GCSE level Physics exam revision notes: Types of energy store

Gravitational potential energy defined and examples explained

(d) Gravitational potential energy (GPE) stores and calculations

This page will help you with ... How to solve numerical problems involving gravitational potential energy, how to explain examples of GPE, how to use the GPE formula and do calculations involving gravitational potential energy

Gravitational potential energy (GPE) and gravitational potential energy stores

• Any object that has mass and is in a gravitational field has gravitational potential energy.

• Any object that can 'fall' under the influence of a gravitational field has gravitational potential energy which on falling is converted to kinetic energy (KE).

• Here an object or material possesses a gravitational potential energy store by virtue of its higher position and can then fall or flow down to release the GPE - usually most converted to KE.

• Gravitational potential energy is an example of a mechanical energy store.

  • e.g. winding up the weights on a clock, water stored behind a dam that can flow down through a turbine generator.

  • In other words, if any object/material is raised above the Earth's surface, energy is transferred to increase its gravitational potential energy store.

  • Conversely, as the object/material is reduced in height (falling clock weight, water flow) the GPE is transferred to some other energy store and can do useful work e.g. hydroelectric power stations.

  • When an object is raised above the ground, work must be done in lifting it up, so energy is transferred from one energy store to the GPE store of the object.

  • If you assume there is no friction or air resistance etc. when the object has stopped moving/rising, the work done equals the gain in the object's gravitational potential energy store (see calculations further down).

• Any object falling is converting its GPE store into kinetic energy (eg skier) and any object raised in height gains GPE (eg cable car). You doing work against the weight of an object or material due to the attraction of the gravitational field of the Earth.

• Since gravitational energy is a form of stored energy, it does nothing until it is released and converted into another form of energy.

• The greater the weight of an object or material or the greater the height it is raised, the greater is the gravitational potential energy store created.

• GPE is also greater, the greater the strength of the gravitational field.

  • Raising the same object to the same height on Mars or the Moon will not increase the GPE of the object as much as on Earth because the strength of gravity is less on these smaller bodies of smaller mass than Earth.

• The amount of gravitational potential energy gained by an object raised above ground level can be calculated using the equation:

  • gravitational potential energy (GPE) = mass × gravitational field strength × height

    • E_gpe = m x g x h   (∆GPE = mgh)

  • gravitational potential energy (gpe), E_gpe, in joules, J

  • mass, m, in kilograms, kg;

  • gravitational field strength, g, in newtons per kilogram, N/kg

  • height, h, in metres, m

  • In any calculation the value of the gravitational field strength (g) will be given, on the Earth's surface it is 9.8 N/kg.

  • Formula connection:

    • Work done (J) = force (N) x distance (m)

    • Weight (N) = mass (kg) x gravitational field strength constant (g in N/kg or m/s²)

    • So, m(kg) x g(N/kg) = weight(N) = force(N)

    • Therefore raising an object vertically is the same as operating a force through a distance h (height).

    • So, E_gpe = m g h = work done = F x h

  • When an object or material falls its gravitational potential energy store is decreased as the GPE is converted into kinetic energy.

  • When an object in air or flowing water is falling you always get some heat loss from friction forces, thereby increasing the thermal energy store of the surroundings.

  • If you ignore the frictional heat losses due to air resistance the sum of GPE + KE is a constant as the object falls.

  • So, when a stationary object is then dropped from a height, at the point of impact on the ground, the maximum amount of kinetic energy equals the original amount of gravitational potential energy (based on the height from which the object is dropped). This is quite handy to know in solving certain kinds of problems.

  • See also FORCES 2. Mass and the effect of gravity force on it - weight, (mention of work done and GPE)

  • Hydroelectric power generation relies on water flowing from a greater to a lower height, losing gravitational potential energy in the process.

    • The greater the height the water is stored at, the greater its gravitational potential energy store.

  • GPE as stored water behind a dam Hydroelectric power gcse physics notes

• See also

• FORCES 2. Mass and the effect of gravity force on it - weight, (mention of work done and GPE)

ANSWERS to questions on gravitational potential energy store problems

Q1 A grandfather clock weight of mass 5500g is raised 135 cm when fully wound up.

