Questions on elastic potential energy

Question 1

Write down the formula for the energy stored in a stretched elastic string, and explain where the factor of one half comes from.

Accepted Answer

EPE = λx²/(2l). The tension grows linearly from 0 to λx/l as the string stretches, so the work done stretching it is the area of the triangle under the Hooke's law graph, half the base x times the height λx/l. B1 for the formula, B1 for the triangle under the Hooke's law graph. Using the final tension throughout would double the answer.

Question 2

An elastic string of natural length 2 m and modulus 80 N is stretched by 0.5 m. Find the energy stored.

Accepted Answer

EPE = 80 × 0.5²/(2 × 2) = 80 × 0.25/4 = 5 J. M1 for substituting into λx²/(2l), A1 for 5 J. Checking by area: the tension at that extension is 20 N, and ½ × 0.5 × 20 = 5 J.

Question 3

A spring of natural length 0.6 m and modulus 30 N is compressed to a length of 0.4 m. Find the energy stored in it.

Accepted Answer

The compression is 0.6 − 0.4 = 0.2 m, and it takes the place of the extension: EPE = 30 × 0.2²/(2 × 0.6) = 1.2/1.2 = 1 J. M1 for λx²/(2l) with the compression, A1 for 1 J. The formula is the same for a compression because the thrust grows linearly from zero just as a tension does, so the area under the graph is again a triangle.

Question 4

An elastic string has natural length 1 m and modulus 40 N. Find the work done in stretching it from an extension of 0.2 m to an extension of 0.5 m.

Accepted Answer

The work done is the increase in the stored energy. At extension 0.2 m the string holds 40 × 0.2²/(2 × 1) = 0.8 J, and at 0.5 m it holds 40 × 0.5²/2 = 5 J. M1 for λx²/(2l) at each extension, A1 for 0.8 J and 5 J. The work done is 5 − 0.8 = 4.2 J. A1. Putting the change in extension, 0.3 m, into the formula gives 1.8 J, which is the energy of a stretch from zero to 0.3 m and not of this one: the tension is already 8 N when the stretch begins, so the later 0.3 m costs more than the first 0.3 m would.

Question 5

A particle of mass 0.2 kg on a smooth horizontal table is attached to an elastic string of natural length 1 m and modulus 20 N. It is pulled out to an extension of 0.4 m and released. Find its speed when the string reaches its natural length.

Accepted Answer

EPE = 20 × 0.4²/(2 × 1) = 20 × 0.16/2 = 1.6 J. The table is smooth and horizontal, so all of it becomes kinetic energy: ½(0.2)v² = 1.6 gives v² = 16 and v = 4 m/s. M1 for the elastic energy, A1 for 1.6 J, M1 for equating it to kinetic energy, A1 for 4 m/s. Beyond the natural length the string goes slack and the particle keeps that speed.

Question 6

A particle of mass 0.5 kg hangs from an elastic string of natural length 1 m and modulus 39.2 N attached to a ceiling. It is pulled down until the string is 1.5 m long and released. Find its speed when the string returns to its natural length, taking g = 9.8 m/s².

Accepted Answer

Elastic energy released = 39.2 × 0.5²/(2 × 1) = 4.9 J. The particle rises 0.5 m, gaining 0.5(9.8)(0.5) = 2.45 J of potential energy. What remains is kinetic: ½(0.5)v² = 2.45, so v² = 9.8 and v = 3.13 m/s. M1 for the elastic energy, A1 for 4.9 J, B1 for the gain in potential energy, M1 for the energy equation, A1 for 3.13 m/s. Forgetting the height change would give 4.43 m/s.

Question 7

State two errors that make an energy equation with an elastic string wrong, and state how to avoid each.

Accepted Answer

First, using the total length in place of the extension. Write l and x down separately before substituting anything. Second, measuring heights from different levels at the two instants. Choose one zero level, mark it on the diagram and use it in both energy totals. B1 B1 for the two slips, B1 for the remedies. A third slip, less common, is forgetting that a string stores nothing once it goes slack.

Question 8

A particle of mass 0.5 kg lies on a smooth horizontal table, attached to one end of an elastic string of natural length 1 m and modulus 32 N. The other end of the string is fixed to the table at O. The particle is projected from O, directly away from it, at 4 m/s. Find the extension of the string when the particle comes instantaneously to rest, and state the speed of the particle when it next passes through O.

Accepted Answer

For the first metre the string is slack and stores nothing, so the particle arrives at the natural length still carrying its full ½ × 0.5 × 4² = 4 J. At the instant of rest all of it is elastic: 32x²/(2 × 1) = 4, so 16x² = 4, x² = 0.25 and x = 0.5 m. B1 for the string staying slack, and the speed unchanged, until the natural length, M1 for equating the kinetic energy to λx²/(2l), A1 for x = 0.5 m, B1 for 4 m/s at O. The string then pulls the particle back, returns its 4 J by the natural length, and goes slack again, so with the table smooth the particle passes through O at 4 m/s, the speed it started with. The string is slack for the first metre, so the extension at the instant of rest is not 1.5 m.

