## SAT Physics Subject Test

**Chapter 4 ****Work, Energy, and Power**

**POTENTIAL ENERGY**

Kinetic energy is the energy an object has by virtue of its motion, but potential energy is independent of motion and arises from the object’s position. For example, a ball at the edge of a tabletop has energy that could be transformed into kinetic energy if it falls off. An arrow in an archer’s pulled-back bow has energy that could be transformed into kinetic energy if the archer releases the arrow. Both of these examples illustrate the concept of **potential energy** (symbolized as *U* or *PE*), the energy an object or a system has by virtue of its position. In each case, work was done on the object to put it in the given position (the ball was lifted to the tabletop, the arrow was pulled back), and since work is the means of transferring energy, these things have *stored energy* *that can be retrieved*, as kinetic energy. When an object falls, gravity does positive work, thereby giving the object kinetic energy. We can think of this situation differently by imagining that the kinetic energy came from a “storehouse” of energy. This energy is called potential energy.

Because there are different types of forces, there are different types of potential energy. The ball at the edge of the tabletop provides an example of **gravitational potential energy**, *U*_{grav}, which is the energy stored by virtue of an object’s position in a gravitational field. This energy would be converted to kinetic energy as gravity pulled the ball down to the floor. For now, let’s concentrate on gravitational potential energy.

Assume the ball has a mass *m* of 2 kg, and that the tabletop is *h* = 1.5 m above the floor. How much work did gravity do as the ball was lifted from the floor to the table? The strength of the gravitational force on the ball is **F**_{w} = *mg* = (2 kg)(10 N/kg) = 20 N. The force **F**_{w} points downward,and the ball’s motion was upward, so the work done by gravity during the ball’s ascent was

*W*_{by gravity} = –**F**_{w}*h* = –*mgh* = –(20 N)(1.5 m) = –30 J

Someone performed +30 J of work to raise the ball from the floor to the tabletop. That energy is now stored, and if someone gave the ball a push to send it over the edge, by the time the ball reached the floor it would acquire a kinetic energy of 30 J. So we’d say that the change in the ball’s gravitational potential energy in moving from the floor to the table was +30 J. That is

∆ *U*_{grav} = –*W*_{by gravity}

Notice that potential energy, like work (and kinetic energy), is expressed in joules.

In general, if an object of mass *m* is raised a height *h* (which is small enough that *g* stays essentially constant over this altitude change), then the increase in the object’s gravitational potential energy is

∆ *U*_{grav} = *mgh*

An important fact that makes the above equation possible is that the work done by gravity as the object is raised does not depend on the path taken by the object. The ball could be lifted straight upward or on some curvy path—it would make no difference. Gravity is said to be a **conservative**force because of this property.

If we decide on a reference level to call *h* = 0, then we can say that the gravitational potential energy of an object of mass *m* at a height *h* is *U*_{grav} = *mgh*. To use this last equation, it’s essential that we choose a reference level for height. For example, consider a passenger in an airplane reading a book. If the book is 1 m above the floor of the plane then, to the passenger, the gravitational potential energy of the book is *mgh*, where *h* = 1 m. However, to someone on the ground looking up, the floor of the plane may be, say, 9,000 m above the ground. So, to this person, the gravitational potential energy of the book is *mgH*, where *H* = 9,001 m. What both would agree on, though, is that the difference in potential energy between the floor of the plane and the position of the book is *mg* × (1 m), since the airplane passenger would calculate the difference as *mg* × (1 m – 0 m), while the person on the ground would calculate it as *mg* × (9,001 m – 9,000 m).

13. A stuntwoman (mass = 60 kg) scales a 20-meter-tall rock face. What is her gravitational potential energy (relative to the ground)?

Here’s How to Crack It

Calling the ground *h* = 0, we find

*U*_{grav} = *mgh* = (60 kg)(10 N/kg)(20 m) = 12,000 J