa/atm)(50.0 m 2 ) = 9.1 × 106 N.
EVALUATE: This is a very large net force. 11-12 Chapter 11 11.30. Apply Eq.(11.13).
V Δp
SET UP: ΔV = − 0
. Δp is positive when the pressure increases.
B
EXECUTE: (a) The volume would increase slightly.
(b) The volume change would be twice as great.
(c) The volume change is inversely proportional to the bulk modulus for a given pressure change, so the volume
change of the lead ingot would be four times that of the gold.
EVALUATE: For lead, B = 4.1 × 1010 Pa , so Δp / B is very small and the fractional change in volume is very
small.
IDENTIFY: p = F / A 11.31. IDENTIFY: 1 cm 2 = 1 × 10−4 m 2 SET UP: 250 N
= 3.33 × 106 Pa.
0.75 × 10−4 m 2
(b) (3.33 × 106 Pa)(2)(200 × 10−4 m 2 ) = 133 kN.
EVALUATE: The pressure in part (a) is over 30 times larger than normal atmospheric pressure.
IDENTIFY: Apply Eq.(11.13). Density = m / V .
EXECUTE: 11.32. (a) At the surface the pressure is 1.0 × 105 Pa , so Δp = 1.16 × 108 Pa . V0 = 1.00 m3 . At the surface SET UP: 1.00 m3 of water has mass 1.03 × 103 kg .
EXECUTE: (a) B = − (Δp )V0
(1.16 × 108 Pa)(1.00 m 3 )
(Δp )V0
gives ΔV = −
=−
= −0.0527 m3
ΔV
B
2.2 × 109 Pa (b) At this depth 1.03 × 103 kg of seawater has volume V0 + ΔV = 0.9473 m3 . The density is 11.33. 1.03 × 103 kg
= 1.09 × 103 kg/m3 .
0.9473 m3
EVALUATE: The density is increased because the volume is compressed due to the increased pressure.
IDENTIFY and SET UP: Use Eqs.(11.13) and (11.14) to calculate B and k.
EXECUTE: 11.34. B=− Δp
(3.6 × 106 Pa)(600 cm3 )
=−
= +4.8 × 109 Pa
ΔV / V0
(−0.45 cm3 ) k = 1/ B = 1/ 4.8 × 109 Pa = 2.1 × 10−10 Pa −1
EVALUATE: k is the same as for glycerine (Table 11.2).
IDENTIFY: Apply Eq.(11.17).
SET UP: F" = 9.0 × 105 N . A = (0.100 m)(0.500 × 10−2 m) . h = 0.100 m . From Table 11.1, S = 7.5 × 1010 Pa for
steel.
EXECUTE: (a) Shear strain = F||
AS = (9 × 105 N)
= 2.4 × 10−2.
[(0.100 m)(0.500 × 10−2 m)][7.5 × 1010 Pa] (b) Using Eq.(11.16), x = (Shear strain) ⋅ h = (0.024)(0.100 m) = 2.4 × 10−3 m .
EVALUATE: This very large force produces a small displacement; x / h = 2.4% .
11.35. IDENTIFY: The forces on the cube must balance. The deformation x is related to the force by S = F" h
.
Ax F" = F since F is applied parallel to the upper face.
SET UP: A = (0.0600 m) 2 and h = 0.0600 m . Table 11.1 gives S = 4.4 × 1010 Pa for copper and 0.6 × 1010 Pa for
lead.
EXECUTE: (a) Since the horizontal forces balance, the glue exerts a force F in the opposite direction.
(b) F = Fh
(6.6 × 105 N)(0.0600 m)
=
= 1.8 mm
AS (0.0600 m) 2 (0.6 × 1010 Pa)
EVALUATE: Lead has a smaller S than copper, so the lead cube has a greater deformation than the copper cube.
IDENTIFY and SET UP: Use Eq.(11.17). Same material implies same S
stress F" / A
stress
EXECUTE: S =
so strain =
=
and same forces implies same F" .
strain
S
S
(c) x = 11.36. AxS (0.0600 m)2 (0.250 × 10−3 m)(4.4 × 1010 Pa)
=
= 6.6 × 105 N
h
0.0600 m Equilibrium and Elasticity 11-13 For the smaller object, (strain)1 = F" / A1S
For the larger object, (strain) 2 = F" / A2 S
(strain) 2 ⎛ F" ⎞ ⎛ A1S ⎞ A1
=⎜
⎟=
⎟⎜
(strain)1 ⎝ A2 S ⎠ ⎜ F" ⎟ A2
⎝
⎠
Larger solid has triple each edge length, so A2 = 9 A1 , and
11.37. 11.38. (strain) 2 1
=
(strain)1 9 EVALUATE: The larger object has a smaller deformation.
IDENTIFY and SET UP: Use Eq.(11.8).
F
F
90.8 N
EXECUTE: Tensile stress = ⊥ = ⊥2 =
= 3.41 × 107 Pa
A πr
π (0.92 × 10−3 m) 2
EVALUATE: A modest force produces a very large stress because the cross-sectional area is small.
IDENTIFY: The proportional limit and breaking stress are values of the stress, F⊥ /A . Use Eq.(11.10) to
calculate Δl .
SET UP: For steel, Y = 20 × 1010 Pa . F⊥ = w .
EXECUTE: (a) w = (1.6 × 10−3 )(20 × 1010 Pa)(5 × 10−6 m 2 ) = 1.60 × 103 N. ⎛ F ⎞l
(b) Δl = ⎜ ⊥ ⎟ 0 = (1.6 × 10−3 )(4.0 m) = 6.4 mm
⎝ A ⎠Y 11.39. (c) (6.5 × 10−3 )(20 × 1010 Pa)(5 × 10−6 m 2 ) = 6.5 × 103 N.
EVALUATE: At the proportional limit, the fractional change in the length of the wire is 0.16%.
!
!
IDENTIFY: The elastic limit is a value of the stress, F⊥ / A . Apply ∑ F = ma to the elevator in order to find the tension in the cable.
F⊥ 1
SET UP:
= (2.40 × 108 Pa) = 0.80 × 108 Pa . The free-body diagram for the elevator is given in Figure 11.39.
A3
F⊥ is the tension in the cable.
EXECUTE: F⊥ = A(0.80 × 108 Pa) = (3.00 × 10−4 m 2 )(0.80 × 108 Pa) = 2.40 × 104 N . ∑F y = ma y applied to the F⊥
2.40 × 10 N
−g=
− 9.80 m/s 2 = 10.2 m/s 2
m
1200 kg
EVALUATE: The tension in the cable is about twice the weight of the elevator. elevator gives F⊥ − mg = ma and a = 4 Figure 11.39
11.40. IDENTIFY: The breaking stress of the wire is the value of F⊥ / A at which the wire breaks. SET UP: From Table 11.3, the breaking stress of brass is 4.7 × 108 Pa . The area A of the wire is related to its
diameter by A = π d 2 / 4 .
350 N
EXECUTE: A =
= 7.45 × 10−7 m 2 , so d = 4 A π = 0.97 mm.
