Labs from Chicago, Fall 1993 :
Gravitational Potential Energy.

Dr. Rich Kron, Dr. Heidi Newberg, and Luisa Rebull
Labs written for the CARA Space Explorers, Fall 1993

This is meant to be handed out to the students.

I. Introduction

Two weeks ago we illustrated forms of kinetic energy; this week we will 1) introduce the concept of potential energy; 2) show how to calculate its value in one simple case; and 3) show how it is related to kinetic energy in one simple case.

If you wind up the rubber band of a toy airplane, you are storing energy that can run the airplane (the energy then appears as kinetic energy). This stored energy is called more generally potential energy because it is potentially available to do stuff. Another example is the potential energy stored in a battery: the battery can run an electrical motor for some period of time which can then run a toy, golf cart, or other gadget, again producing kinetic energy. You can think of potential energy as stored energy that will be converted into kinetic energy.

We used the following formula to calculate the kinetic energy of a moving object of mass m:

K.E. = 1/2 m * v^2

Presumably there is some other formula that will allow us to calculate the potential energy - we know that we worked to wind up the rubber band of the airplane, and we ought to be able to measure that effort.

In this lab, we will work with a very simple case, namely the potential energy of a falling object. If you raise a weight above the floor, the weight has potential energy because it can potentially be dropped; when it is dropped, it acquires kinetic energy.

The amount of potential energy of the weight before it is dropped depends on how hard it is to raise it to that height. (A physicist would say: the amount of potential energy depends on the amount of work done on it.) Maybe from this simple idea we can figure out what the formula is for calculating potential energy. That is. we are looking for an expression like

P.E. = stuff related to difficulty of lifting weights

Obviously the mass m of the weight must be a factor here - the more mass, the harder it is to lift, and the more potential energy it will have. Also, it is harder to lift something six feet than it is six inches, so the distance the weight is lifted (we'll call it the height, or h) must also be a factor.

The last factor depends on the strength of gravity at the surface of the Earth. The reason you have to work at all to lift a weight is because the force of gravity pulls all weights towards the center of the Earth. If we were on the Moon or on Mars, it would be easier to lift the same mass the same height, and the weight would have less potential energy. If we were standing Jupiter or on the Sun, the opposite would be the case. For present purposes, we will measure the "strength of gravity" by a number called the acceleration due to gravity, or g. This number g depends on the mass and size of the Earth. Since the Earth is approximately round, g has very nearly the same value anywhere on the surface of the Earth (but if you were to tunnel far into the Earth, you would experience a different weight). The value of g is

g = 981 cm / sec^2 .

Note the units of g: they are the units of acceleration.

We can now write down the full expression for the potential energy of an object of mass m that has the potential to fall a distance h:

P.E. = m * g * h .

Since you can measure m and h, and since you have memorized the value of g, you can now calculate the energy that can obtained from falling matter, such as the energy stored in the reservoir of a hydroelectric power plant.

II. The Experiment

In today's experiment, we will demonstrate (again) that energy is conserved. The “system” is the falling weight, and the statement that energy is conserved means that the total energy, which is the sum of the kinetic energy and the kinetic energy, is the same when the weight is falling. That is,

K.E. + P.E = total E = constant .

For example, suppose you drop a weight of mass 1 gram from a height h = 100 centimeters. At the instant the mass is dropped, it is not yet falling, the velocity v = 0, and it has no kinetic energy. All the energy is in the form of potential energy, which can be calculated with the formula. Later, just before the mass hits the floor, the height above the floor h = 0, there is no potential energy, and all the energy is in the form of kinetic energy. In other words, the potential energy at the beginning is equal to the kinetic energy at the end.

We can write this as an equation,

m * g * h = 1/2 m * v^2 .

We are going to check this formula by dropping a weight from a known height and measuring its velocity as it passes that height. We want to measure the velocity accurately, but this is not necessarily easy to do for something that is accelerating as it falls. Our technique will be to video tape a falling ball, and use the fact that video “frames” are taken every 1/30th of a second. The VCR can play back one frame at a time, and we can then see how far the ball has moved from one frame to the next.

First, set up the video camera so that it is facing the blackboard, and draw a set of horizontal lines that are spaced every 5 centimeters (you can try putting a meter stick up against the chalk board to provide a scale, but the numerals may not be visible on the video tape). Drop a ball in front of the vertical scale while taping, making sure that the release position is visible to the video camera. Play back the tape, and find the last two frames that show the falling ball. The distance traveled in 1/30th of a second can be measured by interpolating between the horizontal lines.

position of ball on last frame: _____________ cm

position of ball on second-to-last frame: _____________ cm

difference in position = distance traveled: ____________ cm

Now, calculate the velocity. Remember that

v = interval of distance / interval of time

What is the "interval of distance"? What is the "interval of time"?

What is the speed of the ball? _____________________cm/s

Now, calculate potential energy of the ball at the beginning.

P.E. (beginning) = _________________ ergs

Calculate the kinetic energy at the end.

K.E. (end) = _________________ ergs .

Is energy conserved (are these values equal) to within your estimate of the experimental uncertainty? ______________

III. Possible Quiz Questions

A constant you might need: g = 981 cm/s^2

1. How much potential energy does a mass of 15 grams at a height of 2 meters have? Don't forget units.

2. How much kinetic energy does a mass of 15 grams have if it is moving at 10 cm/s? Don't forget units.

3. If I measure the kinetic energy of a ball right before it hits the floor to be 512 ergs, what was its gravitational potential energy when I let it go at some height above the floor? Don't forget units. (Hint - You don't need to know the height of the ball to figure this out. Think about conservation of energy.)

4. If the mass of the ball in problem 3 was 20 grams, what was the height of the ball when I let it go? Don't forget units.

Important Disclaimers and Caveats

Go back to the Chicago Fall 1993 Energy home page.