Question 9

A ball of mass 0.2 kg is held on top of a vertical spring of natural length 0.5 m and modulus 49 N, compressing it by 0.2 m. The ball is released. Find the speed of the ball as it leaves the spring, and the greatest height it reaches above the point of release. Take g = 9.8 m/s².

Accepted Answer

Energy stored = 49 × 0.2²/(2 × 0.5) = 1.96 J. The ball leaves the spring at the natural length, 0.2 m above release, and that rise already costs potential energy: ½(0.2)v² = 1.96 − 0.2 × 9.8 × 0.2 = 1.568 J, so v² = 15.68 and v = 3.96 m/s. In free flight it rises a further 15.68/(2 × 9.8) = 0.8 m, so the greatest height is 0.2 + 0.8 = 1.0 m above release. M1 for the stored energy, A1 for 1.96 J, M1 for the energy equation including the 0.2 m rise while in contact, A1 for 3.96 m/s, M1 for the free-flight stage, A1 for 1.0 m. The height alone comes faster from the whole journey, rest to rest: 1.96 = 0.2 × 9.8 × h gives h = 1.0 m in one line, and the agreement is the check that the two-stage working is right. Forgetting the potential energy gained during contact gives 4.43 m/s for the launch speed and a two-stage height of 1.2 m, which disagrees with the one-line 1.0 m and so exposes the slip.

Question 10

A particle of mass 0.5 kg is attached to one end of an elastic string of natural length 1 m and modulus 49 N, whose other end is fixed to a point O on a ceiling. The particle is held 1 m vertically below O, with the string just taut, and released from rest. Find the speed of the particle when the extension of the string is 0.1 m, and the extension at its lowest point. Take g = 9.8 m/s².

Accepted Answer

Take the release point as the zero level. When the extension is x the particle has fallen x, so 0.5(9.8)x = ½(0.5)v² + 49x²/(2 × 1). At x = 0.1: 0.49 = 0.25v² + 0.245, so v² = 0.98 and v = 0.99 m/s. M1 for the energy equation with all three terms, A1 for 0.99 m/s. At the lowest point v = 0, so 4.9x = 24.5x², giving x = 4.9/24.5 = 0.2 m, the root x = 0 being the release point. M1 for setting the kinetic energy to zero, A1 for 0.2 m. The extension 0.1 m is where the tension 49(0.1)/1 = 4.9 N equals the weight, so the particle is passing through its equilibrium position and 0.99 m/s is its greatest speed; the lowest point is at exactly twice the equilibrium extension, where the tension is twice the weight and the particle starts back up.

Question 11

A particle of mass 2 kg lies on a smooth plane inclined at 30° to the horizontal, held against a light spring of natural length 0.5 m and modulus 98 N that lies along a line of greatest slope with its lower end fixed. The spring is compressed by 0.3 m and the particle is released from rest. Find the speed of the particle at the instant the spring reaches its natural length, taking g = 9.8 m/s².

Accepted Answer

The energy stored in the compressed spring is 98 × 0.3²/(2 × 0.5) = 8.82 J. M1 for λx²/(2l), A1 for 8.82 J. As the spring extends to its natural length the particle moves 0.3 m up the slope and rises 0.3 sin30° = 0.15 m, gaining 2(9.8)(0.15) = 2.94 J of potential energy. B1 for the potential energy with the vertical rise. The plane is smooth, so the rest is kinetic: ½(2)v² = 8.82 − 2.94 = 5.88, giving v = 2.42 m/s. M1 for the energy equation with the elastic, potential and kinetic terms, A1 for 2.42 m/s. Ignoring the rise gives 2.97 m/s, and using the 0.3 m itself as the height gives a potential energy of 5.88 J; the height is the vertical component of the distance moved along the slope. The normal reaction does no work because it is perpendicular to the motion.

Question 12

A particle of mass 1 kg lies against a spring of natural length 0.5 m and modulus 50 N, compressed by 0.1 m, on a rough horizontal surface with coefficient of friction 0.2. Find the speed of the particle as it leaves the spring and the total distance it travels before coming to rest. Take g = 9.8 m/s².

