SAT Physics Subject Test
Chapter 4 Work, Energy, and Power
THE WORK–ENERGY THEOREM
Kinetic energy is expressed in joules just like work, since in the case at which we just looked, W = K. In fact, the total work done on an object—or the work done by the net force—is equal to the object”s change in kinetic energy; this is known as the work–energy theorem.
Wtotal = ∆K
Kinetic energy, like work, is a scalar quantity.
9. What is the kinetic energy of a baseball (mass = 0.15 kg) moving with a speed of 20 m/s ?
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From the definition,
10. How much work would it take to stop an object that has 30 J of kinetic energy?
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To stop an object means to change its kinetic energy to zero. So, if the initial kinetic energy is 30 J, then the change in kinetic energy has to be 0 – 30 = –30 J. By the work–energy theorem, Wtotal = ΔK, the total amount of work that would be required is –30 J.
Kinematics vs.
Work–Kinetic Energy
For objects moving in
a straight line with a
constant force, you can
use the work–kinetic
energy theorem or Big
Five #5 and Fnet = ma for
problems where time is
not involved.
11. An object initially has 10 J of kinetic energy. Two forces act on it, one performing 40 J of work and the other (friction) performing –20 J. What is the final kinetic energy of this object?
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The total work done is (40 J) + (–20 J) = 20 J. So, by the work–energy theorem, Wtotal = ΔK, we have 20 J = ΔK. Since ΔK = Kf – Ki, we find that Kf = Ki + ΔK = 10 J + 20 J = 30 J.
12. A pool cue striking a stationary billiard ball (mass = 0.25 kg) gives the ball a speed of 2 m/s. If the force of the cue on the ball was 25 N, over what distance did this force act?
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The kinetic energy of the ball as it leaves the cue is
K = 
mv2 = (0.25 kg)(2 m/s)2 = 0.5 J
The work W done by the cue gave the ball this kinetic energy, so
Note that this could have been solved by using Fnet = ma to find the acceleration, and then using Big Five #5 to find the displacement.