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Unformatted text preview: pposes the motion, it is a force that is always directed
opposite to the displacement. Therefore, negative work is done.
13. (d) The velocity is constant. Therefore, the speed is also constant, and so is the kinetic
energy. However, the total mechanical energy is the kinetic plus the gravitational potential
energy, and the car moves up the hill, gaining potential energy as it goes. Thus, in the
circumstance described mechanical energy cannot be conserved.
14. (c) The ball comes to a halt at B, when all of its initial kinetic energy is converted into
potential energy.
15. (a) The stone’s motion is an example of projectile motion, in which the final vertical height
is the same as the initial vertical height. Because friction and air resistance are being
ignored, mechanical energy is conserved. Since the initial and final vertical heights are the
same, ∆PE = 0 J. Since energy is conserved, it follows that ∆KE = 0 J also.
16. (b) The total mechanical energy is the kinetic energy plus the gravitational potential energy:
2
E = 1 mv 2 + mgh = 1 ( 88.0 kg )(19.0 m/s ) + ( 88.0 kg ) 9.80 m/s 2 ( 55.0 m ) . Since friction
2
2 ( ) and air resistance are being ignored, the total mechanical energy is conserved, which means
that it has the same value, no matter what the height above sea level is.
17. (d) Since the track is frictionless, the conservation of mechanical energy applies. We take
the initial vertical level of the car as the zero level for gravitational potential energy. The
highest vertical level that the car can attain occurs when the final speed of the car is zero and
all of the initial kinetic energy is converted into gravitational potential energy. According to
KE 0
2.2 J
=
= 0.45 m . Thus,
the conservation principle, KE0 = mgh, or h =
mg ( 0.50 kg ) 9.80 m/s 2 ( ) the car can pass over hills A, B, C, and D but not hill E.
18. (a) Using the ground as the zero level for measuring height and applying the energy
conservation principle gives the following result:
1
2 m ( 7.00 m/s ) + 1 m ( 9.00 m/s ) + mg (11.0 m ) = 1 mvf2
2
2
2 2 The mass m of the ball can be eliminated algebraically. Chapter 6 Answers to Focus on Concepts Questions 283 19. 7.07 m/s
20. (d) The work done by the net nonconservative force is Wnc = KEf + PEf −KE0 − PE0.
21. (b) The principle of conservation of mechanical energy applies only if the work done by the
net nonconservative force is zero, Wnc = 0 J. When only a single nonconservative force is
present, it is the net nonconservative force, and a force that is perpendicular to the
displacement does no work.
22. (a) A single nonconservative force is the net nonconservative force, and the work it does is
given by Wnc = ∆KE + ∆PE. Since the velocity is constant, ∆KE = 0 J. Therefore,
Wnc = ∆PE = mg(hf − h0). But hf − h0 = −325 m, since hf is smaller than h0. Thus,
Wnc = (92.0 kg)(9.80 m/s2)(−325 m).
23. 71.5 J
24. 12 370 J
25. 344 W
26. 130 N
27. 376 J 284 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY
PROBLEMS
______________________________________________________________________________
1. SSM REASONING The work W done by the tension in the tow rope is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the tension, s is the magnitude
of the skier’s displacement, and θ is the angle between the tension and the displacement
vectors. The magnitude of the displacement (or the distance) is the speed v of the skier
multiplied by the time t (see Equation 2.1), or s = v t .
SOLUTION Substituting s = v t into the expression for the work, W = ( F cos θ ) s , we have
W = ( F cos θ ) v t . Since the skier moves parallel to the boat and since the tow rope is
parallel to the water, the angle between the tension and the skier’s displacement is θ = 37.0°.
Thus, the work done by the tension is
W = ( F cos θ ) vt = [(135 N ) cos 37.0°] ( 9.30 m/s )(12.0 s ) = 1.20 × 104 J
____________________________________________________________________________________________ 2. REASONING
a. We will use W = ( F cos θ ) s (Equation 6.1) to determine the work W. The force
(magnitude = F) doing the work is the force exerted on you by the elevator floor. This force
is the normal force and has a magnitude FN, so F = FN in Equation 6.1. To determine the
normal force, we will use the fact that the elevator is moving at a constant velocity and
apply Newton’s second law with the acceleration set to zero. Since the force exerted by the
elevator and the displacement (magnitude = s) are in the same direction on the upward part
of the trip, the angle between them is θ = 0°, with the result that the work done by the force
is positive.
b. To determine the normal force, we will again use the fact that the elevator is moving at a
constant velocity and apply Newton’s second law with the acceleration set to zero. Since
the force exerted by the elevator and the displacement are in opposite directions on the
downward part of the trip, the angle between them is θ = 180°, and so the work done by the
force is negative.
SOLUTION
a. The fr...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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