Accepted Answer

Energy stored = 50 × 0.1²/(2 × 0.5) = 0.5 J, and the frictional force is 0.2 × 1 × 9.8 = 1.96 N. The particle leaves the spring at the natural length, after 0.1 m. By then friction has taken 1.96(0.1) = 0.196 J, leaving 0.5 − 0.196 = 0.304 J as kinetic energy, so ½(1)v² = 0.304 and v = 0.780 m/s. For the total distance, all the elastic energy is eventually spent against friction. So 1.96d = 0.5 and d = 0.255 m, of which the last 0.155 m is travelled clear of the spring. M1 for the stored energy, A1 for 0.5 J, B1 for the frictional force, M1 for the energy equation over the 0.1 m of contact, A1 for 0.780 m/s, M1 for spending all the stored energy against friction, A1 for 0.255 m. Friction acts over the whole 0.255 m, including the 0.1 m of contact, so it must not be applied only to the free part. A spring is used rather than a string because only a spring can push.

Question 13

A particle of mass 0.5 kg is attached to one end of an elastic string of natural length 1 m and modulus 19.6 N. The other end is fixed to a point O on a ceiling. The particle is released from rest at O and falls vertically. Find the extension of the string at its lowest point, and the depth of the lowest point below O. Take g = 9.8 m/s².

Accepted Answer

Draw the two instants: at O the particle is at rest with the string slack, and at the lowest point it is at rest with extension x, having fallen 1 + x in total. Rest to rest means no kinetic term, so the potential energy given up equals the elastic energy stored: 0.5 × 9.8 × (1 + x) = 19.6x²/(2 × 1). That is 4.9(1 + x) = 9.8x², and dividing by 4.9 gives 2x² − x − 1 = 0, so (2x + 1)(x − 1) = 0 and x = 1 m, rejecting the negative root. The lowest point is 1 + 1 = 2 m below O. B1 for the total fall of 1 + x with both instants at rest, M1 for the energy equation, A1 for 2x² − x − 1 = 0, M1 for solving the quadratic, A1 for x = 1 m with the negative root rejected, A1 for the depth of 2 m. The lowest point is not equilibrium: the tension there is 19.6 × 1/1 = 19.6 N against a weight of 4.9 N, so the particle is momentarily at rest but accelerating upwards, which is why it bounces back rather than hanging there.

Question 14

A particle of mass m hangs from a fixed point by a light elastic string of natural length l and modulus λ. It is released from rest with the string at its natural length. Prove that the extension of the string at the lowest point of the motion is twice the extension in the equilibrium position, and explain why the particle does not stay at the lowest point.

Accepted Answer

In equilibrium the tension equals the weight: λx/l = mg, so the equilibrium extension is x_e = mgl/λ. M1 for T = mg with Hooke's law, A1 for mgl/λ.
From the release point to the lowest point the particle is at rest at both instants, so the potential energy given up in falling the extension x is all stored in the string: mgx = λx²/(2l). M1 for the rest-to-rest energy equation. Dividing by x, which is not zero at the lowest point, gives x = 2mgl/λ = 2x_e, as required. A1.
At the lowest point the tension is λ(2x_e)/l = 2mg, twice the weight, so the resultant force on the particle is mg upwards and it accelerates back up: it is instantaneously at rest but not in equilibrium. B1 for the tension being 2mg with the conclusion.
The particle passes through the equilibrium position with its greatest speed and overshoots by the same distance it fell to get there, which is the symmetry of an oscillation about equilibrium.

Question 15

A particle of mass 2 kg lies on a rough horizontal table, attached to one end of an elastic string of natural length 1 m and modulus 80 N whose other end is fixed to a point O on the table. The coefficient of friction between the particle and the table is 0.25. The particle is pulled away from O until the extension is 0.5 m and released from rest. Determine whether the particle reaches O, and find where it comes to rest. Take g = 9.8 m/s².

Accepted Answer

The energy stored at release is 80 × 0.5²/(2 × 1) = 10 J, and friction is 0.25 × 2 × 9.8 = 4.9 N throughout the motion. M1 for the stored energy, B1 for the friction.
By the time the string reaches its natural length the particle has moved 0.5 m and friction has taken 4.9 × 0.5 = 2.45 J, leaving 10 − 2.45 = 7.55 J as kinetic energy. M1 for the energy equation over the stretched part, A1 for 7.55 J.
From there the string is slack and only friction acts. Reaching O needs a further 1 m, costing 4.9 J, and 7.55 > 4.9, so the particle does reach O, passing through it with 7.55 − 4.9 = 2.65 J, that is ½(2)v² = 2.65 and v = 1.63 m/s. M1 for comparing the energy in hand with the work needed to reach O. Beyond O the string is still slack, since the particle is again less than 1 m from O, so it slides on until friction has spent the 2.65 J: 4.9d = 2.65 gives d = 0.541 m, and the particle comes to rest 0.541 m beyond O. A1.
The decision is whether the string is taut: taut for the first 0.5 m, slack for the next 1.54 m. Keeping the elastic term on after the natural length would have the string pushing the particle, which a string cannot do. As a check, the total distance 0.5 + 1 + 0.541 = 2.04 m is 10/4.9, the whole stored energy spent against friction.