8
4.7 × 10 Pa
EVALUATE: The maximum force a wire can withstand without breaking is proportional to the square of its
diameter. 11-14 Chapter 11 11.41. IDENTIFY: Apply the conditions of equilibrium to the climber. For the minimum coefficient of friction the static
friction force has the value f s = μs n . The free-body diagram for the climber is given in Figure 11.41. f s and n are the vertical and horizontal SET UP: components of the force exerted by the cliff face on the climber. The moment arm for the force T is (1.4 m)cos10° .
EXECUTE: (a) ∑τ z = 0 gives T (1.4 m)cos10° − w(1.1 m)cos35.0° = 0 . T= (1.1 m)cos35.0°
(82.0 kg)(9.80 m/s 2 ) = 525 N
(1.4 m)cos10° (b) ∑F x = 0 gives n = T sin 25.0° = 222 N . ∑F y = 0 gives f s + T cos 25° − w = 0 and f s = (82.0 kg)(9.80 m/s ) − (525 N)cos 25° = 328 N .
2 f s 328 N
=
= 1.48
n 222 N
EVALUATE: To achieve this large value of μs the climber must wear special rough-soled shoes.
(c) μs = Figure 11.41
11.42. IDENTIFY: Apply ∑τ z = 0 to the bridge. SET UP: Let the axis of rotation be at the left end of the bridge and let counterclockwise torques be positive.
EXECUTE: If Lancelot were at the end of the bridge, the tension in the cable would be (from taking torques about
the hinge of the bridge) obtained from T (12.0 N) = (600 kg)(9.80 m s 2 )(12.0 m) + (200 kg)(9.80 m s 2 )(6.0 m) ,
so T = 6860 N . This exceeds the maximum tension that the cable can have, so Lancelot is going into the drink. To
find the distance x Lancelot can ride, replace the 12.0 m multiplying Lancelot’s weight by x and the tension
T by Tmax = 5.80 × 103 N and solve for x; x= 11.43. (5.80 × 10 3 N)(12.0 m) − (200 kg)(9.80 m s 2 )(6.0 m)
= 9.84 m.
(600 kg)(9.80 m s 2 ) EVALUATE: Before Lancelot goes onto the bridge, the tension in the supporting cable is
(6.0 m)(200 kg)(9.80 m/s 2 )
T=
= 9800 N , well below the breaking strength of the cable. As he moves along the
12.0 m
bridge, the increase in tension is proportional to x, the distance he has moved along the bridge.
IDENTIFY: For the airplane to remain in level flight, both ∑ Fy = 0 and ∑τ z = 0 .
SET UP:
EXECUTE: The free-body diagram for the airplane is given in Figure 11.43. Let + y be upward.
Ftail − W + Fwing = 0 . Taking the counterclockwise direction as positive, and taking torques about the point where the tail force acts, −(3.66 m)(6700 N) + (3.36 m) Fwing = 0. This gives Fwing = 7300 N(up) and Ftail = 7300 N − 6700 N = 600 N(down). Equilibrium and Elasticity 11-15 EVALUATE: We assumed that the wing force was upward and the tail force was downward. When we solved for
these forces we obtained positive values for them, which confirms that they do have these directions. Note that the
rear stabilizer provides a downward force. It does not hold up the tail of the aircraft, but serves to counter the
torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading. Figure 11.43
11.44. IDENTIFY: Apply the first and second conditions of equilibrium to the truck.
SET UP: The weight on the front wheels is nf , the normal force exerted by the ground on the front wheels. The weight on the rear wheels is nr , the normal force exerted by the ground on the rear wheels. When the front wheels
come off the ground, nf → 0 . The free-body diagram for the truck without the box is given in Figure 11.44a and
with the box in Figure 11.44b. The center of gravity of the truck, without the box, is a distance x from the rear
wheels.
EXECUTE: ∑ Fy = 0 in Fig.11.44a gives w = nr + nf = 8820 N + 10,780 N = 19,600 N ∑τ = 0 in Fig.11.44a, with the axis at the rear wheels and counterclockwise torques positive, gives nf (3.00 m) ⎛ 10,780 N ⎞
=⎜
⎟ (3.00 m) = 1.65 m .
w
⎝ 19,600 N ⎠
(a) ∑τ = 0 in Fig.11.44b, with the axis at the rear wheels and counterclockwise torques positive, gives
nf (3.00 m) − wx = 0 and x = wbox (1.00 m) + nf (3.00 m) − w(1.65 m) = 0 .
nf = ∑F y −(3600 N)(1.00 m) + (19,600 N)(1.65 m)
= 9,580 N
3.00 m = 0 gives nr + nf = wbox + w and nr = 3600 N + 19,600 N − 9580 N = 13,620 N . There is 9,580 N on the front wheels and 13,620 N on the rear wheels.
(b) nf → 0 . ∑τ = 0 gives wbox (1.00 m) − w(1.65 m) = 0 and wbox = 1.65w = 3.23 × 104 N .
EVALUATE: Placing the box on the tailgate in part (b) reduces the normal force exerted at the front wheels. Figure 11.44a, b
11.45. IDENTIFY: In each case, to achieve balance the center of gravity of the system must be at the fulcrum. Use
Eq.(11.3) to locate xcm , with mi replaced by wi .
SET UP: Let the origin be at the left-hand end of the rod and take the + x axis to lie along the rod. Let
w1 = 255 N (the rod) so x1 = 1.00 m , let w2 = 225 N so x2 = 2.00 m and let w3 = W . In part (a) x3 = 0.500 m and in part (b) x3 = 0.750 m .
EXECUTE: (a) xcm = 1.25 m . xcm = w1 x1 + w2 x2 + w3 x3
( w + w2 ) xcm − w1 x1 − w2 x2
gives w3 = 1
and
w1 + w2 + w3
x3 − xcm (480 N)(1.25 m) − (255 N)(1.00 m) − (225 N)(2.00 m)
= 140 N .
0.500 m − 1.25 m
(b) Now w3 = W = 140 N and x3 = 0.750 m . W= (255 N)(1.00 m) + (225 N)(2.00 m) + (140 N)(0.750 m)
= 1.31 m . W must be moved
255 N + 225 N + 140 N
1.31 m − 1.25 m = 6 cm to the right.
EVALUATE: Moving W to the right means xcm for the system moves to the right. xcm = 11-16 Chapter 11 11.46. IDENTIFY: The center of gravity of the object must have the same x coordinate as the hook. Use Eq.(11.3) for
xcm . The mass of a segment is proportional to its length. Define α to be the mass per unit length, so mi = α li , where li is the length of a piece that has mass mi .
SET UP: Use coordinates with the origin at the right-hand edge of the object and + x to the left. xcm = L . The
mass of each piece can be taken at its center of gravity, which is at its geometrical center. Let 1 be the horizontal
piece of length L, 2 be the vertical piece of length L and 3 be the horizontal piece with length x.
m x + m2 x2 + m3 x3
α L( L / 2) + α x( x / 2)
. α divides out and the equation reduces to
EXECUTE: xcm = 1 1
gives L =
αL +αL +αx
m1 + m2 + m3 x 2 − 2 xL − 3L2 = 0 . x = 1 (2 L ± 4 L) . x must be positive, so x = 3L .
