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How to solve gravitational potential energy store problems

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

E_gpe = m g h

On the Earth's surface the gravitational field strength is quoted as g = 9.8 N/kg

This section of calculations also includes problems based on using E_gpe = m g h = E_KE = 0.5 m v²

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)

 

 

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

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Energy resources, energy transfers, work done and electrical power supply revision notes index

Types of energy & stores - examples compared/explained, calculations of mechanical work done and power

Chemical energy stores  * Elastic potential energy stores, calculations  * Electrical & electrostatic energy stores

Gravitational potential energy and calculations  *  Kinetic energy stores and calculations  *  Nuclear energy store

Thermal energy stores  *  Light energy  * Sound energy  * Magnetic energy stores

Conservation of energy, energy transfers-conversions, efficiency - calculations and Sankey diagrams gcse physics

See also More on methods of reducing heat transfer eg in a house, insulating properties of materials

Energy resources: uses, general survey & trends, comparing renewables, non-renewables, generating electricity

Renewable energy (1) Wind power and solar power, advantages and disadvantages gcse physics revision notes

Renewable energy (2) Hydroelectric power and geothermal power, advantages and disadvantages gcse physics

Renewable energy (3) Wave power and tidal barrage power, advantages and disadvantages gcse physics

See also Renewable energy - biomass - biofuels & alternative fuels, hydrogen, biogas, biodiesel gcse chemistry notes

Greenhouse effect, global warming, climate change, carbon footprint from fossil fuel burning gcse chemistry

The absorption and emission of radiation by materials - temperature & surface factors including global warming

The Usefulness of Electricity gcse physics electricity revision notes

and The 'National Grid' power supply, mention of small scale supplies, transformers gcse physics notes

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Doc Brown's School Physics Revision Notes

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