Physics Q3: HELP: Potential Energy Additional Resources

The Physics Classroom Explanations

Shooting an arrow from a bow requires work done on the bow by the shooter's arm to bend the bow and thus produce potential energy. The release of the bow converts the potential energy of the bent bow into the kinetic energy of the flying arrow.

Potential Energy

When an object is held above the earth, it has the ability to make matter move because all you have to do is let go of the object and it will fall of its own accord. Since energy is defined as the ability to make matter move, this object has energy. This type of energy is stored energy and is called potential energy. An object held in a stretched rubber band also contains this stored energy. If the stretched rubber band is released, the object will move. A pebble on a flexed ruler has potential energy because if the ruler is released, the pebble will fly. If you hold two positive charges near each other, they have potential energy because they will move if you release them. Potential energy is also the energy stored in chemical bonds. If the chemical bonds are broken and allowed to form lower potential energy chemical bonds, the excess energy is released and can make matter move. This is frequently seen as increased molecular motion or heat.

If a cannon ball is fired straight up into the air, it begins with a high kinetic energy. As the cannon ball rises, it slows down due to the force of gravity pulling it toward the earth. As the ball rises, its gravitational potential energy is increasing and its kinetic energy is decreasing. When the cannon ball reaches the top of its arc, it kinetic energy is zero and its potential energy is maximum. Then gravity continues to pull the cannon ball toward the earth and the ball will fall toward the earth. As it falls, its speed increases and its height decreases. Therefore, its kinetic energy increases as it falls and its potential energy decreases. When the ball returns to its original height, its kinetic energy will be the same as when it started upward.

When work is done on an object, the work may be converted into either kinetic or potential energy. If the work results in motion, the work was converted into kinetic energy and if the work done resulted in change in position, the work was converted into potential energy. Work could also be spent overcoming friction and that work would be converted into heat but usually we will consider frictionless systems.

If we consider the potential energy of a bent stick or a stretched rubber band, the potential energy can be calculated by multiplying the force exerted by the stick or rubber band times the distance over which the force will be exerted. The formula for calculating this potential energy looks exactly like the formula for calculating work done, that is $W = Fd$ . The only difference is that work is calculated when the object actually moves under the force and potential energy is calculated when the system is at rest before any motion actually occurs.

In the case of gravitational potential energy, the force exerted by the object is its weight and the distance it could travel would be its height above the earth. Since the weight of an object is calculated by $W = mg$ , then gravitational potential energy can be calculate by $PE = mgh$ , where $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height the object will fall.

Example Problem: A 3.00 kg object is lifted from the floor and placed on a shelf that is 2.50 m above the floor.

(a) What was the work done in lifting the object?

(b) What is the gravitational potential energy of the object sitting on the shelf?

Solution: $\text{weight of the object} = mg = (3.00 \ kg)(9.80 \ m/s^2) = 29.4 \ N$

(a) $W = Fd = (29.4 \ N)(2.50 \ m) = 73.5 \ J$

(b) $PE = mgh = (3.00 \ kg)(9.80 \ m/s^2)(2.50 \ m) = 73.5 \ J$

$v=\sqrt{\frac{(2)(73.5 \ J)}{3.00 \ kg}}=7.00 \ m/s$

Example Problem: A pendulum is constructed from a 7.58 kg bowling ball hanging on a 3.00 m long rope. The ball is pulled back until the rope makes an angle of $45^\circ$ with the vertical.

(a) What is the potential energy of the ball?

(b) If the ball is released, how fast will it be traveling at the bottom of its arc?

Solution: You can use trigonometry to find the vertical height of the ball in the pulled back position. This vertical height is found to be 0.877 m.

$PE = mgh = (7.58 \ kg)(9.80 \ m/s^2)(0.877 \ m) = 65.1 \ J$

When the ball is released, the $PE$ will be converted into $KE$ as the ball swings through the arc.

$KE = \frac{1}{2} \ mv^2 = 65.1 \ J$

$v=\sqrt{\frac{(2)(65.1 \ kg \cdot m^2/s^2)}{7.58 \ kg}}=4.14 \ m/s$

Summary

Stored energy is called potential energy.
Energy may be stored by holding an object elevated in a gravitational field or by holding it while a force is attempting to move it.
Potential energy may be converted to kinetic energy.
The formula for gravitational potential energy is $PE = mgh$ .
In the absence of friction or bending, work done on an object must become either potential energy or kinetic energy or both.

Practice

The following video discusses types of energy. Use this resource to answer the questions that follow.

What is the definition of energy?
Name two types of potential energy.
How is energy transferred from one object to another?

Potential and kinetic energy practice problems with solutions:

http://www.physicsclassroom.com/Class/energy/U5L2bc.cfm

Review

A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top?
A 50.0 kg shell was fired from a cannon at earth’s surface to a maximum height of 400. m. What is the potential energy at maximum height?
If the shell in problem #3 then fell to a height of 100. m, what was the loss of $PE$ ?
A person weighing 645 N climbs up a ladder to a height of 4.55 m.
1. What work does the person do?
2. What is the increase in gravitational potential energy?
3. Where does the energy come from to cause this increase in $PE$ ?

Learn what potential energy is and how to calculate the potential energy of gravity and also that of a spring.

Key Equations

Gravitational potential energy

$U_g= m g h \begin{cases} h & \text{height above the ground in meters(m)}\\g & \text{acceleration due to gravity,}\ 9.8\ \text{m/s}^{2}\\ U_{g} & \text{Potential energy of gravity (in Joules; 1 J} = 1 \text{kg} \ \text{m}^{2}/\text{s}^{2}\text{)}\end{cases}$

Spring potential energy

$U_{sp} = \tfrac{1}{2}k{\Delta x}^2 \begin{cases}k & \text{spring constant measured in Newtons(N) per meters(m)}\\x & \text{amount spring is displaced from resting position}\\ U_{sp} & \text{potential energy of a spring (in Joules; 1 J} = 1 \text{kg} \ \text{m}^{2}/\text{s}^{2}\text{)}\end{cases}$

Guidance

Example 1

A 2 kg block of wood is suspended 5m above a spring of spring constant 3000 N/m. When the block is dropped on the spring, how far will the spring be compressed from it's equilibrium position.

Solution

We can solve for the distance the spring will be compressed using conservation of energy. In this problem, the gravitational potential energy of the block will be turned into spring potential energy.

$U_g&=U_{sp} && \text{start with by setting the gravitational potential energy equal to the spring potential energy}\\mgh&=\frac{1}{2}k\Delta x^2 && \text{substitute the equations for gravitational and spring potential energy}\\\Delta x&=\sqrt{\frac{2mgh}{k}} && \text{solve for } \Delta x\\\Delta x&=\sqrt{\frac{2*2\;\text{kg} * 9.8\;\text{m/s}^2 * 5\;\text{m}}{3000\;\text{N/m}}} && \text{plug in the known values}\\\Delta x&=.26\;\text{m}\\$

Time for Practice

If you lift a 30 kg weight 0.5 meters, how much Potential energy has it gained?
A spring has a spring constant k equal to 200 N/m. If it is stretched 0.4 m, what is its potential energy?
A 12 kg box is resting on a table 1.5 m off the ground. It is then lifted up to 3.2 m off the ground. What is its increase in potential energy?

Answers

(using g = 10 m/s2)

150 J
16 J
204 J

Last modified: Tuesday, 31 January 2017, 7:33 AM