Questions on Hooke's law and elastic strings

Question 1

State Hooke's law, defining every symbol, and say what the modulus of elasticity is measured in.

Accepted Answer

T = λx/l, where T is the tension, x the extension beyond the natural length, l the natural length and λ the modulus of elasticity. The modulus is measured in newtons, since x/l is a ratio. B1 for the formula, B1 for the symbols, B1 for the units.

Question 2

An elastic string of natural length 2 m and modulus 80 N is stretched to a length of 2.5 m. Find the tension.

Accepted Answer

Extension = 2.5 − 2 = 0.5 m. T = 80 × 0.5/2 = 20 N. Using the length 2.5 in place of the extension gives 100 N, and that is the standard error in this topic. M1 for using the extension, A1 for 20 N.

Question 3

A spring of natural length 0.8 m and modulus 40 N is compressed to a length of 0.5 m. Find the thrust it exerts.

Accepted Answer

The compression is 0.8 − 0.5 = 0.3 m, and it plays the part of x: thrust = 40 × 0.3/0.8 = 15 N, pushing the two ends apart. M1 for using the compression in Hooke's law, A1 for 15 N. A string could not do this, since a string shorter than its natural length is slack and exerts nothing.

Question 4

An elastic string of natural length 0.75 m is stretched to a length of 1.05 m by a force of 30 N. Find its modulus of elasticity.

Accepted Answer

The extension is 1.05 − 0.75 = 0.3 m. Rearranging T = λx/l gives λ = Tl/x = 30 × 0.75/0.3 = 75 N. B1 for the extension, M1 for rearranging Hooke's law with the natural length on top, A1 for 75 N. Putting the stretched length 1.05 in place of l gives 105 N, and dividing by the extension the wrong way round gives 12 N; the modulus is a force, and 75 N is the tension the string would carry if it were stretched to twice its natural length.

Question 5

A particle of mass 3 kg hangs in equilibrium from an elastic string of natural length 1.2 m and modulus 147 N. Find the extension and the total length, taking g = 9.8 m/s².

Accepted Answer

Resolving vertically for the particle, the tension equals the weight, so T = 3(9.8) = 29.4 N. Hooke's law then gives 147x/1.2 = 29.4, so 122.5x = 29.4 and x = 0.24 m. The string is 1.44 m long. State T = W before using Hooke's law, since equilibrium is what fixes the tension. M1 for T = W, A1 for 29.4 N, M1 for Hooke's law, A1 for the extension and the total length.

Question 6

Two points A and B are 3 m apart on a smooth horizontal table. A particle is joined to A by a string of natural length 1.2 m and modulus 24 N, and to B by a string of natural length 1.2 m and modulus 36 N. Find the tension in equilibrium.

Accepted Answer

The two natural lengths total 2.4 m, so the extensions add to 3 − 2.4 = 0.6 m. The particle is in equilibrium between two strings that both pull, so the tensions are equal. Then 24x₁/1.2 = 36x₂/1.2, so 24x₁ = 36x₂ and x₁ = 1.5x₂. Hence 2.5x₂ = 0.6, giving x₂ = 0.24 m and x₁ = 0.36 m. The tension is 24(0.36)/1.2 = 7.2 N, and the other string gives the same. B1 for the extensions totalling 0.6 m, M1 for equating the tensions, A1 for x₁ = 1.5x₂, A1 for the two extensions, A1 for the tension.

Question 7

State two ways in which a spring differs from a string in these problems.

Accepted Answer

A spring can be compressed as well as stretched, giving a thrust with the same formula and the compression in place of the extension, whereas a string goes slack and exerts nothing when its ends come closer than the natural length. A spring also stays straight in compression, so it can hold a particle away from a support, which a string cannot. B1 B1 for the two differences.

Question 8

A particle of mass 2 kg rests on a smooth plane inclined at 30° to the horizontal, held by an elastic string of natural length 1 m and modulus 49 N running up the line of greatest slope to a fixed point. Find the extension.