2
11.47. EVALUATE: xcm = L is equivalent to saying that the net torque is zero for an axis at the hook.
IDENTIFY: Apply the conditions of equilibrium to the horizontal beam. Since the two wires are symmetrically
placed on either side of the middle of the sign, their tensions are equal and are each equal to Tw = mg / 2 = 137 N .
SET UP: The free-body diagram for the beam is given in Figure 11.47. Fv and Fh are the horizontal and vertical
forces exerted by the hinge on the sign. Since the cable is 2.00 m long and the beam is 1.50 m long,
1.50 m
cosθ =
and θ = 41.4° . The tension Tc in the cable has been replaced by its horizontal and vertical
2.00 m
components.
EXECUTE: (a) ∑τ z = 0 gives Tc (sin 41.4°)(1.50 m) − wbeam (0.750 m) − Tw (1.50 m) − Tw (0.60 m) = 0 . Tc =
(b) (18.0 kg)(9.80 m/s 2 )(0.750 m) + (137 N)(1.50 m + 0.60 m)
= 423 N .
(1.50 m)(sin 41.4°) ∑F y = 0 gives Fv + Tc sin 41.4° − wbeam − 2Tw = 0 and Fv = 2Tw + wbeam − Tc sin 41.4° = 2(137 N) + (18.0 kg)(9.80 m/s 2 ) − (423 N)(sin 41.4°) = 171 N . The hinge must be
able to supply a vertical force of 171 N.
EVALUATE: The force from the two wires could be replaced by the weight of the sign acting at a point 0.60 m to
the left of the right-hand edge of the sign. Figure 11.47
11.48. IDENTIFY: Apply ∑τ z = 0 to the hammer. SET UP: Take the axis of rotation to be at point A.
!
EXECUTE: The force F1 is directed along the length of the nail, and so has a moment arm of (0.800 m)sin 60° .
!
The moment arm of F2 is 0.300 m, so F2 = F1 11.49. (0.0800 m)sin 60°
= (500 N)(0.231) = 116 N.
(0.300 m) EVALUATE: The force F2 that must be applied to the hammer handle is much less than the force that the hammer
applies to the nail, because of the large difference in the lengths of the moment arms.
IDENTIFY: Apply the first and second conditions of equilibrium to the bar.
SET UP: The free-body diagram for the bar is given in Figure 11.49. n is the normal force exerted on the bar by
the surface. There is no friction force at this surface. H h and H v are the components of the force exerted on the Equilibrium and Elasticity 11-17 bar by the hinge. The components of the force of the bar on the hinge will be equal in magnitude and opposite in
direction.
EXECUTE:
∑ Fx = max F = H h = 120 N ∑F y = ma y n − Hv = 0
H v = n, but we don’t know
either of these forces.
Figure 11.49 ∑τ 11.50. B = 0 gives F (4.00 m) − n(3.00 m) = 0 4
n = (4.00 m/3.00 m)F = 3 (120 N) = 160 N and then H v = 160 N
Force of bar on hinge:
horizontal component 120 N, to right
vertical component 160 N, upward
!
EVALUATE: H h / H v = 120 /160 = 3.00 / 4.00, so the force the hinge exerts on the bar is directed along the bar. n
!
!
and F have zero torque about point A, so the line of action of the hinge force H must pass through this point
also if the net torque is to be zero.
IDENTIFY: Apply ∑τ z = 0 to the piece of art. The free-body diagram for the piece of art is given in Figure 11.50.
⎛ 1.02 m ⎞
EXECUTE: ∑τ z = 0 gives TB (1.25 m) − w(1.02 m) = 0 . TB = (358 N) ⎜
⎟ = 292 N .
⎝ 1.25 m ⎠
SET UP: ∑F = 0 gives y TA + TB − w = 0 and TA = w − TB = 358 N − 292 N = 66 N .
EVALUATE: If we consider the sum of torques about the center of gravity of the piece of art, TA has a larger moment arm than TB , and this is why TA < TB . Figure 11.50
11.51. IDENTIFY: Apply the conditions of equilibrium to the beam.
SET UP: The free-body diagram for the beam is given in Figure 11.51. Let Tφ and Tθ be the tension in the two cables. Each tension has been replaced by its horizontal and vertical components.
EXECUTE: (a) The center of gravity of the beam is a distance L / 2 from each end and ∑τ z the center of gravity of the beam gives −Tφ sin φ ( L / 2) + Tθ sin θ ( L / 2) = 0 . Tφ sin φ = Tθ sin θ . = 0 with the axis at ∑F x = 0 gives Tφ cos φ = Tθ cosθ . Dividing the first equation by the second gives tan φ = tan θ and φ = θ . Then the equations also
say Tφ = Tθ .
(b) The center of gravity of the beam is a distance 3L / 4 from the left-hand end so a distance L / 4 from the righthand end. ∑τ z = 0 with the axis at the center of gravity of the beam gives −Tφ sin φ (3L / 2) + Tθ sinθ ( L / 2) = 0 and 3Tφ sin φ = Tθ sin θ .
3tan φ = tan θ . ∑F x = 0 gives Tφ cos φ = Tθ cosθ . Dividing the first equation by the second gives 11-18 Chapter 11 EVALUATE: 3 tan φ = tan θ requires θ > φ . The cable closest to the center of gravity must be closer to the vertical ⎛ cos φ ⎞
direction. Tθ = Tφ ⎜
⎟ and θ > φ means the tension is greater in the wire that is closest to the center of gravity.
⎝ cosθ ⎠ Figure 11.51
11.52. IDENTIFY: Apply the first and second conditions for equilibrium to the bridge.
SET UP: Find torques about the hinge. Use L as the length of the bridge and wT and wB for the weights of the
truck and the raised section of the bridge. Take + y to be upward and + x to be to the right.
EXECUTE: (a) TL sin70° = wT ( 3 L ) cos30° + wB ( 1 L ) cos30° , so
4
2 T= ( 3 mT + 12 mB ) (9.80 m s 2 )cos30° = 2.57 × 105 N.
4
sin 70° (b) Horizontal: T cos ( 70° − 30° ) = 1.97 × 10 N (to the right). Vertical: wT + wB − T sin 40° = 2.46 × 105 N (upward).
5 EVALUATE:
11.53. If φ is the angle of the hinge force above the horizontal, tan φ = 2.46 × 105 N
and φ = 51.3° . The
1.97 × 105 N hinge force is not directed along the bridge.
IDENTIFY: Apply the conditions of equilibrium to the cylinder.
SET UP: The free-body diagram for the cylinder is given in Figure 11.53. The center of gravity of the cylinder is
at its geometrical center. The cylinder has radius R.
EXECUTE: (a) T produces a clockwise torque about the center of gravity so there must be a friction force, that
produces a counterclockwise torque about this axis.
(b) Applying ∑τ z = 0 to an axis at the center of gravity gives −TR + fR = 0 and T = f . ∑τ z = 0 applied to an axis at the point of contact between the cylinder and the ramp gives −T (2R ) + MgR sin θ = 0 . T = ( Mg / 2)sin θ .
EVALUATE: We can show that ∑F x = 0 and ∑F y = 0 , for x and y axes parallel and perpendicular to the ramp, or for x and y axes that are horizontal and vertical. Figure 11.53 Equilibrium and Elasticity 11.54. IDENTIFY: Apply the first and second conditions of equilibrium to the ladder.
