Hard Further Maths Elastic strings and springs Questions

Challenging, exam-style Further Maths Elastic strings and springs questions with worked solutions. Stretch yourself on the hardest energy-methods, quadratic-equation, gravitational-potential-energy, slack-string problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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}.
Question 2
9 markschallenging
A particle of mass 1 kg1\ \text{kg} is attached to one end of a light elastic string of natural length 2 m2\ \text{m} and modulus of elasticity 60 N60\ \text{N}. The other end of the string is attached to a fixed point OO on a smooth horizontal table. The particle is held at rest on the table at a distance 2.8 m2.8\ \text{m} from OO and is then released. Consider the instant at which the particle is at a distance 1.4 m1.4\ \text{m} from OO. Suppose instead that the string were replaced by a light elastic spring of the same natural length and the same modulus of elasticity. Which of the following correctly compares the energy stored in the elastic at this instant?
Show worked solution

Worked solution

  1. Compare the distance from OO with the natural length

    1.4 m<2 m1.4\ \text{m}<2\ \text{m}

    The elastic is shorter than its natural length at this instant.

  2. State what a string does in this situation

    string: T=0,EPE=0\text{string}:\ T=0,\quad\text{EPE}=0

    A string can only pull, so it simply goes slack and stores nothing.

  3. State what a spring does in this situation

    spring: compressed by 0.6 m\text{spring}:\ \text{compressed by}\ 0.6\ \text{m}

    A spring can be compressed, and it then pushes the particle away from OO.

  4. Compute the energy stored by the compressed spring

    EPE=60×0.622×2=5.4 J\text{EPE}=\frac{60\times0.6^{2}}{2\times2}=5.4\ \text{J}

    Hooke's law and the energy formula apply to a compression exactly as to an extension.

  5. Contrast the two results

    5.4 J (spring)versus0 J (string)5.4\ \text{J}\ \text{(spring)}\quad\text{versus}\quad0\ \text{J}\ \text{(string)}

    This is the single most important distinction in the topic.

  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. State the value of gg

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

    All numerical work in this topic uses g=9.8g=9.8.

  16. State the principle of conservation of mechanical energy

    KE+GPE+EPE=constant\text{KE}+\text{GPE}+\text{EPE}=\text{constant}

    With no resistance, the total mechanical energy is unchanged.

  17. Recall the kinetic energy formula

    KE=12mv2\text{KE}=\frac{1}{2}mv^{2}

    The kinetic energy of a particle of mass mm moving with speed vv.

  18. Select the correct option

    spring=5.4 J,string=0 J\text{spring}=5.4\ \text{J},\quad\text{string}=0\ \text{J}

    Only the spring stores energy when the elastic is shorter than its natural length.

Answer
The spring would store 5.4 J5.4\ \text{J}, whereas the string stores 0 J0\ \text{J}.
Question 3
9 markschallenging
A particle of mass 0.5 kg0.5\ \text{kg} is attached to one end of a light elastic string of natural length 1 m1\ \text{m} and modulus of elasticity 40 N40\ \text{N}. The other end of the string is attached to a fixed point OO on a smooth horizontal table. The particle is held at rest on the table at a distance 1.6 m1.6\ \text{m} from OO and is then released. Find the speed of the particle when it is at a distance 0.5 m0.5\ \text{m} from OO. Give your answer to 33 significant figures. Select the correct value from the options below.
Show worked solution

Worked solution

  1. Write down the information given in the question

    m=0.5 kg,l=1 m,λ=40 N,start=1.6 m,finish=0.5 mm=0.5\ \text{kg},\quad l=1\ \text{m},\quad \lambda=40\ \text{N},\quad \text{start}=1.6\ \text{m},\quad \text{finish}=0.5\ \text{m}

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

  2. Note that the table is smooth and horizontal

    ΔGPE=0\Delta\text{GPE}=0

    No energy is lost to friction and no height is gained.

  3. Check that the string is taut in the starting position

    1.6 m>1 m  ⟹  taut1.6\ \text{m}>1\ \text{m}\implies\text{taut}

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

  4. Evaluate the elastic potential energy from the extension

    EPE1=40×(0.6)22×1=7.2 J\text{EPE}_{1}=\frac{40\times\left(0.6\right)^{2}}{2\times1}=7.2\ \text{J}

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

  5. Check whether the string is taut in the final position

    0.5 m<1 m  ⟹  slack0.5\ \text{m}<1\ \text{m}\implies\text{slack}

    The length is below the natural length, so the string is SLACK: its tension and its elastic potential energy are both zero.

  6. Set the elastic potential energy to zero because the string is slack

    EPE2=0 J\text{EPE}_{2}=0\ \text{J}

    A slack string stores no energy; it must NOT be given a negative extension.

  7. Apply conservation of energy

    EPE1=12mv2+EPE2\text{EPE}_{1}=\frac{1}{2}mv^{2}+\text{EPE}_{2}

    The particle starts from rest, so all of the kinetic energy comes from the elastic.

