## SAT Physics Subject Test

## Chapter 4 Work, Energy, and Power

### CONSERVATION OF MECHANICAL ENERGY

We have seen energy in its two basic forms: kinetic energy (*K*) and potential energy (*U*). The sum of an object”s kinetic and potential energies is called its **mechanical energy**, *E*.

*E* = *K* + *U*

(Notice that because *U* is relative, so is *E*.) Assuming that no nonconservative forces (friction, for example) act on an object or a system as it undergoes some change, mechanical energy is conserved. That is, the initial mechanical energy, *E*_{i}, is equal to the final mechanical energy, *E*_{f}, or

*K*_{i} + *U*_{i} = *K*_{f} + *U*_{f}

This is the simplest form of the law of conservation of total energy, which we mentioned at the beginning of this section.

15. A ball of mass 2 kg is gently pushed off the edge of a tabletop that is 1.8 m above the floor. Find the speed of the ball as it strikes the floor.

Here”s How to Crack It

Ignoring the friction due to the air, we can apply conservation of mechanical energy. Calling the floor our *h* = 0 reference level, we write

Notice that the ball”s potential energy decreased, while its kinetic energy increased. This is the basic idea behind conservation of mechanical energy: One form of energy decreases while the other increases. (Also, notice that although the question gives you the mass of the ball, it wasn”t necessary since the mass *m* cancelled out of the equation.)

**Simplify**

The formula *v* =

is useful and worth

remembering for

the test.

16. A box is projected up a long ramp (incline angle with the horizontal = 30°) with an initial speed of 8 m/s. If the surface of the ramp is very smooth (essentially frictionless), how high up the ramp will the box go? What distance along the ramp will it slide?

Here”s How to Crack It

Because friction is negligible, we can apply conservation of mechanical energy. Calling the bottom of the ramp our *h* = 0 reference level, we write

Since the incline angle is *θ* = 30°, the distance, *d,* it slides up the ramp is found in the following way.

17. Wile E. Coyote (mass = 40 kg) falls off a 50-meter-high cliff. On the way down, the force of air resistance has an average strength of 40 N. Find the speed with which he crashes into the ground.

**Simplify**

When a

nonconservative force

does work, an alternate

equation is W_{TOTAL} = Δ*K*,

where the work done

by gravity replaces the

change in potential energy.

Here”s How to Crack It

The force of air resistance opposes the downward motion, so it does negative work on the coyote as he falls: *W*_{r} = –*F*_{r}*h*. Calling the ground *h* = 0, we find that

18. Find an expression for the minimum speed at which an object of mass m must be launched in order to escape Earth”s gravitational field. (This is called **escape speed**.)

Here”s How to Crack It

When launched, the object is at the surface of the earth (*r _{0}* = r

_{E}) and has an upward, initial velocity of magnitude

*v*. To get it far away from the earth, we want to bring its gravitational potential energy to zero, but to find the minimum launch speed, we want the object”s final speed to be zero by the time it gets to this distant location. So, by conservation of energy

_{0}which gives

**POWER**

Simply put, **power** is the rate at which work is done (or energy is transferred, which is the same thing). Suppose you and I each do 1,000 J of work, but I do the work in 2 minutes while you do it in 1 minute. We both did the same amount of work, but you were quicker; you were more powerful. Here”s the definition of power.

The unit of power is the joule per second (J/s), which is renamed the **watt** and symbolized W (not to be confused with the symbol for work, *W*). One watt is 1 joule per second: 1 W = 1 J/s.

19. A mover pushes a large crate (mass *m* = 75 kg) from the inside of the truck to the back end (a distance of 6 m), exerting a steady push of 300 N. If he moves the crate this distance in 20 s, what is his power output during this time?

Here”s How to Crack It

The work done on the crate by the mover is *W* = **F***d* = (300 N)(6 m) = 1,800 J. If this much work is done in 20 s, then the power delivered is *P* = *W*/*t* = (1,800 J)/(20 s) = 90 W. Note that *P* = *W/t* = F*d*/*t* = *Fv*; the formula *P* = F*v* is often useful.