SET UP: Take torques about the pivot. Let + y be upward.
EXECUTE: 11.55. 11-19 (a) The force FV that the ground exerts on the ladder is given to be vertical, so ∑τ z =0 gives FV (6.0 m)sin θ = (250 N)(4.0 m)sin θ + (750 N)(1.50 m)sin θ , so FV = 354 N.
(b) There are no other horizontal forces on the ladder, so the horizontal pivot force is zero. The vertical force that
the pivot exerts on the ladder must be (750 N) + (250 N) − (354 N) = 646 N, up, so the ladder exerts a downward
force of 646 N on the pivot.
(c) The results in parts (a) and (b) are independent of θ.
EVALUATE: All the forces on the ladder are vertical, so all the moment arms are vertical and are proportional to
sin θ . Therefore, sin θ divides out of the torque equations and the results are independent of θ .
IDENTIFY: Apply the first and second conditions for equilibrium to the strut.
SET UP: Denote the length of the strut by L .
EXECUTE: (a) V = mg + w and H = T . To find the tension, take torques about the pivot point.
mg ⎞
⎛2 ⎞
⎛2 ⎞
⎛L⎞
⎛
T ⎜ L ⎟ sin θ = w ⎜ L ⎟ cos θ + mg ⎜ ⎟ cos θ and T = ⎜ w +
⎟ cot θ .
4⎠
3⎠
3⎠
6⎠
⎝
⎝
⎝
⎝
(b) Solving the above for w , and using the maximum tension for T ,
mg
= (700 N) tan 55.0° − (5.0 kg)(9.80 m s 2 ) = 951 N.
w = T tan θ −
4 mg
= 0.700, so θ = 4.00°.
4T
EVALUATE: As the strut becomes closer to the horizontal, the moment arm for the horizontal tension force
approaches zero and the tension approaches infinity.
IDENTIFY: Apply the first and second conditions of equilibrium to each rod.
SET UP: Apply ∑ Fy = 0 with + y upward and apply ∑τ = 0 with the pivot at the point of suspension for each
(c) Solving the expression obtained in part (a) for tan θ and letting ω → 0, tan θ = 11.56. rod.
EXECUTE: (a) The free-body diagram for each rod is given in Figure 11.56.
(b) ∑τ = 0 for the lower rod: (6.0 N)(4.0 cm) = wA (8.0 cm) and wA = 3.0 N . ∑F y = 0 for the lower rod: S3 = 6.0 N + wA = 9.0 N ∑τ = 0 for the middle rod:
∑F y = 0 for the middle rod: S 2 = 9.0 N + S3 = 24.0 N ∑τ = 0 for the upper rod:
∑F y ⎛ 5.0 ⎞
wB (3.0 cm) = (5.0 cm)S3 and wB = ⎜
⎟ (9.0 N) = 15.0 N .
⎝ 3.0 ⎠ ⎛ 2.0 ⎞
S 2 (2.0 cm) = wC (6.0 cm) and wC = ⎜
⎟ (24.0 N) = 8.0 N .
⎝ 6.0 ⎠ = 0 for the upper rod: S1 = S 2 + wC = 32.0 N . In summary, wA = 3.0 N , wB = 15.0 N , wC = 8.0 N . S1 = 32.0 N , S 2 = 24.0 N , S3 = 9.0 N .
(c) The center of gravity of the entire mobile must lie along a vertical line that passes through the point where S1 is
located.
EVALUATE: For the mobile as a whole the vertical forces must balance, so S1 = wA + wB + wC + 6.0 N . Figure 11.56 11-20 Chapter 11 11.57. IDENTIFY: Apply ∑τ z = 0 to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.57.
EXECUTE: Στ z = 0, axis at hinge , gives T (6.0 m)(sin 40°) − w(3.75 m)(cos30°) = 0 and T = 7600 N .
EVALUATE: The tension in the cable is less than the weight of the beam. T sin 40° is the component of T that is
perpendicular to the beam. Figure 11.57
11.58. IDENTIFY: Apply the first and second conditions of equilibrium to the drawbridge.
SET UP: The free-body diagram for the drawbridge is given in Figure 11.58. H v and H h are the components of
the force the hinge exerts on the bridge.
EXECUTE: (a) ∑τ z = 0 with the axis at the hinge gives − w(7.0 m)(cos37°) + T (3.5 m)(sin 37°) = 0 and cos37° (45,000 N)
=
= 1.19 × 105 N
sin 37°
tan 37°
(b) ∑ Fx = 0 gives H h = T = 1.19 × 105 N .
T = 2w 2
2
H = H h + H v = 1.27 × 105 N . tan θ = ∑F y = 0 gives H v = w = 4.50 × 104 N . Hv
and θ = 20.7° . The hinge force has magnitude 1.27 × 105 N and is
Hh directed at 20.7° above the horizontal.
EVALUATE: The hinge force is not directed along the bridge. If it were, it would have zero torque for an axis at
the center of gravity of the bridge and for that axis the tension in the cable would produce a single, unbalanced
torque. Figure 11.58
11.59. IDENTIFY: Apply the first and second conditions of equilibrium to the beam.
SET UP: The free-body diagram for the beam is given in Figure 11.59. Figure 11.59 Equilibrium and Elasticity EXECUTE: (a) ∑τ z 11-21 = 0, axis at lower end of beam Let the length of the beam be L.
⎛L⎞
T (sin 20°) L = −mg ⎜ ⎟ cos 40° = 0
⎝2⎠
1
mg cos 40°
T=2
= 2700 N
sin 20°
(b) Take + y upward. ∑F
∑F y = 0 gives n − w + T sin 60° = 0 so n = 73.6 N x = 0 gives f s = T cos60° = 1372 N 11.60. f s 1372 N
=
= 19
n 73.6 N
EVALUATE: The floor must be very rough for the beam not to slip. The friction force exerted by the floor is to
the left because T has a component that pulls the beam to the right.
IDENTIFY: Apply ∑τ z = 0 to the beam. 11.61. SET UP: The center of mass of the beam is 1.0 m from the suspension point.
EXECUTE: (a) Taking torques about the suspension point,
w(4.00 m)sin 30° + (140.0 N)(1.00 m)sin 30° = (100 N)(2.00 m)sin 30° . The common factor of sin 30° divides out,
from which w = 15.0 N.
(b) In this case, a common factor of sin 45° would be factored out, and the result would be the same.
EVALUATE: All the forces are vertical, so the moments are all horizontal and all contain the factor sin θ , where
θ is the angle the beam makes with the horizontal.
IDENTIFY: Apply ∑τ z = 0 to the flagpole. f s = μs n, μs = SET UP: The free-body diagram for the flagpole is given in Figure 11.61. Let clockwise torques be positive. θ is
the angle the cable makes with the horizontal pole.
EXECUTE: (a) Taking torques about the hinged end of the pole
(200 N)(2.50 m) + (600 N)(5.00 m) − Ty (5.00 m) = 0 . Ty = 700 N . The x-component of the tension is then Tx = (1000 N) 2 − (700 N) 2 = 714 N . tan θ = T
h
= y . The height above the pole that the wire must be
5.00 m Tx attached is (5.00 m) 700 = 4.90 m .
