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Unformatted text preview: CHAPTER FOUR SOLUTIONS CHAPTER FOUR SOLUTIONS Chapter Four Readings Brancazio, P., "The Physics of Kicking a Football," The Physics Teacher,
October 1985, p. 403 Cohen, B.I., "Isaac Newton," Scientiﬁc American, December 1955, p. 73. Crane, H.R., "Digital Electronic Balances: Mass or Weight?," The Physics
Teacher, 29, 142, 1991 McCloskey, M., "Intuitive Physics," Scientiﬁc American, April 1983, p. 122.
Palmer, F., "Friction," Scientiﬁc American, February 1051, p. 54. Zimmerer, R., "The Measurement of Mass," The Physics Teacher, September
1983, p. 354 4.1 (a) FR=ma = (6.0 kg)(2.0 m/sz) = 12 N. (b) a=ﬂ= 12 N 3.0m/32. m 4.0 kg‘ 4.2 The acceleration given to the football is found from v= V0 + aras VVo_10m/S0_ 2
r ' 0.205 ‘50m/5° Then, from Newton's 211d law, we ﬁndF= ma = (0.50 kg)(SO m/sZ) = 25 N. a= 4.3 Onearth w=mg. If g changes to g', the new weight would be w' mg‘. L“ in“ g: g.
I f _. z w _ ,
here ore, w 1m g g , or g w
On the moon, g'_— 6 g. Thus, w'zIE w= "65 lb. This is equivalent to 3.71 N. On Jupiter, g' = 2.64 g. The techniques above give the weight on Jupiter
to be 58.7 N. The mass is the same at all three locations. The weight on
w 22.2 N the earth is 22.2 N, and E=m= 2.27 kg,
4.4 33 L— OOOON =0.05 m/sZ, and a=m'”1.5 x 107 kg
V=Vo+at=0+(0.05)t with v=80kmfh=22.22 m/sgives
t=444.4s=7.4min 4.5 Summing the forces on the plane a = 2 m/sz shOWn gives '—'.'*
‘ E'FX=F_f=10Nf %
= 2
(0.20 kg)(2.0 m/s ). f F=10N From which
1‘  9.6 N 4.6 First, ﬁnd the acceleration of the bullet from the equations of motion
with constant acceleration, as 38 CHAPTER FOUR SOLUTIONS 2  2 2
_V V9 5320 m/31  0_ 4 2
a— .= 2(0.82 ) —6.24x10 m/s . Then, F= ma = (5.0 x 103 kg)(6.24 x 104 m/sZ) = 310 N 4.7 (a) From the second law, the acceleration of the boat is PR 2000 N  1800 N
a=E=—————"‘—1000 kg =020m/82 r=1soon F=2000N (b) The distance moved is found from x= vot+%at2 =0 +% (0.2 m/sz)(10.0 s)2 = 10111. (c) Its velocity is found as v= v0+ at= 0+ (0.2 m/sz)(10.0 s) = 2 m/s. ‘  4.8 (a) We resolve the forces shown into
their components as
man: 31.591119
400 N: 200. N 346. N
450 N: M m
FR: 122 N 789 N u
The magnitude of the resultant force is found from the gorean theorem as FR= (213,02 + (2Py)2 = (122 N)2 + (789 N)? = 798 N,
2
and tantLe2 =§—3* = 6.47, from which 9 = 81.2”. Thus, the resul
tant force is at an angle of 8.8“ to the right of the forward direction.
(13) The acceleration is in the same direction as F3 and is given by a=PR/m= 798 N/ 3000 kg= 0.266 m/sz. 4.9 The resultant of the two forces is found from the Pythagorean theorem
as FR=\I(390 N)2 + (180 N)? =430 N, and tan 9=i% 2 2.166, ,, ﬁg 430 N
or 8—652. Thus, a=m=270 kg = 1.59 m/s2 at 65.2” north of east. 4.10 Using v2 = v02  2gb with v0; 0, we ﬁnd that at the moment the ball hits the ground, its velocity v1 is v1 =V(2)(9.80)(30 = 24.25 m/s
in the downward direction. If the ball rebounds to a height of 20 m,
then its velocity V2 as it leaves the ground is V2 =V(—2)(—9.80)(20) = 19.80 m/s in the upward direction. The
average acceleration between when the ball hits the ground and when
it leaves the ground is then V2 ' V1 19.8 — 24.25 2 a= At = 0.002 s = 22025 m/s upward. Therefore, the average force is F= ma = (0.50 kg)(22025 111/82) = 1.10 x 104 N upward. CHAPTER FOUR SOLUTIONS 4.17 The forces on the bucket are the tension in the rope and the
weight, 49 N, of the bucket We call the positive direction
upward, and use the second law. EFy = may
T 49 N = (5.0 kg)(3.0 m/sz)
T= 64 N. 4.18 (a) From the second law, we ﬁnd the acceleration as =55: :3 ﬁg 0.333 m/sZ.
_ —)
To ﬁnd the distance moved, we use F = 10 N
x= vot+ %at2 =0+ % (0.333 m/s2)(3.0 s)2=
1.5 m (b) If the shopper places her 30 N (3.06 kg) child in the cart, the new F 10 N __ , 2
Moral33.06 kg — 0.302 m/s , and the new distance traveled in 3.0 s will be x' = 0 + %(0.302 m/sz)(3.0 s)2 = 1.4 m. acceleration will be a = 4.19 (a) The average acceleration is given by
3= Av/Ata (Vf  vi)/At= (5.00 m/s  20.0 m/s)/4.000 s =  3.75 m/sz.