Accepted Answer

Resolving along the slope, the tension balances the component of weight down it: T = 2(9.8)sin30° = 9.8 N. The normal reaction is perpendicular to the slope and does not enter this equation. Then 49x/1 = 9.8, so x = 0.2 m. M1 for resolving along the slope, A1 for the tension, M1 for Hooke's law, A1 for the extension. On a smooth plane no friction term appears. On a rough one the extension would depend on which way the particle was about to slip.

Question 9

A light spring of natural length 0.5 m and modulus 98 N stands vertically with its lower end fixed to a horizontal floor. A particle of mass 2 kg is placed on its upper end and rests in equilibrium. Find the length of the compressed spring, taking g = 9.8 m/s².

Accepted Answer

Resolving vertically for the particle, the thrust in the spring balances the weight: T = 2(9.8) = 19.6 N. Hooke's law with the compression as x gives 98x/0.5 = 19.6, so 196x = 19.6 and x = 0.1 m. The spring is 0.5 − 0.1 = 0.4 m long. M1 for thrust equals weight, A1 for 19.6 N, M1 for Hooke's law, A1 for the length.

Question 10

Two points A and B on a horizontal ceiling are 1.2 m apart. A particle of mass 3 kg hangs from two identical light elastic strings, each of natural length 0.8 m, one attached to A and the other to B, and rests in equilibrium 0.8 m vertically below the midpoint of AB. Find the modulus of elasticity of the strings, taking g = 9.8 m/s².

Accepted Answer

Each string runs from a point 0.6 m horizontally and 0.8 m vertically from the particle, so its stretched length is √(0.6² + 0.8²) = 1 m and its extension is 1 − 0.8 = 0.2 m. Each string makes an angle with the vertical whose cosine is 0.8/1 = 0.8. By symmetry the two tensions are equal, and resolving vertically gives 2T(0.8) = 3(9.8), so T = 29.4/1.6 = 18.375 N. Hooke's law then gives λ = Tl/x = 18.375 × 0.8/0.2 = 73.5 N. B1 for the stretched length and extension, M1 for resolving vertically with the correct cosine, A1 for the tension, M1 for Hooke's law, A1 for 73.5 N. The horizontal components cancel by symmetry and need not be written down. Using the 0.6 in place of the 0.8 gives a tension of 24.5 N, which is the sine of the angle rather than its cosine.

Question 11

Two elastic strings are made of the same material and have the same modulus of elasticity, 60 N. One has natural length 0.5 m and the other has natural length 2 m. Each is stretched by 0.1 m. Find the tension in each string and explain why they differ.

Accepted Answer

For the short string T = 60 × 0.1/0.5 = 12 N; for the long one T = 60 × 0.1/2 = 3 N. B1 B1 for the two tensions. They differ because Hooke's law relates the tension to the extension as a fraction of the natural length, x/l, and 0.1 m is a fifth of the short string but only a twentieth of the long one: the same modulus means the same tension at the same proportional stretch, not at the same extension. B1 for the explanation in terms of x/l. The gradient of the tension-extension line is λ/l, so the long string is a quarter as stiff as the short one.

Question 12

One end of an elastic string of natural length 1.5 m and modulus 49 N is attached to a fixed point O on a rough horizontal table. A particle of mass 2 kg is attached to the other end and rests in equilibrium on the table. The coefficient of friction between the particle and the table is 0.2. Find the set of possible distances of the particle from O, taking g = 9.8 m/s².

Accepted Answer

The normal reaction is R = 2(9.8) = 19.6 N, so limiting friction is μR = 0.2(19.6) = 3.92 N.
The string pulls the particle towards O and friction is all that opposes it, so equilibrium requires T ≤ 3.92. Hooke's law gives 49x/1.5 ≤ 3.92, that is x ≤ 0.12 m.
A string exerts no force at all until it is stretched, so any distance up to the natural length also gives equilibrium. The particle can rest anywhere with OP ≤ 1.62 m, and the string is taut only for 1.5 < OP ≤ 1.62.
B1 for R = 19.6 N, M1 A1 for the limiting friction, M1 for T ≤ 3.92, A1 for x ≤ 0.12 m, B1 for the slack case.

