Work and Energy • Topic 2 of 3

Kinetic & Potential Energy

Energy is the capacity to do work. An object that can do work is said to possess energy, and the more work it can do, the more energy it has. Because energy is measured by the work it can perform, it has the same SI unit as work, the joule (J). Energy comes in many forms — mechanical, heat, light, sound, chemical, electrical and nuclear. In this chapter we focus on mechanical energy, which is the sum of two kinds: kinetic energy and potential energy.

Kinetic energy is the energy an object possesses because of its motion. A moving cricket ball can knock down stumps, flowing water can turn a turbine, and wind can drive a windmill — all because moving objects carry kinetic energy. The kinetic energy of an object of mass $m$ moving with speed $v$ is:

$E_k=\frac{1}{2}mv^2$

Notice that kinetic energy depends on the square of the speed. If the speed doubles, the kinetic energy becomes four times as large; if it triples, the energy becomes nine times as large. This is why high-speed collisions are so much more dangerous than slow ones.

The work–energy theorem connects work and kinetic energy: the work done by the net force on an object equals the change in its kinetic energy. In symbols, $W=\Delta E_k=\frac{1}{2}mv^2-\frac{1}{2}mu^2$, where $u$ is the initial speed and $v$ the final speed. If you do positive work on a body it speeds up; if friction does negative work it slows down.

Potential energy is the energy stored in an object because of its position or configuration. A stone held high above the ground, a stretched rubber band, and a wound-up spring all store potential energy that can later be released to do work. The most common type at this level is gravitational potential energy — the energy a body has due to its height above the ground.

When an object of mass $m$ is raised to a height $h$ above a reference level, the work done against gravity is stored as gravitational potential energy:

$E_p=mgh$

where $g$ is the acceleration due to gravity (about $9.8\,\text{m/s}^2$, often taken as $10\,\text{m/s}^2$). The potential energy depends on the height measured from a chosen reference level — usually the ground. Lifting a heavier object, or raising it higher, stores more potential energy, which is why water stored high in a dam can generate so much electricity.

A falling ball converting potential energy into kinetic energyKinetic and Potential Energy of a Falling BallGround (reference level)Top: max PE = mgh, KE = 0Middle: PE and KE sharedBottom: max KE, PE = 0gAs the ball falls, PE = mgh decreases and KE = (1/2)mv^2 increases
1
Worked Example
Find the kinetic energy of a 2 kg ball moving with a speed of 5 m/s.
Solution
  1. Given: $m=2\,\text{kg}$, $v=5\,\text{m/s}$.
  2. $E_k=\frac{1}{2}mv^2$.
  3. $E_k=\frac{1}{2}\times2\times5^2=\frac{1}{2}\times2\times25=25\,\text{J}$.

Answer: Kinetic energy $=25\,\text{J}$.

2
Worked Example
A 1500 kg car travelling at 20 m/s slows down to 10 m/s. Find the change in its kinetic energy.
Solution
  1. Initial KE $=\frac{1}{2}\times1500\times20^2=\frac{1}{2}\times1500\times400=300000\,\text{J}$.
  2. Final KE $=\frac{1}{2}\times1500\times10^2=\frac{1}{2}\times1500\times100=75000\,\text{J}$.
  3. Change $=75000-300000=-225000\,\text{J}$.

Answer: Kinetic energy decreases by $225000\,\text{J}$ ($-2.25\times10^5\,\text{J}$).

3
Worked Example
How much potential energy is gained by a 5 kg block raised to a height of 3 m? Take $g=10\,\text{m/s}^2$.
Solution
  1. Given: $m=5\,\text{kg}$, $h=3\,\text{m}$, $g=10\,\text{m/s}^2$.
  2. $E_p=mgh$.
  3. $E_p=5\times10\times3=150\,\text{J}$.

Answer: Potential energy gained $=150\,\text{J}$.

4
Worked Example
If the speed of a moving body is doubled, what happens to its kinetic energy?
Solution
  1. $E_k=\frac{1}{2}mv^2$, so $E_k\propto v^2$.
  2. New speed $=2v$, so new $E_k=\frac{1}{2}m(2v)^2=\frac{1}{2}m\times4v^2$.
  3. This is $4\times\frac{1}{2}mv^2=4E_k$.

Answer: The kinetic energy becomes four times the original value.

5
Worked Example
A net force does 80 J of work on a stationary 4 kg trolley. Using the work-energy theorem, find the final speed of the trolley.
Solution
  1. Work-energy theorem: $W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$, with $u=0$.
  2. $80=\frac{1}{2}\times4\times v^2=2v^2$.
  3. $v^2=40$, so $v=\sqrt{40}\approx6.32\,\text{m/s}$.

Answer: Final speed $\approx6.3\,\text{m/s}$.

6
Worked Example
An object of mass 10 kg is at a height of 6 m. Find its potential energy, and its kinetic energy just before hitting the ground. Take $g=10\,\text{m/s}^2$.
Solution
  1. Potential energy at top: $E_p=mgh=10\times10\times6=600\,\text{J}$.
  2. Ignoring air resistance, all PE converts to KE during the fall.
  3. So KE just before hitting the ground $=600\,\text{J}$.

Answer: $E_p=600\,\text{J}$ at the top; KE $=600\,\text{J}$ at the bottom.

Key Points

  • Energy is the capacity to do work; its SI unit is the joule (J), the same as work.
  • Kinetic energy of a moving body is $E_k=\frac{1}{2}mv^2$ and depends on the square of the speed.
  • Work-energy theorem: work done by the net force equals the change in kinetic energy, $W=\Delta E_k$.
  • Gravitational potential energy of a body at height $h$ is $E_p=mgh$.
  • Potential energy is stored due to position; height is measured from a chosen reference level.
Tap an option to check your answer0 / 4
Q1.The kinetic energy of a body of mass $m$ moving with speed $v$ is
Explanation: Kinetic energy is $E_k=\frac{1}{2}mv^2$.
Q2.If the speed of a body is doubled, its kinetic energy becomes
Explanation: Since $E_k\propto v^2$, doubling $v$ makes $E_k$ four times larger.
Q3.Gravitational potential energy of a body of mass $m$ at height $h$ is
Explanation: $E_p=mgh$ is the gravitational potential energy.
Q4.The work done by the net force on an object equals the change in its
Explanation: This is the work-energy theorem: $W=\Delta E_k$.