714
(b) The y-component of the tension remains 700 N. Now tan θ = Ty 4.40 m
and θ = 41.35° , so
5.00 m 700 N
= 1060 N , an increase of 60 N.
sin 41.35°
EVALUATE: As the wire is fastened closer to the hinged end of the pole, the moment arm for T decreases and T
must increase to produce the same torque about that end.
T= 11.62. sin θ = IDENTIFY: Figure 11.61
!
Apply ∑ F = 0 to each object, including the point where D, C and B are joined. Apply ∑τ z = 0 to the rod.
SET UP: To find TC and TD , use a coordinate system with axes parallel to the cords.
EXECUTE: A and B are straightforward, the tensions being the weights suspended:
Τ A = (0.0360 kg)(9.80 m/s 2 ) = 0.353 N and TB = (0.0240 kg + 0.0360 kg)(9.80 m s 2 ) = 0.588 N . Applying ∑F x = 0 and ∑F y = 0 to the point where the cords are joined, TC = TB cos36.9° = 0.470 N and TD = TB cos53.1° = 0.353 N . To find TE , take torques about the point where string F is attached. 11-22 Chapter 11 TE (1.00 m) = TD sin 36.9° (0.800 m) + TC sin 53.1° (0.200 m) + (0.120 kg)(9.80 m s 2 )(0.500 m) and TE = 0.833 N.
TF may be found similarly, or from the fact that TE + TF must be the total weight of the ornament. (0.180kg)(9.80m s 2 ) = 1.76 N, from which TF = 0.931 N. 11.63. EVALUATE: The vertical line through the spheres is closer to F than to E, so we expect TF > TE , and this is
indeed the case.
IDENTIFY: Apply the equilibrium conditions to the plate. τ = Fr sin φ .
SET UP: The free-body diagram for the plate is sketched in Figure 11.63. For the force T (tension in the cable), τ T = Tr sin φ = T h 2 + d 2 sin φ .
EXECUTE: (a) ∑τ z ⎛h⎞
⎝⎠ = 0 gives T h 2 + d 2 sin φ − W θ = tan −1 ⎜ ⎟ . Then T = W
d d 2 h + d2
2 d
h
= 0 . T is least for φ = 90° , and in that case tan θ = so
d
2 . ⎛
⎞
h
Whd
. ∑ Fy = 0 gives Fv + T cosθ − W = 0 and
⎜2
⎟=
2
2(h 2 + d 2 )
2 h +d ⎝ h +d ⎠
⎛
⎞⎛
⎞
⎛
⎞
Wd
d
d2
2h 2 + d 2
Fv = W − ⎜
⎟⎜ 2
⎟ = W ⎜1 −
⎟ =W
2
2
2
2
2
2(h 2 + d 2 )
⎝ 2(h + d ) ⎠
⎝ 2 h + d ⎠⎝ h + d ⎠
EVALUATE: The angle α that the net force exerted by the hinge makes with the horizontal is given by
F W (2h 2 + d 2 ) 2(h 2 + d 2 ) 2h 2 + d 2
tan α = v =
=
. This force does not lie along the diagonal of the plate.
2(h 2 + d 2 )
Fh
Whd
hd
(b) ∑F x = 0 gives Fh = T sin θ = Wd
2 2 Figure 11.63
11.64. Apply Eq.(11.10) and the relation Δw / w0 = −σΔl / l0 that is given in the problem. IDENTIFY: The steel rod in Example 11.5 has Δl / l0 = 9.0 × 10−4 . For nickel, Y = 2.1 × 1011 Pa . The width w0 is SET UP: w0 = 4 A/π .
EXECUTE: (a) Δw = −σ ( Δl l ) w0 = −(0.23)(9.0 × 10−4 ) 4(0.30 × 10−4 m 2 ) π = 1.3 μm. (2.1 × 1011 Pa) (π (2.0 × 10−2 m) 2 ) 0.10 × 10−3 m
Δl
1 Δw
= 3.1 × 106 N .
= AY
and F⊥ =
l
σw
0.42
2.0 × 10−2 m
EVALUATE: For nickel and steel, σ < 1 and the fractional change in width is less than the fractional change in length.
IDENTIFY: Apply the equilibrium conditions to the crate. When the crate is on the verge of tipping it touches the
floor only at its lower left-hand corner and the normal force acts at this point. The minimum coefficient of static
friction is given by the equation f s = μs n .
SET UP: The free-body diagram for the crate when it is ready to tip is given in Figure 11.65.
⎛
⎞
1.10 m
3
EXECUTE: (a) ∑τ z = 0 gives P (1.50 m)sin 53.0° − w(1.10 m) = 0 . P = w ⎜
⎟ = 1.15 × 10 N
[1.50 m][sin53.0°] ⎠
⎝
(b) F⊥ = AY 11.65. ∑ F = 0 gives n − w − P cos53.0° = 0 . n = w + P cos53.0° = 1250 N + (1.15 × 10
(c) ∑ F = 0 gives f = P sin 53.0° = (1.15 × 10 N)sin 53.0° = 918 N .
(b) y 3 x (d) μs = s fs
918 N
=
= 0.473
n 1.94 × 103 N 3 N)cos53° = 1.94 × 103 N Equilibrium and Elasticity EVALUATE: 11-23 The normal force is greater than the weight because P has a downward component. Figure 11.65
11.66. IDENTIFY: Apply ∑τ z = 0 to the meter stick. SET UP: The wall exerts an upward static friction force f and a horizontal normal force n on the stick. Denote the
length of the stick by l. f = μs n .
EXECUTE: (a) Taking torques about the right end of the stick, the friction force is half the weight of the
stick, f = w / 2 . Taking torques about the point where the cord is attached to the wall (the tension in the cord and
the friction force exert no torque about this point), and noting that the moment arm of the normal force is l tan θ ,
n tan θ = w / 2 ⋅ Then, ( f / n) = tan θ < 0.40, so θ < arctan (0.40) = 22°. l
l
(b) Taking torques as in part (a), fl = w + w(l − x) and nl tan θ = w + wx. In terms of the coefficient of friction
2
2
l 3 tan θ − μs
f l / 2 + (l − x)
3l − 2 x
= 30.2 cm.
tan θ =
tan θ. Solving for x, x >
μs , μs > =
n
l/2+ x
l + 2x
2 μs + tan θ
(c) In the above expression, setting x = 10 cm and solving for μs gives μs >
EVALUATE:
11.67. (3 − 20 l ) tan θ
= 0.625.
1 + 20 l For θ = 15° and without the block suspended from the stick, a value of μs ≥ 0.268 is required to prevent slipping. Hanging the block from the stick increases the value of μs that is required.
IDENTIFY: Apply the first and second conditions of equilibrium to the crate.
SET UP: The free-body diagram for the crate is given in Figure 11.67. lw = (0.375 m)cos 45°
l2 = (1.25 m)cos 45°
!
!
