Grant Morby, Homework 14 Due: Feb 24 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A 78.3 g bullet is red from a rie having a barrel 0.783 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + b x c x2 , where a = 15400 N, b = 5720 N/m, c = 36300 N/m2 , with x in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. Correct answer: 8003.05 J. Explanation: The work is found by integrating the force over the distance.
xf
n 1 1 = a x + b x2 c x 3 2 3 0 = (15400 N)(0.953 m) 1 + (5720 N/m) (0.953 m)2 2 1 (36300 N/m2 ) (0.953 m)3 3 = 6800.85 J .
1
003 (part 1 of 1) 10 points A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in the gure below). It leaves the track horizontally, ies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.81 m/s2 . 9.81 m/s2
W=
xi
F ds
451 g v h 2.1 m
For the force in this problem we have, W=
0
(a + b x c x2 ) dx
1.4 m =0.3
1 1 = a x + b x2 c x 3 2 3 0 = (15400 N)(0.783 m) 1 + (5720 N/m) (0.783 m)2 2 1 (36300 N/m2 ) (0.783 m)3 3 = 8003.05 J . 002 (part 2 of 2) 10 points If the barrel is 0.953 m long, how much work is done? Correct answer: 6800.85 J. Explanation: For the barrel of length n = 0.953 m m, we need only change the limits of integration,
n
3.53 m At what height h above the ground is the block released? Correct answer: 4.00344 m. Explanation:
Let :
W=
0
(a + b x c x2 ) dx
x = 3.53 m , g = 9.81 m/s2 , m = 451 g , = 0.3 , = 1.4 m , h2 = 2.1 m , h = h1 h2 , and vx = v .
Morby, Grant Homework 14 Due: Feb 24 2006, noon Inst: Drummond m h1
2
2 Using Eq. 6 and substituting vx from Eq. 8, we have
h2
g
v
h
h1 =
x Basic Concepts: chanical Energy Conservation of Me(1)
Ui = U f + Kf + W . since vi = 0 m/s. K= 1 m v2 2
g x2 + 2 h2 2 g x2 = + 4 h2 (3.53 m)2 = + (0.3) (1.4 m) 4 (2.1 m) = 1.90344 m , and h = h 1 h2 = (1.90344 m) (2.1 m) = 4.00344 m .
Ug = m g h W = mg .
Choosing the point where the block leaves the track as the origin of the coordinate system, x = vx t 1 h2 = g t2 2 (5) (6)
since axi = 0 m/s2 and vyi = 0 m/s. Solution: From energy conservation Eqs. 1, 2, 3, and 4, we have 1 2 m vx = m g (h h2 ) m g 2
2 vx = 2 g h 1 2 g 2 vx + 2g 1 h 2 = g t2 2 x = vx t .
h1 =
Using Eq. 6 and substituting t = Eq. 5, we have x 1 h2 = g 2 vx 2 gx 2 . vx = 2 h2
2
,
so (8)
(9)
(2) (3) (4)
004 (part 1 of 1) 10 points A 61 g block is released from rest and slides down a frictionless track that begins 1 m above the horizontal, as shown in the gure. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 32 N/m. The acceleration of gravity is 9.8 m/s2 . 61 g 1m 32 N/m
Find the maximum distance the spring is compressed. Correct answer: 0.193294 m. Explanation: (7) (6) (5) x from vx Let : m = 61 g = 0.061 kg , hi = 1 m , and k = 32 N/m . From conservation of mechanical energy, we have m g hi = 1 k x2 2 2 m g hi x= k = 2 (0.061 kg) (9.8 m/s2 ) (1 m) 32 N/m
Morby, Grant Homework 14 Due: Feb 24 2006, noon Inst: Drummond = 0.193294 m . 005 (part 1 of 3) 10 points The gure is a graph of the gravitational potential energy and kinetic energy of a 76 g yo-yo as it moves up and down on its string. The acceleration of gravity is 9.81 m/s2 .
Potential energy Kinetic energy Mechanical energy
3
007 (part 3 of 3) 10 points c) What is the maximum height of the yo-yo? Correct answer: 0.804764 m. Explanation: Basic Concept: Umax = m g hmax Given: m = 76 g Umax = 600 mJ g = 9.81 m/s2 . Solution:
Energy (mJ)
600 400 200 0 0 1 2 3 4 5 6 7 8
Time (s)
hmax = =
Umax mg
a) By what amount does the mechanical energy of the yo-yo change after 6.0 s? Correct answer: 0.1 J. Explanation: Basic Concept: Conservation of Mechanical Energy E = Ef Ei . Given: Ei = 600 mJ . Solution: At t = 6 s, Ef = 0.6 J , so = E 0.5 J 0.6 J = 0.1 J . 006 (part 2 of 3) 10 points b) What is the speed of the yo-yo after 1.5 s? Correct answer: 3.1414 m/s. Explanation: Basic Concept: 1 K = m v2 2 Solution: At t = 1.5 s, K = 0.375 J , so v= = 2K m 2 (0.375 J) 0.076 kg
0.6 J (0.076 kg) (9.81 m/s2 ) = 0.804764 m
008 (part 1 of 1) 10 points Two blocks (with masses 32.8 kg and 70 kg) are connected by a string as shown. The pulley is frictionless and of negligible mass. The coecient of kinetic friction between block and machine is 0.235 . The acceleration of gravity is 9.8 m/s2 .