Question 13

A particle of mass 5 kg rests on a rough plane inclined at an angle α to the horizontal, where sin α = 0.6. It is held by a light elastic string of natural length 1 m and modulus 98 N that runs up the line of greatest slope to a fixed point. The coefficient of friction between the particle and the plane is 0.25. Show that the particle is in equilibrium if and only if the extension of the string lies between 0.2 m and 0.4 m, taking g = 9.8 m/s².

Accepted Answer

Perpendicular to the plane, R = 5(9.8)(0.8) = 39.2 N, so limiting friction is 0.25(39.2) = 9.8 N. The weight has component 5(9.8)(0.6) = 29.4 N down the slope. B1 for R and the limiting friction.
Friction can act either way, up to 9.8 N. If the particle is on the point of slipping down, friction acts up the slope and T + 9.8 = 29.4, so T = 19.6 N. If it is on the point of slipping up, friction acts down and T = 29.4 + 9.8 = 39.2 N. M1 for resolving along the slope with friction in one direction, A1 for 19.6 N, A1 for 39.2 N.
Hooke's law gives x = T/98, so the tension range 19.6 ≤ T ≤ 39.2 is the extension range 0.2 ≤ x ≤ 0.4. M1 for Hooke's law, A1 for the range.
The 'only if' needs the slack case ruled out: with no tension the most friction can supply is 9.8 N, which is less than the 29.4 N needed, so the particle cannot rest with the string slack or with any extension below 0.2 m.

Question 14

Two light elastic strings, of natural lengths l₁ and l₂ and moduli λ₁ and λ₂, are tied end to end and the combined string is stretched between two fixed points so that its total extension is x. Explain why the tension is the same in both strings, show that it is T = x/(l₁/λ₁ + l₂/λ₂), and deduce that two identical strings of natural length l and modulus λ tied end to end behave as a single string of natural length 2l and modulus λ.

Accepted Answer

The knot joining the strings is light and in equilibrium, so the two tensions pulling it in opposite directions must be equal; call the common value T. B1 for the argument at the knot.
Hooke's law for each string gives its extension: x₁ = Tl₁/λ₁ and x₂ = Tl₂/λ₂. M1 for writing both extensions in terms of T. The extensions add to the total, so x = Tl₁/λ₁ + Tl₂/λ₂ = T(l₁/λ₁ + l₂/λ₂), and hence T = x/(l₁/λ₁ + l₂/λ₂). M1 for adding the extensions, A1 for the result.
With l₁ = l₂ = l and λ₁ = λ₂ = λ the bracket is 2l/λ, so T = λx/(2l), which is Hooke's law for a single string of natural length 2l and modulus λ. B1 for the deduction. Writing T = λ₁x/l₁ + λ₂x/l₂ treats the strings as side by side sharing the same extension, which is a different arrangement.

Question 15

A room has a horizontal ceiling 2.5 m above a horizontal floor. A particle of mass 3 kg is attached to the ceiling by a light elastic string of natural length 1 m and modulus 49 N, and to the floor directly below by a second light elastic string of natural length 1 m and modulus 98 N. The particle hangs in equilibrium on the vertical line through the two fixed points. Determine whether the lower string is taut, and find the distance of the particle below the ceiling, taking g = 9.8 m/s².

Accepted Answer

Let y be the extension of the upper string, so the particle is 1 + y below the ceiling and the lower string has length 2.5 − (1 + y) = 1.5 − y, an extension of 0.5 − y if it is taut. B1 for the two extensions in terms of y.
Suppose both strings are taut. The upper string pulls up and the lower pulls down, so 49y = 3(9.8) + 98(0.5 − y), giving 147y = 78.4 and y = 8/15 = 0.533 m. M1 for the equilibrium equation with both tensions, A1 for y = 8/15. But then the lower string's extension 0.5 − 8/15 is negative: its ends would be closer than its natural length, and a string cannot push. The assumption fails, so the lower string is slack. B1 for rejecting the assumption for this reason.
With the lower string slack the upper string alone supports the weight: 49y = 29.4, so y = 0.6 m and the particle is 1.6 m below the ceiling. The lower string is then 2.5 − 1.6 = 0.9 m long, less than 1 m, so it is slack as assumed. M1 for the single-string equation, A1 for 1.6 m with the check.
A spring in place of the lower string would push, and 8/15 would then be correct.