Let F1 and F2 be the vertical
forces exerted by you and your
friend. Take the origin at the
lower left-hand corner of the
crate (point A). Figure 11.67 11-24 Chapter 11 EXECUTE: ∑F y = ma y gives F1 + F2 − w = 0 F1 + F2 = w = (200 kg)(9.80 m/s 2 ) = 1960 N ∑τ A = 0 gives F2l2 − wlw = 0 ⎛l ⎞
⎛ 0.375 mcos 45° ⎞
F2 = w ⎜ w ⎟ = 1960 N ⎜
⎟ = 590 N
⎝ 1.25 mcos 45° ⎠
⎝ l2 ⎠
Then F1 = w − F2 = 1960 N − 590 N = 1370 N.
EVALUATE: The person below (you) applies a force of 1370 N. The person above (your friend) applies a force of
!
!
590 N. It is better to be the person above. As the sketch shows, the moment arm for F1 is less than for F2 , so must
11.68. have F1 > F2 to compensate.
IDENTIFY: Apply the first and second conditions for equilibrium to the forearm.
SET UP: The free-body diagram is given in Figure 11.68a, and when holding the weight in Figure 11.68b. Let
+ y be upward.
EXECUTE: (a) Στ Elbow = 0 gives FB (3.80 cm) = (15.0 N)(15.0 cm) and FB = 59.2 N . (b) Στ E = 0 gives FB (3.80 cm) = (15.0 N)(15.0 cm) + (80.0 N)(33.0 cm) and FB = 754 N . The biceps force has a
short lever arm, so it must be large to balance the torques.
(c) ΣFy = 0 gives − FE + FB − 15.0 N − 80.0 N = 0 and FE = 754N − 15.0 N − 80.0 N = 659 N .
EVALUATE: (d) The biceps muscle acts perpendicular to the forearm, so its lever arm stays the same, but those of
the other two forces decrease as the arm is raised. Therefore the tension in the biceps muscle decreases. Figure 11.68a, b
11.69. IDENTIFY: Apply ∑τ z = 0 to the forearm. The free-body diagram for the forearm is given in Fig. 11.10 in the textbook.
h
hD
EXECUTE: (a) Στ z = 0, axis at elbow gives wL − (T sin θ ) D = 0 . sin θ =
.
so w = T
2
2
h +D
L h2 + D 2
hD
.
wmax = Tmax
L h2 + D 2
SET UP: dwmax
Tmax h ⎛
D2 ⎞
=
; the derivative is positive
⎜1 − 2
2⎟
2
2
dD
L h +D ⎝ h +D ⎠
EVALUATE: (c) The result of part (b) shows that wmax increases when D increases, since the derivative is positive.
(b) 11.70. wmax is larger for a chimp since D is larger.
IDENTIFY: Apply the first and second conditions for equilibrium to the table.
SET UP: Label the legs as shown in Figure 11.70a. Legs A and C are 3.6 m apart. Let the weight be placed closest
to legs C and D. By symmetry, A = B and C = D . Redraw the table as viewed from the AC side. The free-body
diagram in this view is given in Figure 11.70b.
EXECUTE: ∑ τ z (about right end) = 0 gives 2 A(3.6 m) = ( 90.0 N ) (1.8 m) + (1500 N ) (0.50 m) and
A = 130 N = B . ∑F y = 0 gives A + B + C + D = 1590 N . Using A = B = 130 N and C = D gives C = D = 670 N . By Newton’s third law of motion, the forces A, B, C, and D on the table are the same magnitude as the forces the
table exerts on the floor. Equilibrium and Elasticity EVALUATE: 11-25 As expected, the legs closest to the 1500 N weight exert a greater force on the floor. Figure 11.70a, b
11.71. Apply IDENTIFY:
(a) SET UP: ∑τ z = 0 first to the roof and then to one wall. Consider the forces on the roof; see Figure 11.71a.
V and H are the vertical
and horizontal forces each
wall exerts on the roof.
w = 20,000 N is the total
weight of the roof.
2V = w so V = w / 2
Figure 11.71a Apply ∑τ z = 0 to one half of the roof, with the axis along the line where the two halves join. Let each half have length L.
EXECUTE: ( w / 2)( L / 2)(cos35.0°) + HL sin 35.0° − VL cos35° = 0
L divides out, and use V = w / 2
H sin 35.0° = 1 w cos35.0°
4
w
= 7140 N
4 tan 35.0°
EVALUATE: By Newton’s 3rd law, the roof exerts a horizontal, outward force on the wall. For torque about an
axis at the lower end of the wall, at the ground, this force has a larger moment arm and hence larger torque the
taller the walls.
(b) SET UP: The force diagram for one wall is given in Figure 11.71b.
H= Consider the torques
on this wall.
Figure 11.71b H is the horizontal force exerted by the roof, as considered in part (a). B is the horizontal force exerted by the
w
= 5959 N
buttress. Now the angle is 40°, so H =
4 tan 40°
EXECUTE: ∑τ z = 0, axis at the ground 11.72. H (40 m) − B(30 m) = 0 and B = 7900 N.
EVALUATE: The horizontal force exerted by the roof is larger as the roof becomes more horizontal, since for
torques applied to the roof the moment arm for H decreases. The force B required from the buttress is less the
higher up on the wall this force is applied.
IDENTIFY: Apply ∑τ z = 0 to the wheel.
SET UP: Take torques about the upper corner of the curb.
!
EXECUTE: The force F acts at a perpendicular distance R − h and the weight acts at a perpendicular distance R 2 − ( R − h ) = 2 Rh − h 2 . Setting the torques equal for the minimum necessary force, F = mg
2 2 Rh − h 2
.
R−h 11-26 Chapter 11 !
(b) The torque due to gravity is the same, but the force F acts at a perpendicular distance 2 R − h, so the minimum
force is (mg ) 2 Rh − hv /(2 R − h). 11.73. #
EVALUATE: (c) Less force is required when the force is applied at the top of the wheel, since in this case F has a
larger moment arm.
IDENTIFY: Apply the first and second conditions of equilibrium to the gate.
SET UP: The free-body diagram for the gate is given in Figure 11.73. Figure 11.73
!
!
Use coordinates with the origin at B. Let H A and H B be the forces exerted by the hinges at A and B. The problem
!
!
states that H A has no horizontal component. Replace the tension T by its horizontal and vertical components.
EXECUTE: (a) ∑τ B = 0 gives + (T sin 30.0°)(4.00 m) + (T cos30.0°)(2.00 m) − w(2.00 m) = 0 T ( 2sin 30.0° + cos30.0°) = w
w
500 N
=
= 268 N
2sin 30.0° + cos30.0° 2sin 30.0° + cos30.0°
(b) ∑ Fx = max says H Bh − T cos30.0° = 0
T= H Bh = T cos30.0° = (268 N)cos30.0° = 232 N
(c) ∑F y = ma y says H Av + H Bv + T sin 30.0° − w = 0 H Av + H Bv = w − T sin 30.0° = 500 N − (268 N)sin 30.0° = 366 N
EVALUATE:
11.74. T has a horizontal component to the left so H Bh must be to the right, as these are the only two horizontal forces. Note that we cannot determine H Av and H Bv separately, only their sum.
IDENTIFY: Use Eq.(11.3) to locate the x-coordinate of the center of gravity of the block combinations.
SET UP: The center of mass and the center of gravity are the same point. For two identical blocks, the center of
gravity is midway between the center of the two blocks.
EXECUTE: (a) The center of gravity of top block can be as far out as the edge of the lower block. The center of
gravity of this combination is then 3L 4 to the left of the right edge of the upper block, so the overhang is 3L 4.
