School Physics notes: (b) Elastic potential energy stores and calculations

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TYPES OF ENERGY STORE - examples explained

(b) Elastic potential energy stores & calculations

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What is a elastic potential energy store?  How solve numerical problems involving elastic potential energy?  How do you explain examples of elastic potential energy  ?  How to use the formula and do calculations involving elastic potential energy

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This is a BIG website, you need to take time to explore it [Website Search Box] Elastic potential energy and elastic potential energy stores

• This is energy stored when some material is stretched or compressed and the energy released when the constriction is released - usually as kinetic energy.

• eg the wound up compressed spring of a clockwork clock - provides the energy store to drive the mechanical motor,

• a pulled elastic rubber band - a stretched catapult stores elastic potential energy, on release the projectile is fired forward as its kinetic energy store is increased,

• stretched coiled metal spring,

• the compressed spring in a an animal trap

• the springs on a car suspension absorb the impact energy of the wheels on the road to give a smoother ride,

• stretched bow before the arrow is released - the 'twang' increases the kinetic energy store of the arrow..

• Since elastic potential energy is a form of stored energy, it does nothing until it is released and converted into another form of energy - often converting to kinetic energy.

• When the forces causing the stretching is removed, the spring or elastic returns to its original length (shape)

• The more an elastic material is stretched, the greater the elastic potential energy store.

• The amount of elastic potential energy stored in a stretched spring can be calculated using the equation:

• elastic potential energy = 0.5 × spring constant × (extension)2,

• Ee = 1/2 k e2

• (assuming the limit of proportionality has not been exceeded)

• elastic potential energy, Ee, in joules, J

• spring constant, k, in newtons per metre, N/m

• extension or compression, e, in metres, m

For more on elastic potential energy see Elasticity and energy stored in a spring - experiments and calculations

How to solve elastic potential energy store problems

elastic potential energy = 0.5 × spring constant × (extension)2

Ee = 1/2 k e2

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.

Eepe = 1/2 k e2,  10.0 cm ≡ 10.0 / 100 = 0.10 m

Eepe = 0.5 x 5.0 x 0.102

Eepe = 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?

Eepe = 1/2 k e2,  rearranging gives e2 = 2Eepe / k,  e = √(2Eepe / k)  and  e = the compression

e = √(2Eepe / 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

Eepe = 1/2 k e2,  rearranging gives k = 2Eepe / e2

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.

Eepe = 1/2 k e2 = 0.5 x 250 x 0.402 = 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.

Eepe = 1/2 k e2, rearranging:

k = Eepe x 2 / e2 = 20 x 2 / 0.402 = 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?

Eepe = 1/2 k e2, rearranging and 60 cm = 0.60 m

e = √(2Eepe / 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: Ee = 1/2 k e2   (dealt with in detail already on this page)

Ee = 1/2 k e2  = 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: Eepe = 1/2 k e2

rearranging: extension e = √(2Eepe / 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/s2 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: Eepe = 1/2 k e2

rearranging gives: k = Eepe x 2 / e2 = 0.147 x 2 / (5.0 /100)2 = 118 N/m (3 sf)

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