The gravitational potential energy store
Definition
Gravitational potential energy
Gravitational potential energy is energy stored by an object because of its position in a gravitational field.
Definition
Gravitational field strength
Gravitational field strength is the force per unit mass acting on an object placed in a gravitational field, measured in newtons per kilogram (N/kg).
- Gravitational potential energy is stored whenever an object is raised in a gravitational field, and the store belongs to the object and the Earth together.
- The change in this store depends on the object's mass, the gravitational field strength and the change in vertical height.
- A heavier object has a larger weight, so more energy has to be transferred to raise it through the same height.
- Raising the same object twice as high transfers twice as much energy into the store.
- Gravitational field strength at the Earth's surface is taken as g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg unless a question gives a different value.
- The value of ggg on the Moon is much smaller, so the same lift through the same height stores far fewer joules there.
- Only changes in this store are calculated, because the height counted as zero is chosen rather than fixed by nature.
The change in GPE equation
- The change in the store is calculated from ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta hΔGPE=m×g×Δh.
- Here ΔGPE\Delta GPEΔGPE is the change in gravitational potential energy in joules, mmm is the mass in kilograms, ggg is the gravitational field strength in N/kg\text{N/kg}N/kg and Δh\Delta hΔh is the change in vertical height in metres.
- The symbol Δ\DeltaΔ means change in, so Δh\Delta hΔh is the final height minus the starting height.
- Rearranging gives m=ΔGPEg×Δhm = \dfrac{\Delta GPE}{g \times \Delta h}m=g×ΔhΔGPE for the mass and Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}Δh=m×gΔGPE for the height.
- Convert grams into kilograms and centimetres into metres first, so 250 g=0.250 kg250\ \text{g} = 0.250\ \text{kg}250 g=0.250 kg and 40 cm=0.40 m40\ \text{cm} = 0.40\ \text{m}40 cm=0.40 m.
- The height Δh\Delta hΔh is measured straight upwards, so an object pushed up a ramp gains the store set by the vertical rise and not by the length of the slope.
- The store increases while the object is being raised and decreases while it falls, so a falling object is emptying this store.
Linking the store to work done
- The weight of an object is its mass multiplied by the gravitational field strength, measured in newtons, and this is the force a steady lift has to balance.
- The work done in that lift is E=F×d=m×g×ΔhE = F \times d = m \times g \times \Delta hE=F×d=m×g×Δh, which is the same expression as the change in the store.
- Lifting a mass therefore transfers the work done straight into the gravitational potential energy store, which is why the two equations agree.
- Work done against friction on a ramp is extra to m×g×Δhm \times g \times \Delta hm×g×Δh, so a ramp needs more energy in total than a straight lift to the same height.
Example
Lifting a box onto a shelf
- A shelf stacker raises a box of mass 25 kg25\ \text{kg}25 kg through a vertical height of 3.0 m3.0\ \text{m}3.0 m, where g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
- The equation is ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta hΔGPE=m×g×Δh.
- Substituting gives ΔGPE=25×10×3.0\Delta GPE = 25 \times 10 \times 3.0ΔGPE=25×10×3.0.
- The change in the store is ΔGPE=750 J\Delta GPE = 750\ \text{J}ΔGPE=750 J, and the store increases because the box has been raised.
Example
Pushing a wheelbarrow up a ramp
- A wheelbarrow of mass 40 kg40\ \text{kg}40 kg is pushed 5.0 m5.0\ \text{m}5.0 m up a ramp that rises 1.2 m1.2\ \text{m}1.2 m vertically, with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
- The vertical rise is the value needed, so Δh=1.2 m\Delta h = 1.2\ \text{m}Δh=1.2 m and the 5.0 m5.0\ \text{m}5.0 m measured along the slope is not used.
- Substituting gives ΔGPE=40×10×1.2\Delta GPE = 40 \times 10 \times 1.2ΔGPE=40×10×1.2.
- The gravitational potential energy store increases by ΔGPE=480 J\Delta GPE = 480\ \text{J}ΔGPE=480 J.
Example
Finding a height from the energy stored
- A crane transfers 36 000 J36\,000\ \text{J}36000 J into the gravitational store of a 300 kg300\ \text{kg}300 kg girder, with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
- Rearranging gives Δh=ΔGPEm×g\Delta h = \dfrac{\Delta GPE}{m \times g}Δh=m×gΔGPE.
- Substituting gives Δh=36 000300×10\Delta h = \dfrac{36\,000}{300 \times 10}Δh=300×1036000.
- The girder has been raised through Δh=12 m\Delta h = 12\ \text{m}Δh=12 m.
Exam technique
Getting the GPE values right
- Write ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta hΔGPE=m×g×Δh first, then list the three values you are about to substitute.
- Take the value of ggg from the question, and use 10 N/kg10\ \text{N/kg}10 N/kg only when no value is given.
- Read the diagram carefully for the vertical height, because ramp lengths and slope distances are printed there to catch you out.
- State whether the store increases or decreases when a description is wanted as well as a number.
- Answer in joules, and convert to kilojoules only when the question asks for them.
Common Mistake
- Do not put a weight in newtons into the mmm position, because that position needs a mass in kilograms.
- Do not use the distance measured along a slope as Δh\Delta hΔh, because only the vertical rise changes this store.
- Do not call ggg gravity, because the quantity is gravitational field strength measured in N/kg\text{N/kg}N/kg.
- Do not leave a mass in grams, because the equation needs kilograms.
Self review
- State what gravitational potential energy is and which two objects share the store.
- Write the equation for the change in gravitational potential energy.
- Give the unit of each quantity in that equation.
- Explain why the vertical rise is used for a load pushed up a ramp.
- Calculate the change in the store when a 2.0 kg2.0\ \text{kg}2.0 kg book is raised 1.5 m1.5\ \text{m}1.5 m with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