(b) Take the two-block combination from part (a), and place it on top of the third block such that the overhang of
3L 4 is from the right edge of the third block; that is, the center of gravity of the first two blocks is above the right
edge of the third block. The center of mass of the three-block combination, measured from the right end of the
bottom block, is − L 6 and so the largest possible overhang is (3L 4) + ( L 6) = 11 L 12. Similarly, placing this
three-block combination with its center of gravity over the right edge of the fourth block allows an extra overhang
of L 8, for a total of 25 L 24.
(c) As the result of part (b) shows, with only four blocks, the overhang can be larger than the length of a single
block.
18 L 22 L 25 L
EVALUATE: The sequence of maximum overhangs is
,
,
,…. The increase of overhang when one
24
24
24
more block is added is decreasing. Equilibrium and Elasticity 11.75. 11-27 IDENTIFY: Apply the first and second conditions of equilibrium, first to both marbles considered as a composite
object and then to the bottom marble.
(a) SET UP: The forces on each marble are shown in Figure 11.75. EXECUTE:
FB = 2 w = 1.47 N
sin θ = R / 2 R so θ = 30°
∑τ z = 0, axis at P FC (2 R cosθ ) − wR = 0
mg
= 0.424 N
2cos30°
FA = FC = 0.424 N FC = Figure 11.75
(b) Consider the forces on the bottom marble. The horizontal forces must sum to zero, so
FA = n sin θ FA
= 0.848 N
sin 30°
Could use instead that the vertical forces sum to zero
FB − mg − n cosθ = 0
n= FB − mg
= 0.848 N, which checks.
cos30°
EVALUATE: If we consider each marble separately, the line of action of every force passes through the center of
the marble so there is clearly no torque about that point for each marble. We can use the results we obtained to
show that ∑ Fx = 0 and ∑ Fy = 0 for the top marble.
n= 11.76. IDENTIFY:
SET UP: Apply ∑τ z = 0 to the right-hand beam. Use the hinge as the axis of rotation and take counterclockwise rotation as positive. If Fwire is the tension in each wire and w = 200 N is the weight of each beam, 2 Fwire − 2 w = 0 and Fwire = w . Let L be the length of each
beam.
θ
θ
L
Lθ
EXECUTE: (a) ∑τ z = 0 gives Fwire L sin − Fc cos − w sin = 0 , where θ is the angle between the beams
2
2
2
2
2
and Fc is the force exerted by the cross bar. The length drops out, and all other quantities except Fc are known, so
Fc = Fwire sin(θ /2) − 1 w sin(θ /2)
θ
53°
2
= 130 N
= (2 Fwire − w) tan . Therefore F = (260 N) tan
1
2
cos(θ /2)
2
2 (b) The crossbar is under compression, as can be seen by imagining the behavior of the two beams if the crossbar
were removed. It is the crossbar that holds them apart.
(c) The upward pull of the wire on each beam is balanced by the downward pull of gravity, due to the symmetry
of the arrangement. The hinge therefore exerts no vertical force. It must, however, balance the outward push of
the crossbar. The hinge exerts a force 130 N horizontally to the left for the right-hand beam and 130 N to the
right for the left-hand beam. Again, it’s instructive to visualize what the beams would do if the hinge were
removed.
EVALUATE: The force exerted on each beam increases as θ increases and exceeds the weight of the beam for
θ ≥ 90° . 11-28 Chapter 11 11.77. IDENTIFY: Apply the first and second conditions of equilibrium to the bale.
(a) SET UP: Find the angle where the bale starts to tip. When it starts to tip only the lower left-hand corner of the
bale makes contact with the conveyor belt. Therefore the line of action of the normal force n passes through the
left-hand edge of the bale. Consider ∑τ A = 0 with point A at the lower left-hand corner. Then τ n = 0 and τ f = 0, so it must be that τ mg = 0 also. This means that the line of action of the gravity must pass through point A. Thus
the free-body diagram must be as shown in Figure 11.77a EXECUTE:
0.125 m
tan β =
0.250 m
β = 27°, angle where tips Figure 11.77a
SET UP: At the angle where the bale is ready to slip down the incline f s has its maximum possible value, f s = μs n. The free-body diagram for the bale, with the origin of coordinates at the cg is given in Figure 11.77b
EXECUTE:
∑ Fy = ma y n − mg cos β = 0
n = mg cos β
f s = μs mg cos β ( fs has maximum value when
bale ready to slip)
∑ Fx = max
f s − mg sin β = 0
μs mg cos β − mg sin β = 0
tan β = μs
μs = 0.60 gives that β = 31°
Figure 11.77b β = 27° to tip; β = 31° to slip, so tips first
(b) The magnitude of the friction force didn’t enter into the calculation of the tipping angle; still tips at β = 27°.
For μs = 0.40 tips at β = arctan(0.40) = 22°
Now the bale will start to slide down the incline before it tips.
EVALUATE: With a smaller μs the slope angle β where the bale slips is smaller.
11.78. IDENTIFY: Apply ∑τ z = 0 and ∑F x = 0 to the bale. SET UP: Let + x be horizontal to the right. Take the rotation axis to be at the forward edge of the bale, where it
contacts the horizontal surface. When the bale just begins to tip, the only point of contact is this point and the
normal force produces no torque.
EXECUTE: (a) F = f = μ k n = μ k mg = (0.35)(30.0 kg)(9.80 m s 2 ) = 103 N
(b) With respect to the forward edge of the bale, the lever arm of the weight is 0.250 m = 0.125 m and the lever
2
mg
1 = 0.125 m = 0.36 m .
arm h of the applied force is then h = (0.125 m)
= (0.125 m)
μk
F
0.35
EVALUATE: As μ k increases, F must increase and the bale tips at a smaller h. Equilibrium and Elasticity 11.79. 11-29 IDENTIFY: Apply the first and second conditions of equilibrium to the door.
(a) SET UP: The free-body diagram for the door is given in Figure 11.79. Figure 11.79 Take the origin of coordinates at the center of the door (at the cg). Let n A f kA , nB , and f kB be the normal and
friction forces exerted on the door at each wheel.
EXECUTE: ∑ Fy = ma y
n A + nB − w = 0
n A + nB = w = 950 N ∑F x = max f kA + f kB − F = 0
F = f kA + f kB
f kA = μ k nA , f kB = μ k nB , so F = μ k (nA + nB ) = μ k w = (0.52)(950 N) = 494 N ∑τ B =0 nB , f kA , and f kB all have zero moment arms and hence zero torque about this point.
Thus + w(1.00 m) − nA (2.00 m) − F (h) = 0
w(1.00 m) − F ( h) (950 N)(1.00 m) − (494 N)(1.60 m)
=
= 80 N
2.00 m
2.00 m
And then nB = 950 N − nA = 950 N − 80 N = 870 N.
(b) SET UP: If h is too large the torque of F will cause wheel A to leave the track. When wheel A just starts to lift
off the track n A and f kA both go to zero.
