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Hooke's law and elastic strings questions
Stretch a string and it pulls back in proportion to how far it has been stretched. The constant of proportionality depends on both the material and the natural length, which is what the modulus of elasticity keeps track of.
7 original questions · 26 marks · the hooke's law and elastic strings notes · Further Mechanics 1
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State Hooke's law, defining every symbol, and say what the modulus of elasticity is measured in.
Worked 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.An elastic string of natural length 2 m and modulus 80 N is stretched to a length of 2.5 m. Find the tension.
Worked 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.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.
Worked 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 that equation carries a mark of its own. M1 for T = W, A1 for 29.4 N, M1 for Hooke's law, A1 for the extension and the total length.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.
Worked 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.State two ways in which a spring differs from a string in these problems.
Worked 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.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.
Worked 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.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.
Worked 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.
Two points earn the marks here. Friction takes its limiting value only at the extreme position, so the condition is an inequality, and the slack case must be mentioned because a string cannot push the particle outwards.
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.
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