Things tend to speed up, or gain kinetic energy, as they move downward, such as when they roll down a hill or fall. When friction can be neglected, conservation of energy is used to relate speed and height. If the mass of the moving object is known, then:

potential energy (mgh) tells you height, and

kinetic energy (½mv^{2}) tells you speed.

How to apply energy conservation

Consider a car that is released from rest at the top of a hilly track. If there is no friction, then we can assume a closed system. At any point, the total mechanical energy of the car rolling down the track is the same as the total mechanical energy it has at the start—i.e., potential energy mgh plus kinetic energy ½mv^{2}. Apply energy conservation between two different points by naming variables, writing down the relevant forms of energy at both points (step 1), and then eliminating terms that are zero (step 2). If the problem is to find the speed of the car at point B, then the resulting equation is solved for the speed at B (step 3).

Problem-solving strategy

Height change determines speed

The reduction in potential energy from A to B is mg(h_{A} − h_{B}); therefore, the speed of the car at point B depends only on the difference in height between A and B and not on the path the car takes. The fact that there is a valley between A and B makes no difference. The gain in kinetic energy equals the loss in potential energy: $${v}_{B}=\sqrt{2g({h}_{A}-{h}_{B})}=\sqrt{2(9.8{\text{m/s}}^{2})(1.0\text{m}-0.50\text{m})}=3.1\text{m/s}$$

Choosing where h = 0

It is usually convenient to define the zero of potential energy (h = 0) to be the lowest point in the problem. At point C the height is defined to be zero, so at this point all of the car’s initial potential energy is converted to kinetic energy. The result is that the car is fastest at C because it has lost more potential energy and gained more kinetic energy: $${v}_{C}=\sqrt{2g{h}_{A}}=\sqrt{2(9.8{\text{m/s}}^{2})(1.0\text{m})}=4.4\text{m/s}$$

In many conservation of energy problems the speed is equal to zero at one position (initial or final) while the height is zero at the other position. In this type of problem the kinetic energy at one position equals the potential energy at the other position. You may find the interactive equation at left to be useful in solving such problems.

Test your knowledge

A ball falls from a height of 8.0 m. How fast is it traveling when it hits the ground?

4.0 m/s

13 m/s

78 m/s

160 m/s

Writing down relevant forms of energy and eliminating zero terms gives $$\begin{array}{ccc}mgh={\scriptscriptstyle \frac{1}{2}}m{v}^{2}& \Rightarrow & gh=\end{array}{\scriptscriptstyle \frac{1}{2}}{v}^{2}$$$$\Rightarrow v=\sqrt{2gh}=\sqrt{2(9.8{\text{m/s}}^{2})(8.0\text{m})}=13\text{m/s}$$