T2 k .8 32 = g 5
39.500000 T1 70 kg
23
m
3 0.2
= 3.1414 m/s .
Morby, Grant Homework 14 Due: Feb 24 2006, noon Inst: Drummond Find the change in the kinetic energy of the block on the incline as it moves a distance of 23 m up the incline if the system starts from rest. Correct answer: 3.10604 kJ. Explanation: Given : m1 = 32.8 kg , m2 = 70 kg , = 0.235 , and = 23 m . The normal force is N = m1 g cos and the friction force is fk = m1 g cos . The vertical distance the block on the incline rises is y1 = sin and the vertical displacement of the suspended block is y2 = . From the work-kinetic energy theorem, Wnc = (K + U )f (K + U )i = K + U g , so that k d = K + m1 g y 1 +m2 g y 2 k m1 g cos = K + m1 g sin m2 g K = m2 g m1 g (sin + k cos ) Since sin + cos = sin(39.5 ) + (0.235) cos(39.5 ) = 0.81741 , so that 1 (m1 + m2 ) vf 2 = m2 g m1 g 2 (sin + k cos )
2 vf =
4
KE 1 =
1 m1 vf 2 2 1 = (32.8 kg) 189.393 m2 /s2 2 1 kJ 1000 J = 3.10604 kJ .
009 (part 1 of 2) 10 points a) What is the kinetic energy of an automobile with a mass of 1262 kg traveling at a speed of 10 m/s? Correct answer: 63100 J. Explanation: Basic Concept: 1 K = mv 2 2 Given: m = 1262 kg v = 10 m/s Solution: 1 K = (1262 kg)(10 m/s)2 2 = 63100 J 010 (part 2 of 2) 10 points b) What speed would a y with a mass of 0.686 g need in order to have the same kinetic energy as the automobile? Correct answer: 13563.4 m/s. Explanation: Given: m = 0.686 g K = 63100 J Solution: v= = 2K m 2(63100 J) 1000 g 0.686 g 1 kg
2g m2 m1 + m 2 m1 [sin + k cos ]
2 (9.8 m/s2 ) (23 m) = (70 kg) (32.8 kg) + (70 kg) (32.8 kg) sin(39.5 ) + (0.235) cos(39.5 ) = 189.393 m2 /s2 .
= 13563.4 m/s
Morby, Grant Homework 14 Due: Feb 24 2006, noon Inst: Drummond 011 (part 1 of 1) 10 points A 2770 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 2.24 m before contacting the beam, and it drives the beam 15.5 cm into the ground before coming to rest. The acceleration of gravity is 9.8 m/s2 . Using the work-energy theorem, calculate the magnitude of the average force the beam exerts on the pile driver while the pile driver is brought to rest. Correct answer: 419449 N. Explanation: The work energy theorem tells us that the change in potential energy of the falling pile driver equals the work done on the I beam. We can express this relation as
5
m1 5.53 m m2
Find the value of heavier mass. Correct answer: 1.91822 kg. Explanation: Let : v = 7.9 m/s , t = 1.4 s , K = 76 J , = 5.53 m , and g = 9.81 m/s2 . The total kinetic energy of the system is 1 (m1 + m2 ) v 2 2 2 (76 J) 2K = 2.43551 kg m1 + m 2 = 2 = v (7.9 m/s)2 K= a= m1 m 2 m1 + m 2 g
mgh = F s,
where h is the distance the pile driver falls, i.e., h = (2.24 m) + (0.155 m). Solving for the average force, F , we obtain
F=
mgh s (2770 kg)(9.8 m/s2 )(2.24 m + 0.155 m) = 0.155 m = 419449 N .
F = (m1 + m2 ) a (m1 m2 ) g = (m1 + m2 ) a m1 m 2 a= g m1 + m 2 The velocity is (m1 m2 ) g t m1 + m 2 (m1 + m2 ) v m1 m 2 = gt (2.43551 kg) (7.9 m/s) = (9.81 m/s2 ) (1.4 s) = 1.40094 kg . v = at =
012 (part 1 of 2) 10 points A simple Atwoods machine uses two masses m1 and m2 . Starting from rest, the speed of the two masses is 7.9 m/s at the end of 1.4 s. At that time, the kinetic energy of the system is 76 J and each mass has moved a distance of 5.53 m. The acceleration of gravity is 9.81 m/s2 .
Morby, Grant Homework 14 Due: Feb 24 2006, noon Inst: Drummond
6
m1 + m2 = 2.43551 kg m1 m2 = 1.40094 kg 2.43551 kg + 1.40094 kg m1 = 2 = 1.91822 kg . 013 (part 2 of 2) 10 points Find the value of lighter mass. Correct answer: 0.517284 kg. Explanation: 2.43551 kg 1.40094 kg 2 = 0.517284 kg .
m2 =