TYPES OF ENERGY STORE  examples explained
(d) Gravitational potential energy and calculations
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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.

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

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/s^{2})

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.

GPE as stored water behind a dam
Hydroelectric power
gcse physics notes
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 v^{2}
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 v^{2} , rearranging: v^{2} = 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 v^{2} , the m's cancel out,
so g x h = ^{1}/_{2}
x v^{2} 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 GPEKE 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 GPEKE 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 'takeoff' (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
takeoff by the pole vaulter? (take gravity strength as 9.8 N/kg).
KE = ^{1}/_{2}mv^{2},
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/PoleVault 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}/_{2}mv^{2},
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}/_{2}mv^{2},
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,
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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
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Gravitational potential
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Kinetic
energy stores
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Nuclear
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Thermal
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Light energy *
Sound energy *
Magnetic
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Conservation of energy,
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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,
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Renewable energy (1) Wind power and
solar power, advantages and disadvantages gcse physics revision
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Renewable energy (2) Hydroelectric power and
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gcse physics
Renewable energy (3) Wave power and tidal barrage power,
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Renewable energy  biomass  biofuels & alternative fuels,
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The absorption and emission of radiation by
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