  8. Find the kinetic energy gained

    12mv2=7.2−0=7.2 J\frac{1}{2}mv^{2}=7.2-0=7.2\ \text{J}

    The difference of the two elastic potential energies.

  9. Solve for the speed

    12×0.5×v2=7.2  ⟹  v=2×7.20.5=5.367 m s−1\frac{1}{2}\times0.5\times v^{2}=7.2\implies v=\sqrt{\frac{2\times7.2}{0.5}}=5.367\ \text{m}\ \text{s}^{-1}

    The kinetic energy gained is converted into a speed.

  10. 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.

  11. 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.

  12. 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.

  13. 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}.

  14. 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.

  15. 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.

  16. 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.

  17. State the final answer

    answer=5.37 m s−1\text{answer}=5.37\ \text{m}\ \text{s}^{-1}

    This is the value asked for in the question.

Answer
5.37 m s−15.37\ \text{m}\ \text{s}^{-1}
Question 4
9 markschallenging
A particle of mass 0.3 kg0.3\ \text{kg} is attached to one end of a light elastic string of natural length 0.8 m0.8\ \text{m} and modulus of elasticity 24 N24\ \text{N}. The other end of the string is attached to a fixed point OO on a smooth horizontal table. The particle is held at rest on the table at a distance 1.4 m1.4\ \text{m} from OO and is then released. Find the speed of the particle at the instant the string becomes slack.
Show worked solution

Worked solution

  1. Write down the information given in the question

    m=0.3 kg,l=0.8 m,λ=24 N,start=1.4 mm=0.3\ \text{kg},\quad l=0.8\ \text{m},\quad \lambda=24\ \text{N},\quad \text{start}=1.4\ \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 starting position

    1.4 m>0.8 m  ⟹  taut1.4\ \text{m}>0.8\ \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=24×(0.6)22×0.8=5.4 J\text{EPE}_{1}=\frac{24\times\left(0.6\right)^{2}}{2\times0.8}=5.4\ \text{J}

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

  4. Identify where the string becomes slack

    length=l=0.8 m\text{length}=l=0.8\ \text{m}

    The string goes slack the instant its length falls back to its natural length.

  5. Set the elastic potential energy to zero at that instant

    EPE2=0 J\text{EPE}_{2}=0\ \text{J}

    The extension is zero there, so no energy remains stored.

  6. State the distance travelled by the particle

    s=1.4−0.8=0.6 ms=1.4-0.8=0.6\ \text{m}

    The particle moves from the starting point to the point where the string is at its natural length.

  7. Apply conservation of energy

    12mv2=EPE1=5.4 J\frac{1}{2}mv^{2}=\text{EPE}_{1}=5.4\ \text{J}

    On a smooth horizontal table all of the stored energy becomes kinetic energy.

  8. Solve for the speed

    12×0.3×v2=5.4  ⟹  v=2×5.40.3=6 m s−1\frac{1}{2}\times0.3\times v^{2}=5.4\implies v=\sqrt{\frac{2\times5.4}{0.3}}=6\ \text{m}\ \text{s}^{-1}

    The kinetic energy gained is converted into a speed.

  9. 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.

  10. 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.

  11. 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.

  12. 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}.

  13. 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.

  14. 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.

  15. 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.

  16. State the final answer

    answer=6 m s−1\text{answer}=6\ \text{m}\ \text{s}^{-1}

    This is the value asked for in the question.

Answer
6 m s−16\ \text{m}\ \text{s}^{-1}
Question 5
9 markschallenging
A particle of mass 2.5 kg2.5\ \text{kg} rests in equilibrium on a smooth plane inclined at 60∘60^{\circ} to the horizontal. The particle is attached to one end of a light elastic string of natural length 1.4 m1.4\ \text{m} and modulus of elasticity 100 N100\ \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.
Show worked solution

Worked solution

  1. Write down the information given in the question

    m=2.5 kg,l=1.4 m,λ=100 N,θ=60∘m=2.5\ \text{kg},\quad l=1.4\ \text{m},\quad \lambda=100\ \text{N},\quad \theta=60^{\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⁡60∘=2.5×9.8×sin⁡60∘=21.22 Nmg\sin60^{\circ}=2.5\times9.8\times\sin60^{\circ}=21.22\ \text{N}

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

  4. Put Hooke's law equal to this component

    100 x1.4=21.22\frac{100\,x}{1.4}=21.22

    The string lies along the line of greatest slope.

  5. Solve for the extension

    x=21.22×1.4100=0.2970 mx=\frac{21.22\times1.4}{100}=0.2970\ \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. 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. State the final answer

    answer=0.297 m\text{answer}=0.297\ \text{m}

    This is the value asked for in the question.

Answer
0.297 m0.297\ \text{m}

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