EXECUTE: The equations in part (a) still apply.
n A + nB − w = 0 gives nB = w = 950 N
nA = Then f kB = μk nB = 0.52(950 N) = 494 N
F = f kA + f kB = 494 N
+ w(1.00 m) − nA (2.00 m) − F (h) = 0
w(1.00 m) (950 N)(1.00 m)
=
= 1.92 m
F
494 N
EVALUATE: The result in part (b) is larger than the value of h in part (a). Increasing h increases the clockwise
torque about B due to F and therefore decreases the clockwise torque that n A must apply.
IDENTIFY: Apply the first and second conditions for equilibrium to the boom.
SET UP: Take the rotation axis at the left end of the boom.
EXECUTE: (a) The magnitude of the torque exerted by the cable must equal the magnitude of the torque due to the
weight of the boom. The torque exerted by the cable about the left end is TL sinθ . For any angle θ ,
sin (180° − θ ) = sin θ , so the tension T will be the same for either angle. The horizontal component of the force that
the pivot exerts on the boom will be T cosθ or T cos(180° − θ ) = −T cosθ .
h= 11.80. 1
and this becomes infinite as θ → 0 or θ → 180°.
sin θ
(c) The tension is a minimum when sin θ is a maximum, or θ = 90°, a vertical cable.
(d) There are no other horizontal forces, so for the boom to be in equilibrium, the pivot exerts zero horizontal force
on the boom.
(b) From the result of part (a), T is proportional to 11-30 11.81. Chapter 11 EVALUATE: As the cable approaches the horizontal direction, its moment arm for the axis at the pivot approaches zero,
so T must go to infinity in order for the torque due to the cable to continue to equal the gravity torque.
IDENTIFY: Apply the first and second conditions of equilibrium to the pole.
(a) SET UP: The free-body diagram for the pole is given in Figure 11.81. n and f are the vertical and
horizontal components of the force
the ground exerts on the pole.
∑ Fx = max
f =0
The force exerted by the ground
has no horizontal component.
Figure 11.81
EXECUTE: ∑τ A =0 +T (7.0 m)cosθ − mg (4.5 m)cosθ = 0
T = mg ( 4.5 m/7.0 m) = (4.5/ 7.0)(5700 N) = 3700 N ∑F y =0 n + T − mg = 0
n = mg − T = 5700 N − 3700 N = 2000 N
The force exerted by the ground is vertical (upward) and has magnitude 2000 N.
EVALUATE: We can verify that ∑τ z = 0 for an axis at the cg of the pole. T > n since T acts at a point closer to
the cg and therefore has a smaller moment arm for this axis than n does.
(b) In the ∑τ A = 0 equation the angle θ divided out. All forces on the pole are vertical and their moment arms
11.82. are all proportional to cos θ .
IDENTIFY: Apply the equilibrium conditions to the pole. The horizontal component of the tension in the wire is
22.0 N.
SET UP: The free-body diagram for the pole is given in Figure 11.82. The tension in the cord equals the weight
W. Fv and Fh are the components of the force exerted by the hinge. If either of these forces is actually in the
opposite direction to what we have assumed, we will get a negative value when we solve for it.
EXECUTE: (a) T sin 37.0° = 22.0 N so T = 36.6 N . ∑τ z = 0 gives (T sin 37.0°)(1.75 m) − W (1.35 m) = 0 .
(22.0 N)(1.75 m)
= 28.5 N .
1.35 m
(b) ∑ Fy = 0 gives Fv − T cos37.0° − W = 0 and Fv = (36.6 N)cos37.0° + 55.0 N = 84.2 N .
W= ∑F x = 0 gives W − T sin 37.0° − Fh = 0 and Fh = 28.5 N − 22.0 N = 6.5 N . The magnitude of the hinge force is
F = Fh2 + Fv2 = 84.5 N .
EVALUATE: If we consider torques about an axis at the top of the plate, we see that Fh must be to the left in order
for its torque to oppose the torque produced by the force W. Figure 11.82 Equilibrium and Elasticity 11.83. IDENTIFY:
SET UP: Apply ∑τ z 11-31 = 0 to the slab. The free-body diagram is given in Figure 11.83a. tan β = 3.75 m
so β = 65.0° . 20.0° + β + α = 90° so
1.75 m
2 α = 5.0° . The distance from the axis to the center of the block is 2 ⎛ 3.75 m ⎞ ⎛ 1.75 m ⎞
⎜
⎟ +⎜
⎟ = 2.07 m .
⎝2⎠⎝2⎠ EXECUTE: (a) w(2.07 m)sin 5.0° − T (3.75 m)sin 52.0° = 0 . T = 0.061w . Each worker must exert a force of
0.012w , where w is the weight of the slab.
(b) As θ increases, the moment arm for w decreases and the moment arm for T increases, so the worker needs to
exert less force.
(c) T → 0 when w passes through the support point. This situation is sketched in Figure 11.83b.
(1.75 m) / 2
and θ = 25.0° . If θ exceeds this value the gravity torque causes the slab to tip over.
tan θ =
(3.75 m) / 2
EVALUATE: The moment arm for T is much greater than the moment arm for w, so the force the workers apply is
much less than the weight of the slab. Figure 11.83a, b
11.84. IDENTIFY:
SET UP: F⊥l0
.
AΔl
F⊥ = F = W and Δl = x . For copper, Y = 11 × 1010 Pa .
For a spring, F = kx . Y = ⎛ YA ⎞
⎛ YA ⎞
YA
(a) F = ⎜ ⎟ Δl = ⎜ ⎟ x . This in the form of F = kx , with k =
.
l0 ⎠
l0 ⎠
l0
⎝
⎝
YA (11 × 1010 Pa)π (6.455 × 10−4 m) 2
(b) k =
=
= 1.9 × 105 N/m
l0
0.750 m
EXECUTE: 11.85. (c) W = kx = (1.9 × 105 N/m)(1.25 × 10−3 m) = 240 N
EVALUATE: For the wire the force constant is very large, much larger than for a typical spring.
IDENTIFY: Apply Newton’s 2nd law to the mass to find the tension in the wire. Then apply Eq.(11.10) to the wire
to find the elongation this tensile force produces.
(a) SET UP: Calculate the tension in the wire as the mass passes through the lowest point. The free-body diagram
for the mass is given in Figure 11.85a. The mass moves in an arc of a circle
with radius R = 0.50 m. It has
!
acceleration arad directed in toward
the center of the circle, so at this point
!
arad is upward.
Figure 11.85a
EXECUTE: ∑F y = ma y T − mg = mRω so that T = m( g + Rω 2 ).
2 11-32 Chapter 11 But ω must be in rad/s:
ω = (120 rev/min)(2π rad/1 rev)(1 min/60 s) = 12.57 rad/s.
Then T = (12.0 kg)(9.80 m/s 2 + (0.50 m)(12.57 rad/s) 2 ) = 1066 N.
Now calculate the elongation Δl of the wire that this tensile force produces:
Fl
(1066 N)(0.50 m)
Fl
= 0.54 cm.
Y = ⊥ 0 so Δl = ⊥ 0 =
A Δl
YA (7.0 × 1010 Pa)(0.014 × 10−4 m 2 )
!
(b) SET UP: The acceleration arad is directed in towards the center of the circular path, and at this point in the
motion this direction is downward. The free-body diagram is given in Figure 11.85b.
EXECUTE:
∑ Fy = ma y mg +