The average force is found from the second law as 1F= an; a (2.000): 103 kg)(3.75 m/sz) = 7500 N
(b) The distance traveled is: x=tz r=;———15'00 m“ a 20'0 m/s (4.000 s)=50.0m. 4.20 Letting m1 = 10.0 kg and 1112 = 5.00 kg, we
have
mla=m1g  T.
m2a= T ngsine.
Adding these gives
(1111 + m2)aam1g  mggsinﬂ. _‘_2'_
Thus, a=(m1 m Sm m1 + 1112
I  I o o
a= (__,._.§___)___10 0 51500 ““40 ) 9.80 = 4.43 m/sZ.
Using this value of a'gives: T= m1(g  a) = 10.0(9.80  4.43) = 53.7 N. g,or 4.21 (a) The resultant external force acting on this system having a total
mass of 6.0 kg is 42 N directed horizontally toward the right. Thus, the . . "I: 42 N 2
acceleration produced is a= m = 6.0 kg = 7.0 m/s . The acceleration of the system is 7.0 m/s2 horizontally toward the right. (b) Draw a free body diagram of the 3.0 kg object and apply Newton‘s
second law to the horizontal forces acting on this object. This gives 41 4.27 4.28 4.29 4.30 CHAPTER FOUR SOLUTIONS From (2), we ﬁnd: T2 = 118 N, and using this in (1), we ﬁnd T1 = 236 N. F= 2795N Apply the 2nd law to the trailer to get: T= (300 kg)(2.15 m/sz) = 645 N. For the car: F r= 1000 kg(2.15 m/sz), which gives: F= 2795 N. (a) The net force on the car = F— T= mcara = 2150 N (directed forward). (b) From above, we know that T= 645 N in forward direction. (c) The force exerted on the trailer by the car is  T= 645 N ( rearward). (d) The force exerted mm by the car is the resultant of the re
action force F and the weight of the car as show above. This force
is found to be PR = 10,190 Nat e = 159" to the left of the vertical. First, consider the 3.00 kg rising mass. The . T T forces on it are the tension, T, and its weight, 29.4 N. With the upward direction . as positive, the second law hec0mes +
T 29.4 N = (3.00 kg)a. (1) The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction doWn for this mass, gives w1 29.4 N
49 N — T= (5.00 kg)a. (2) Equations (1) and (2) can be solved simultaneously to give (a) the tension as T= 36.8 N, (b) and the acceleration as a = 2.45 m/sz. (c) Consider the 3.00 kg mass. We have y= vot+%at2 =0+ % (2.45 m/s2)(1.00 s)2 = 1.23 m. (a) To bring the block to the verge of motion, we exert a force of 75 N,
but from the ﬁrst law, we know: EFX = 0, which gives
75 Nf=0,or f=75 N wherefis the force of friction.
We know, since motion is imminent, that f= usN where its is the
coefﬁcient of static friction and N is the normal force. For our case, N = mg. Thus, f=nsmg, and us =m%g= 17956 1; =0.38. (b) When there is motion, the analysis is virtually the same as that above. We ﬁnd: pk = 16906 I; = 0.31. (a) The static friction force on M1 can have a maximum value of
f= PsN = [1.51113 = 0.5(10)9.80 = 49 N.
This is more than the weight of M2 (39.2 N). Hence, the system does
not move. a = 0
(b) Once in motion the friction force on N11 becomes
f= ukN =ukmg = 0.3(10)9.80 = 29.4 N.
This is less than the gravitational force of 39.2 N on M2. 43 4.39 4.40 4.41 CHAPTER FOUR SOLUTIONS 2
0 = (13.9 m/s)2 +2(—ukg)x, or x= W. (1)
With pk = 0.100, this gives a value for xof X= 98.6m (b) With ilk  0.600, (1) above gives X= 16.4111 First, let us ﬁnd the normal force acting on the box by use of 2B,, =0.
We have N  WCOS30.0° = O, or N =mgcos30.0°.
Now apply Newton's 2nd law to the motion
parallel to the ramp. 213x = mgsin30°  f— max.
But, f=pkN = nkmgcosBOD".
Thus, the 2nd law becomes mgsinSOD"  ukmgcosStLO“ = max, or sin30.0° _ ax
"k“ cos30.0° gc0530.0°
. _._ax__
= “300 ' gcos30.0°
1.20 m/s2
3 0'57? ‘ (9.80 m/sz)(0.866) ‘ (1436’
._________ Using the given distance and time,
X= vot+ % at2 gives 1.00m=0 +%a(16.00 s2). or a = 0.125 111/52. (a) Since the acceleration of the 4.00 kg
mass is then 0.125 m/sz. Newton's
second law gives T— 39.2 N  (4.00 kg)(0.125 m/sZ) or T= 39.7 N. (b) Then the an law for the 9.00 kg mass gives: sz = may = 0, or N = wcos40.0° = (9.00 kg)(9.80 m/sz)cos40.0° = 67.6 N,
and 28x = max , or mgsin40.0°  T f = (9.00 kg)a, which yields
1’2 (9.00 kg)(9.80 m/s2)sin40.0° — 39.7 N — (9.00 kg)(0.125 m/sz) = 15.9 N. Therefore, :1 == §=%3_2—% = 0.235. ( c) First, apply the 211‘:1 law to the 12.0 kg
block: 118N T= (12.0kg)a (1)
For the 7.00 kg block, we have N —.Wcos37.0° = 54.8 N, and f= ukN = 0.25(54.8 N) = 13.7 N
Apply the second law no the 7.00 kg
block. We have:
T— f (68.6 N)sin37“ = (7.00 kg)a. (2)
Solving (1) and (2) simultaneously
gives: a = 3.32 m/sz. ...
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 Fall '01
 Shapiro
 Physics

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