How to solve elastic potential
energy store problems
elastic potential energy = 0.5 × spring constant ×
(extension)^{2}
E_{e} = ^{1}/_{2}
k e^{2}
You may also need the spring relationship: force = a spring constant x extension
F = ke
where F = the
applied force in newtons (N), e = the spring extension in metres (m)
and k is the spring (elastic) constant in N/m.
(overlap with forces 4,
need some graph based questions)
In all questions, unless otherwise
indicated, you assume the limit of proportionality is not exceeded.
Q1 A spring with spring
constant of 5.00 N/m is stretched for an extra 10.0 cm.
How much extra energy is stored in the
elastic potential energy store of the spring by this extension.
E_{epe} = ^{1}/_{2}
k e^{2}, 10.0 cm ≡
10.0 / 100 = 0.10 m
E_{epe} = 0.5 x 5.0 x 0.10^{2}
E_{epe} =
0.025 J
Q2 A
spring has a spring constant of 2000 N/m.
(a) If the elastic potential energy store
of the spring is 50.0 J, how far is the spring compressed?
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging gives e^{2} = 2E_{epe} / k,
e = √(2E_{epe} / k) and e = the compression
e = √(2E_{epe} / k) = √(2 x
50 / 2000) = √0.063 =
0.224 m (22.4 cm, 3 s.f.)
(b) What force is needed to compress
the spring by 15.0 cm in length?
F = ke = 2000 x 15/100 =
300 N
Q3 It takes 5.0 J
of work to stretch a spring 20 cm.
How much extra work must be done to
stretch it another 20 cm?
(i) You need to work out the spring
constant. 20 cm ≡
20 / 100 = 0.20 m
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging gives k = 2E_{epe} / e^{2}
k = (2 x 5.0) / (0.20 x 0.20) =
250 N/m
(ii) Then work out the total work to
stretch the spring a total of 40 cm.
The total work done on the spring
equals its elastic potential energy store when fully stretched 40 cm
(which is 0.40 m). Since you now know the spring constant, you use
the same equation again, but solving for the total elastic potential
energy.
E_{epe} = ^{1}/_{2}
k e^{2} = 0.5 x 250 x 0.40^{2} =
20 J
(iii) You then subtract (i) from (ii)
to get the extra work done.
Therefore the extra work done =
20 - 5 =
15 J
Q4 A spring stores an
extra 20 J of elastic potential energy when stretched an extra 40 cm.
Calculate the spring constant.
40 cm = 0.40 m.
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging:
k = E_{epe} x 2 / e^{2}
= 20 x 2 / 0.40^{2} =
250
N/m
Q5 A stretched string has
a total length of 60 cm and a spring constant of 240 N/m.
If the stretched spring is storing 20
J of energy, what is the length of the unstretched spring to the nearest
cm?
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging and 60 cm = 0.60 m
e = √(2E_{epe} / k) =
√(2 x 20 / 240) = √(1/6) = 0.408 m
0.408 m = 40.8 cm, ~41 cm = extra
length added to the stretched string
Therefore original length of spring =
60 - 41 =
~19
cm
Q6 A
spring is fixed firmly in a vertical position. When a mass of 120.0
g is attached to the spring it extends in length by 3.2 cm.
(a) Assuming the gravitational field strength is 9.8 N/kg,
calculate the spring constant k.
100 g is equivalent to a weight (force) of 9.8 x 120/1000
= 1.176 N
The extension e = 3.2/100 = 0.032 m. Force F
= 1.176 N
(The equation F = ke has already been dealt with in
detail further up this page)
F = ke, so the spring constant k =
F ÷ e = 1.176 ÷ 0.032 =
36.8 N/m (3 sf)
(b) Calculate the extra elastic potential energy
stored in the spring as a result of the added weight.
Using the equation for elastic stored
energy: E_{e} =
^{1}/_{2}
k e^{2}^{ } (dealt with in detail already on this
page)
E_{e} =
^{1}/_{2}
k e^{2 } =
^{1}/_{2}
x 36.75 x 0.032 x 0.032 =
0.019 J
(c) If an extra 200 g mass is placed on
the spring, how much longer will it get?
200 g equates to a weight of 9.8 x
200/1000 = 1.96 N, k = 36.75 N/m
F = ke, so the spring extension e =
F ÷ k = 1.96/36.75 =
0.053 m (5/3 cm)
(d) What force is needed to extend the
spring by 30 cm?
force required = F = ke = 36.75 x
30/100 =
11.0 N (1 dp, 3 sf)
(d) What important assumption have you
made concerning the calculations in (a) to (c)?
You have assumed the spring behaves
truly elastic i.e. within the elastic limit (limit of
proportionality).
Q7 A spring has a spring constant of 20.0 N/m.
What would the extension be in cm,
if 0.50 J of work was done on stretching the spring?
Elastic potential energy formula: E_{epe} = ^{1}/_{2}
k e^{2}
rearranging: extension e = √(2E_{epe}
/ k) = √(2 x 0.50 / 20) = 0.224 m =
22.4
cm
Q8 For this question you
need to know the formula for gravitational potential energy (GPE) and
gravitational field constant g = 9.8 m/s^{2} or 9.8 N/kg). Watch out
for units, remember in the end to work in J, m and kg.
A fun toy consists of a spring and a
funny head on the end.
When the toy spring is compressed 5.0
cm, and released, it leaps vertically up into the air.
The toy spring has a mass of 20 g.
When released it leaps up to a
maximum height of 75 cm.
If we assume all the elastic
potential energy (EPE) is transferred to the GPE store of the toy
spring, deduce the spring constant of the spring.
GPE = mgh = (20 / 1000) x 9.8
x (75 / 100) = 0.147 J
Neglecting air resistance we
can say GPE = EPE for the energy store transfer
Elastic potential energy formula: E_{epe} = ^{1}/_{2}
k e^{2}
rearranging gives: k = E_{epe} x 2 / e^{2}
= 0.147 x 2 / (5.0 /100)^{2} =
118 N/m (3
sf)
Q9
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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
and
The 'National Grid' power supply, mention of small
scale supplies, transformers gcse
physics notes