Calculate the gain in the weight's gravitational potential energy store in J (to 3 sf).

5500 g = 5.50 kg,   135 cm = 1.35 m,  g = 9.8 N/kg

E_gpe = m g h = 5.50 x 9.8 x 1.35 = 72.8 J (3 s.f.)

Always make sure you have the correct units for the numbers in the final line of ANY physics calculation.

Q2 An 80.0 kg person climbs up flights of steps to a vertical gained height of 9.00 m.

(a) Calculate the increase in the person's gravitational energy store (gravitational field strength = 9.8 N/kg).

E_gpe = m g h

GPE store gain = 80 x 9.8 x 9.0 = 7056 = 7060 J (7.06 kJ, 3 s.f.)

(b) What is the work done by the person in climbing the stairs?

work = force x distance, force = weight = 90 x 9.8 = 784 J

work = 784 x 9.0 = 7060 J (3 s.f.)

Note: This is the same answer as (a) because the GPE formula is essentially expressing a force acting through a specified distance.

It does neglect energy lost in friction between the person and the steps.

(c) How high would you have to climb to work off a 45.0 g bar of chocolate with a calorific value of 30.0 kJ/g?

Total energy in the chocolate bar chemical energy store = 45 x 30 x 1000 = 1350000 J

This will be converted to the person's gravitational potential energy store, therefore ..

GPE = m g h = 1350000 J

so h = 1350000 / (m x g) = 1350000 / 784 = 1720 m (to 3 s.f. and higher than Ben Nevis in Scotland)

Note: A typical active adult needs 9000 kJ/day. An elderly person with a sedentary lifestyle might only need 5000 kJ/day. A very active teenager might need 12000 kJ/day. The calculation does make the point that a single bar of chocolate does provide a good 'chunk' of you daily energy needs. Of course if you snack a lot on high calorie foods you will put on weight if it isn't 'burnt off'! The calorific value of fat is ~37 kJ/g and carbohydrates typically ~17 kJ/g and roast potatoes, which I love, are somewhere in between!

Q3 If a 5.00 kg 'weight' is dropped from a height of 10.0 m above the ground, at what speed will it hit the ground?

The GPE of the 'weight' is easily calculated from the formula GPE = m g h     (and g = 9.80 N/kg)

GPE = 5 x 9.8 x 10 = 490 J

As the weight falls its gravitational potential energy store is depleted as its kinetic energy store increases.

Therefore, at the point of impact, all the GPE is converted to kinetic energy, so we use the KE equation to get v.

E_ke = ¹/₂ m v² , rearranging:   v² = 2E_ke / m,  so v = √(2E_ke / m)

v = √(2E_ke / m) = √(2 x 490 / 5.0) = 14.0 m/s (3 s.f.)

Note (i): You can actually simplify the calculation because you don't actually need to know the mass of the weight.

initial E_gpe = final E_ke = m g h = ¹/₂ m v² , the m's cancel out,

so  g x h = ¹/₂ x v²  and  v = √(2 x g x h) = √(2 x 9.8 x 10) = 14.0 m/s (3 s.f.)

Notes

(i): All these calculations ignore air resistance, so some KE is lost as heat, so the final speed is actually just a bit less than the theoretical calculations above.

(ii) The greater the distance an object falls, the greater its speed of descent becomes (acceleration) and the greater its kinetic energy increases.

(iii) As the object falls, its gravitational potential energy store decreases and its kinetic energy store increases.

Some of the GPE-KE will be lost as heat energy due to the friction effects of air resistance.

For dense heavy objects, the air resistance is small, but for something a feather, most of the GPE-KE will be dissipated as heat to the surroundings due to friction between the air molecules and the feather's surface.

At energy given point, and neglecting air resistance, E_total = E_GPE + E_KE

(iii) When an object is thrown up into the air it loses KE and gains GPE, maximising at the greatest height it rises to.

The higher you raise any object or material, the greater its gravitational potential energy store.

