Further Maths Elastic strings and springs Practice Questions

Free Further Maths Elastic strings and springs practice questions with full step-by-step worked solutions. Covers hookes-law, tension, modulus-of-elasticity, elastic-potential-energy. Practise exam-style problems and check your method.

hookes-lawtensionmodulus-of-elasticityelastic-potential-energyslack-stringreasoning
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A light elastic string of natural length 0.8 m0.8\ \text{m} and modulus of elasticity 24 N24\ \text{N} has one end attached to a fixed point AA. The string is stretched until its total length is 1.1 m1.1\ \text{m}. Find the tension in the string.
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Worked solution

  1. Write down the information given in the question

    l=0.8 m,λ=24 N,length=1.1 ml=0.8\ \text{m},\quad \lambda=24\ \text{N},\quad \text{length}=1.1\ \text{m}

    The natural length ll and the modulus of elasticity λ\lambda are the two constants that describe the elastic.

  2. Find the extension by subtracting the natural length from the total length

    x=1.1−0.8=0.3 mx=1.1-0.8=0.3\ \text{m}

    The extension is measured from the natural length, not from the fixed end.

  3. State the final answer

    answer=9 N\text{answer}=9\ \text{N}

    This is the value asked for in the question.

Answer
9 N9\ \text{N}
Question 2
2 markseasy
A light elastic spring of natural length 0.45 m0.45\ \text{m} and modulus of elasticity 18 N18\ \text{N} has one end attached to a fixed point AA. The spring is stretched until the tension in it is 6 N6\ \text{N}. Find the total length of the stretched spring.
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Worked solution

  1. Write down the information given in the question

    l=0.45 m,λ=18 N,T=6 Nl=0.45\ \text{m},\quad \lambda=18\ \text{N},\quad T=6\ \text{N}

    The natural length ll and the modulus of elasticity λ\lambda are the two constants that describe the elastic.

  2. Rearrange Hooke's law to make the extension the subject

    x=Tlλx=\frac{Tl}{\lambda}

    Multiply Hooke's law through by ll and divide by λ\lambda.

  3. Substitute the tension, the natural length and the modulus

    x=6×0.4518=0.15 mx=\frac{6\times0.45}{18}=0.15\ \text{m}

    This is the extension, not the total length.

  4. State the final answer

    answer=0.6 m\text{answer}=0.6\ \text{m}

    This is the value asked for in the question.

Answer
0.6 m0.6\ \text{m}
Question 3
4 marksintermediate
A light elastic string of natural length 2 m2\ \text{m} and modulus of elasticity 50 N50\ \text{N} has one end attached to a fixed point AA. Find the work done in stretching the string from a total length of 2.2 m2.2\ \text{m} to a total length of 2.6 m2.6\ \text{m}. Select the correct value from the options below.
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Worked solution

  1. Write down the information given in the question

    l=2 m,λ=50 N,length1=2.2 m,length2=2.6 ml=2\ \text{m},\quad \lambda=50\ \text{N},\quad \text{length}_{1}=2.2\ \text{m},\quad \text{length}_{2}=2.6\ \text{m}

    The natural length ll and the modulus of elasticity λ\lambda are the two constants that describe the elastic.

  2. Check that the string is taut in the first position

    2.2 m>2 m  ⟹  taut2.2\ \text{m}>2\ \text{m}\implies\text{taut}

    The length exceeds the natural length, so the string is stretched and stores energy.

  3. Evaluate the elastic potential energy from the extension

    EPE1=50×(0.2)22×2=0.5 J\text{EPE}_{1}=\frac{50\times\left(0.2\right)^{2}}{2\times2}=0.5\ \text{J}

    The extension is 0.2 m0.2\ \text{m}.

  4. Check that the string is taut in the second position

    2.6 m>2 m  ⟹  taut2.6\ \text{m}>2\ \text{m}\implies\text{taut}

    The length exceeds the natural length, so the string is stretched and stores energy.

  5. Evaluate the elastic potential energy from the extension

    EPE2=50×(0.6)22×2=4.5 J\text{EPE}_{2}=\frac{50\times\left(0.6\right)^{2}}{2\times2}=4.5\ \text{J}

    The extension is 0.6 m0.6\ \text{m}.

  6. The work done is the increase in elastic potential energy

    W=EPE2−EPE1=4.5−0.5=4 JW=\text{EPE}_{2}-\text{EPE}_{1}=4.5-0.5=4\ \text{J}

    Work done by the stretching force is stored as elastic potential energy.

  7. State the final answer

    answer=4 J\text{answer}=4\ \text{J}

    This is the value asked for in the question.

Answer
4 J4\ \text{J}
Question 4
6 markshard
A particle of mass 3 kg3\ \text{kg} rests in equilibrium on a smooth plane inclined at 25∘25^{\circ} to the horizontal. The particle is attached to one end of a light elastic string of natural length 1 m1\ \text{m} and modulus of elasticity 80 N80\ \text{N}; the other end of the string is attached to a fixed point AA on the plane. The string lies along a line of greatest slope with the particle below AA. Take g=9.8 m s−2g = 9.8\ \text{m}\ \text{s}^{-2}. Find the extension of the string. Give your answer to 33 significant figures.
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Worked solution

  1. Write down the information given in the question

    m=3 kg,l=1 m,λ=80 N,θ=25∘m=3\ \text{kg},\quad l=1\ \text{m},\quad \lambda=80\ \text{N},\quad \theta=25^{\circ}

    The natural length ll and the modulus of elasticity λ\lambda are the two constants that describe the elastic.

