Use the formulas for kinetic energy and gravitational potential energy in simple situations

What this dot point is asking

SEAB wants you to use the kinetic energy formula for a moving object and the gravitational potential energy formula for a raised object, and to apply conservation of energy to a falling or rising object. The big idea is that movement and height both store energy, and that energy moves between these two stores as an object speeds up or slows down.

The answer

Kinetic energy

Kinetic energy is the energy an object has because it is moving. The formula is:

where $m$ is the mass in kilograms and $v$ is the speed in $\text{m s}^{-1}$. The unit is the joule ($\text{J}$).

Notice the speed is squared. This means doubling the speed makes the kinetic energy four times larger, which is why fast vehicles are so much harder to stop and why speed matters so much in crash safety.

Gravitational potential energy

Gravitational potential energy is the energy an object has because it is raised up high. The change in gravitational potential energy when an object is lifted is:

where $m$ is the mass in kilograms, $g$ is the gravitational field strength in $\text{N kg}^{-1}$ (about $10$ on Earth), and $h$ is the height gained in metres. The unit is again the joule.

Lifting a heavier object, or lifting it higher, stores more gravitational potential energy.

Energy changing between the two stores

When an object falls, gravitational potential energy is transferred to kinetic energy, so it speeds up. When it rises, kinetic energy is transferred to gravitational potential energy, so it slows down.

If we ignore air resistance, conservation of energy gives a simple rule: the kinetic energy gained equals the gravitational potential energy lost. So a box dropped from a height $h$ arrives at the bottom with a kinetic energy equal to the $mgh$ it had at the top.

Putting it together

Because the two energies are equal in a free fall (ignoring air resistance), you can find a landing speed without knowing the time: set $\dfrac{1}{2} m v^2 = mgh$, and the mass cancels, leaving $v^2 = 2gh$.

Examples in context

Example 1. A pendulum. At the top of its swing a pendulum bob is momentarily still but high up, so it has maximum gravitational potential energy and no kinetic energy. At the bottom it moves fastest, so the energy is all kinetic. Energy swings back and forth between the two stores, which keeps the pendulum going (until friction slowly drains it).

Example 2. Road safety. Because kinetic energy depends on the square of the speed, a car at $40\ \text{m s}^{-1}$ has four times the kinetic energy of the same car at $20\ \text{m s}^{-1}$. The brakes must remove four times as much energy, so the stopping distance is much longer. This is the physics behind speed limits.

Try this

• Cue. A $2.0\ \text{kg}$ trolley moves at $3.0\ \text{m s}^{-1}$. Find its kinetic energy. [2 marks] $E_k = \dfrac{1}{2} \times 2.0 \times 3.0^2 = \dfrac{1}{2} \times 2.0 \times 9.0 = 9.0\ \text{J}$.

• Cue. A $5.0\ \text{kg}$ mass is raised $4.0\ \text{m}$, with $g = 10\ \text{N kg}^{-1}$. Find the gain in gravitational potential energy. [2 marks] $E_p = mgh = 5.0 \times 10 \times 4.0 = 200\ \text{J}$.

• Cue. A ball is dropped from $2.0\ \text{m}$ with $g = 10\ \text{N kg}^{-1}$. Find its landing speed, ignoring air resistance. [3 marks] $v^2 = 2gh = 2 \times 10 \times 2.0 = 40$, so $v = \sqrt{40} = 6.3\ \text{m s}^{-1}$.