Q4. A pole vaulter has a weight of 800 N and vaults to a height of 4.5 m.

(a) How much work does the pole vaulter do?

work (J) = force (N) x distance of action (m)

work done = 800 x 4.5 = 3600 J

(c) At the maximum height reached, what gravitational potential energy does the pole vaulter possess?

The GPE equals the work done in raising the pole vaulter to a height of 4.5 m, that is 3600 J

(b) What kinetic energy did the pole vaulter impart to its body on 'take-off' (and what assumption are you making?)

If you ignore air resistance i.e. no energy lost - wasted, the initial KE = maximum GPE = 3600 J

(c) What is the kinetic energy of the pole vaulter immediately before impacting on the ground?

maximum KE = maximum GPE gained = 3600 J (neglecting air resistance energy losses)

Extras - but only if you are ready for them! First see kinetic energy calculation page

(d) What was the initial speed of take-off by the pole vaulter? (take gravity strength as 9.8 N/kg).

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

initial KE = 3600 J, mass = 800 / 9.8 = 81.63 kg

v = √(2KE/m) = √(2 x 3600/81.63) = 9.4 m/s (2sf)

(e) Suppose the pole vaulter misses the soft mat and landed on a hard surface.

If the trainers compress 16 mm, calculate the average force acting on the trainers (hence the pole vaulter's feet) on landing and comment on the result.

energy transferred = force x distance, so

average force (N) = energy transferred (J) / distance (m)

The energy transferred equals the kinetic energy lost in impact = 3600 J

Distance = 16 mm = 0.016 m

Therefore impact force = 3600 / 0.016 = 225 000 N

This is an enormous and dangerous impact force that would cause injury to the athlete.

(f) If the pole vaulter lands on a soft mat which depresses by 30 cm on impact, recalculate the impact force and comment on the result and suggest how to land as safely as possible!

30 cm = 0.30 m, impact force = 3600 / 0.3 = 12 000 N

This is a considerably smaller impact force, but still potentially dangerous - its equal to 15 x the pole vaulter's body weight.

What pole vaulters actually do is twist their body to land spreading their body over as large area of the deep soft mat to dissipate the impact force - this reduces the pressure on the body (force/area) - the area of the twisted body is much greater than the area of the soles of the trainers.

See https://www.wikihow.com/Pole-Vault for a picture guide to pole vaulting!

 

Q5 A small bee of mass 0.06 g leaps up 10 cm above a flower. (g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy.

∆E = mgh = (0.0.06/1000) x 10 x (10/100)

= 6 x 10^-5 x 10 x 10^-1 =  6.0 x 10^-5 J  (2 sf)

(b) Assuming there is no further acceleration, and ignoring air resistance, what was the initial take off speed of the bee?

As the bee ascends, the kinetic energy store of the bee is converted into the GPE store of the bee,

Therefore final maximum ∆GPE = initial maximum ∆KE

So, ∆GPE = ∆KE = 6.0 x 10^-5 J

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 10^-5 / 6 x 10^-5) = √0.3333 = 0.58 m/s  (2 sf)

(c) If the same bee leaps from another flower and gains 2.0 x 10^-4 J, how high did the bee rise?

∆GPE = mgh, rearranging gives h = ∆GPE / mg

h = 2.0 x 10^-4 / (6.0 x 10^-5 x 10) = 0.33 m  (33 cm, 2 sf)

Q6 In falling, an initially stationary 2.00 kg weight, loses 500 J of energy before hitting the floor.

(a) At what height was the weight dropped? (assume g = 10 N/kg)

The loss of energy is equal to the loss of GPE as it is converted to KE.

∆GPE = mgh, rearranging gives:

h = ∆GPE / (g x m) = 500 / (10 x 2) = 500 / 20 = 25 m

(b) Neglecting air resistance, at what speed did the 2 kg weight hit the floor?

Assuming maximum ∆GPE = ∆KE on impact

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 500 /2) = √500 = 22.4 m/s  (3 sf)

 INDEX ENERGY: Types, stores, transfers, energy calculations