  2. Resolve along the line of greatest slope

    T=mgsin⁡θT=mg\sin\theta

    The plane is smooth, so only the tension and the component of the weight act along the slope.

  3. Compute the component of the weight down the slope

    mgsin⁡25∘=3×9.8×sin⁡25∘=12.42 Nmg\sin25^{\circ}=3\times9.8\times\sin25^{\circ}=12.42\ \text{N}

    g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}.

  4. Put Hooke's law equal to this component

    80 x1=12.42\frac{80\,x}{1}=12.42

    The string lies along the line of greatest slope.

  5. Solve for the extension

    x=12.42×180=0.1553 mx=\frac{12.42\times1}{80}=0.1553\ \text{m}

    This is the extension of the string in equilibrium.

  6. State Hooke's law

    T=λxlT=\frac{\lambda x}{l}

    λ\lambda is the modulus of elasticity, ll the natural length and xx the extension.

  7. Distinguish the extension from the total length

    x=total length−natural lengthx=\text{total length}-\text{natural length}

    The extension is never the total length; this is the most common slip in the topic.

  8. State the formula for elastic potential energy

    EPE=λx22l\text{EPE}=\frac{\lambda x^{2}}{2l}

    The energy stored in a stretched string or spring.

  9. Derive the elastic potential energy by integrating the tension

    ∫0xλsl ds=λx22l\int_{0}^{x}\frac{\lambda s}{l}\,\mathrm{d}s=\frac{\lambda x^{2}}{2l}

    The work done against a tension that grows linearly is the area of a triangle, hence the factor 12\tfrac{1}{2}.

  10. State the final answer

    answer=0.155 m\text{answer}=0.155\ \text{m}

    This is the value asked for in the question.

Answer
0.155 m0.155\ \text{m}
Question 5
9 markschallenging
A light elastic spring of natural length 0.5 m0.5\ \text{m} and modulus of elasticity 40 N40\ \text{N} has one end attached to a fixed point AA. The spring is stretched until its total length is 0.9 m0.9\ \text{m}. Which of the following correctly gives the extension of the spring and the elastic potential energy stored in it?
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Worked solution

  1. Identify the natural length and the total length

    l=0.5 m,length=0.9 ml=0.5\ \text{m},\quad\text{length}=0.9\ \text{m}

    These are two different lengths and must not be confused.

  2. Subtract to obtain the extension

    x=0.9−0.5=0.4 mx=0.9-0.5=0.4\ \text{m}

    The extension is measured from the natural length.

  3. Apply the elastic potential energy formula

    EPE=λx22l=40×0.422×0.5=6.4 J\text{EPE}=\frac{\lambda x^{2}}{2l}=\frac{40\times0.4^{2}}{2\times0.5}=6.4\ \text{J}

    The factor of 12\tfrac{1}{2} comes from integrating the tension from 00 to xx.

  4. Reject the answer obtained from TxTx

    Tx=12.8 J (wrong)Tx=12.8\ \text{J}\ \text{(wrong)}

    TxTx is twice the correct value because the tension is not constant.

  5. Reject the answer obtained from the total length

    40×0.922×0.5=32.4 J (wrong)\frac{40\times0.9^{2}}{2\times0.5}=32.4\ \text{J}\ \text{(wrong)}

    Using the total length in place of the extension is the other classic error.

  6. State Hooke's law

    T=λxlT=\frac{\lambda x}{l}

    λ\lambda is the modulus of elasticity, ll the natural length and xx the extension.

  7. Distinguish the extension from the total length

    x=total length−natural lengthx=\text{total length}-\text{natural length}

    The extension is never the total length; this is the most common slip in the topic.

  8. State the formula for elastic potential energy

    EPE=λx22l\text{EPE}=\frac{\lambda x^{2}}{2l}

    The energy stored in a stretched string or spring.

  9. Derive the elastic potential energy by integrating the tension

    ∫0xλsl ds=λx22l\int_{0}^{x}\frac{\lambda s}{l}\,\mathrm{d}s=\frac{\lambda x^{2}}{2l}

    The work done against a tension that grows linearly is the area of a triangle, hence the factor 12\tfrac{1}{2}.

  10. Beware the missing factor of one half

    EPE≠Tx\text{EPE}\neq Tx

    TxTx would be the work done by a constant force; the tension is not constant.

  11. Recall that a string can only pull

    T≥0for a stringT\geq0\quad\text{for a string}

    A string can never push, so its tension is never negative.

  12. Recall the slack condition for a string

    length≤l  ⟹  T=0 and EPE=0\text{length}\leq l\implies T=0\ \text{and}\ \text{EPE}=0

    A slack string stores no energy at all.

  13. Contrast a spring with a string

    a spring may be compressed;a string may not\text{a spring may be compressed};\quad\text{a string may not}

    A compressed spring pushes and still stores λx22l\frac{\lambda x^{2}}{2l}.

  14. State the modelling assumptions

    particle,light string,smooth surface\text{particle},\quad\text{light string},\quad\text{smooth surface}

    The mass of the string and any air resistance are neglected.

  15. Select the correct option

    x=0.4 m,EPE=6.4 Jx=0.4\ \text{m},\quad\text{EPE}=6.4\ \text{J}

    The extension is the total length minus the natural length, and the energy carries the factor of one half.

Answer
The extension is 0.4 m0.4\ \text{m} and the elastic potential energy stored is 6.4 J6.4\ \text{J}.

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