## 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**.

*W*_{total} = ∆*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 ?

Here’s How to Crack It

From the definition,

10. How much work would it take to stop an object that has 30 J of kinetic energy?

Here’s How to Crack It

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, *W*_{total} = Δ*K*, 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 *F** _{net}* =

*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?

Here’s How to Crack It

The total work done is (40 J) + (–20 J) = 20 J. So, by the work–energy theorem, *W*_{total} = Δ*K*, we have 20 J = Δ*K*. Since Δ*K* = *K*_{f} – *K*_{i}, we find that *K*_{f} = *K*_{i} + Δ*K* = 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?

Here’s How to Crack It

The kinetic energy of the ball as it leaves the cue is

*K* = *mv*^{2} = (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 *F** _{net}* =

*ma*to find the acceleration, and then using Big Five #5 to find the displacement.