Physics 3.1 - 3.2 - Gravitational and kinetic energy

Physics 3.1 - 3.2 - Gravitational and kinetic energy

A raised object and a moving object require different energy calculations. Use mass, vertical height change and speed to calculate gravitational and kinetic energy, rearrange the equations and check units and scale.

Energy models and quantities

Imagine first a toolbox being lifted from the floor to a shelf. Its position relative to Earth changes, so the gravitational energy store of the Earth-toolbox system increases. Now imagine the toolbox on a trolley moving across the floor. Because it is moving, the toolbox and trolley have energy in their kinetic store.

Energy is not a material that an object contains. It is a measurable quantity used to describe a system. Both gravitational potential energy and kinetic energy are measured in joules (J), but they answer different questions:

SituationEnergy quantityMeasurements needed
An object is raisedchange in gravitational potential energy, ΔGPE\Delta\mathrm{GPE}mass, gravitational field strength, change in vertical height
An object is movingkinetic energy, KE\mathrm{KE}mass and speed

The system boundary matters. For gravitational potential energy, track the object together with Earth and compare a clear starting height with a clear finishing height. For kinetic energy, track the named moving object or group of objects and use its total mass and speed at the stated instant.

The equations require standard units:

QuantitySymbolUnit
energyΔGPE\Delta\mathrm{GPE} or KE\mathrm{KE}joule, J
massmmkilogram, kg
gravitational field strengthggnewton per kilogram, N/kg
change in vertical heightΔh\Delta hmetre, m
speedvvmetre per second, m/s

The equations are provided on the current GCSE Physics equation sheet. Learn what each quantity means, how to choose and rearrange the equation, and how to use it accurately; the calculations below develop those skills.

In these equations, gg tells you the gravitational force per kilogram at that location. Use the value given in the question, often 9.8 N kg−19.8\,\mathrm{N\,kg^{-1}} or 10 N kg−110\,\mathrm{N\,kg^{-1}}; do not swap values during a calculation. Convert grams to kilograms and kilometres per hour to metres per second before substituting.

Change in gravitational potential energy

When an object is raised above the ground, increasing its vertical height increases the gravitational energy store of the Earth-object system. The equation calculates the change, not an absolute amount with a universal zero.

Change in gravitational potential energy

change in gravitational potential energy=mass×gravitational field strength×change in vertical height\begin{gathered}\text{change in gravitational potential energy}\\=\text{mass}\times\text{gravitational field strength}\\\times\text{change in vertical height}\end{gathered} ΔGPE=m×g×Δh\Delta\mathrm{GPE} = m\times g\times\Delta h

The triangle Δ\Delta means change in. For an object raised from one height to another,

Δh=final height−initial height\Delta h=\text{final height}-\text{initial height}

.

Only the vertical rise belongs in the equation. The distance travelled along a ramp, a curved track or any other route is not Δh\Delta h. At fixed mm and gg, twice the vertical rise produces twice the change in gravitational potential energy.

Worked example: lifting a toolbox

A 2500 g2500\,\mathrm{g} toolbox is raised from 0.30 m0.30\,\mathrm{m} above the floor to a shelf 2.10 m2.10\,\mathrm{m} above the floor. The gravitational field strength is 9.8 N kg−19.8\,\mathrm{N\,kg^{-1}}. Calculate the change in its gravitational potential energy.

1. Identify and convert the quantities.

m=2500 g=2.5 kgm = 2500\,\mathrm{g} = 2.5\,\mathrm{kg} Δh=2.10 m−0.30 m=1.80 m\Delta h = 2.10\,\mathrm{m} - 0.30\,\mathrm{m} = 1.80\,\mathrm{m} g=9.8 N kg−1g = 9.8\,\mathrm{N\,kg^{-1}}

2. Select the relationship.

ΔGPE=m×g×Δh\Delta\mathrm{GPE} = m\times g\times\Delta h

3. Substitute with units already compatible.

ΔGPE=2.5×9.8×1.80\Delta\mathrm{GPE} = 2.5\times9.8\times1.80

4. Calculate and report the result.

ΔGPE=44.1 J\Delta\mathrm{GPE} = 44.1\,\mathrm{J}, which is 44 J44\,\mathrm{J} to two significant figures.

The units also check: mass in kilograms multiplied by field strength in newtons per kilogram gives force in newtons. Multiplying by metres gives newton-metres. For energy transfer, 1 N m=1 J1\,\mathrm{N\,m}=1\,\mathrm{J}, so 44.1 N m=44.1 J44.1\,\mathrm{N\,m}=44.1\,\mathrm{J}.

5. Sense-check. The object was raised, so the change should be positive. A few kilograms raised by nearly two metres in Earth's gravitational field gaining a few tens of joules is a plausible order of magnitude.

Worked example: rearranging for height

A 5.0 kg5.0\,\mathrm{kg} display is raised and its gravitational potential energy increases by 294 J294\,\mathrm{J}. The gravitational field strength is 9.8 N kg−19.8\,\mathrm{N\,kg^{-1}}. Determine its change in vertical height.

Start with the same equation and rearrange before substituting:

ΔGPE=m×g×Δh\Delta\mathrm{GPE} = m\times g\times\Delta h Δh=ΔGPEmg\Delta h=\frac{\Delta\mathrm{GPE}}{m g} Δh=2945.0×9.8=6.0 m\Delta h=\frac{294}{5.0\times9.8}=6.0\,\mathrm{m}

The result is a height in metres. Substitution back into the original equation gives 5.0×9.8×6.0=294 J5.0\times9.8\times6.0 = 294\,\mathrm{J}, so the rearrangement is consistent.

Kinetic energy

A moving object has energy in its kinetic store. The amount depends on how much mass is moving and on its speed. Speed has no direction, so two identical objects moving at the same speed in opposite directions have the same kinetic energy.

Kinetic energy

kinetic energy=12×mass×(speed)2\begin{gathered}\text{kinetic energy}\\=\frac12\times\text{mass}\times(\text{speed})^2\end{gathered} KE=12×m×v2\mathrm{KE} = \frac12\times m\times v^2

Only the speed is squared. A safe calculator entry is 0.5×mass×speed×speed0.5\times\text{mass}\times\text{speed}\times\text{speed}. The equation requires mass in kilograms and speed in metres per second; its answer is in joules.

Worked example: a moving car

An electric car of mass 1200 kg1200\,\mathrm{kg} travels at 72 km h−172\,\mathrm{km\,h^{-1}}. Calculate its kinetic energy.

1. Identify and convert the quantities.

m=1200 kgm = 1200\,\mathrm{kg}

To convert kilometres per hour to metres per second, divide by 3.63.6:

v=723.6=20 m s−1v=\frac{72}{3.6}=20\,\mathrm{m\,s^{-1}}

2. Select the relationship.

KE=12×m×v2\mathrm{KE} = \frac12\times m\times v^2

3. Substitute, keeping the square attached to speed.

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