Elastic strings and springs Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Elastic strings and springs questions. See exactly how to solve problems on hookes-law, tension, modulus-of-elasticity, elastic-potential-energy.

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.

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 0.5 m0.5\ \text{m} and modulus of elasticity 30 N30\ \text{N} has one end attached to a fixed point AA. The spring is stretched until its total length is 0.7 m0.7\ \text{m}. Find the tension in the spring.

Worked solution

  1. Write down the information given in the question

    l=0.5 m,λ=30 N,length=0.7 ml=0.5\ \text{m},\quad \lambda=30\ \text{N},\quad \text{length}=0.7\ \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=0.7−0.5=0.2 mx=0.7-0.5=0.2\ \text{m}

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

  3. State Hooke's law for the spring

    T=λxlT=\frac{\lambda x}{l}

    The tension is proportional to the extension.

  4. State the final answer

    answer=12 N\text{answer}=12\ \text{N}

    This is the value asked for in the question.

Answer
12 N12\ \text{N}
Question 3
2 markseasy
A light elastic string of natural length 1.2 m1.2\ \text{m} and modulus of elasticity 45 N45\ \text{N} has one end attached to a fixed point AA. The string is stretched until its total length is 1.6 m1.6\ \text{m}. Find the tension in the string.

Worked solution

  1. Write down the information given in the question

    l=1.2 m,λ=45 N,length=1.6 ml=1.2\ \text{m},\quad \lambda=45\ \text{N},\quad \text{length}=1.6\ \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=1.6−1.2=0.4 mx=1.6-1.2=0.4\ \text{m}

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

  3. State the final answer

    answer=15 N\text{answer}=15\ \text{N}

    This is the value asked for in the question.

Answer
15 N15\ \text{N}
Question 4
2 markseasy
A light elastic string of natural length 1.5 m1.5\ \text{m} has one end attached to a fixed point AA. The string is stretched until its total length is 1.8 m1.8\ \text{m}, and in this position the tension in it is 12 N12\ \text{N}. Find the modulus of elasticity of the string.

Worked solution

  1. Write down the information given in the question

    l=1.5 m,length=1.8 m,T=12 Nl=1.5\ \text{m},\quad \text{length}=1.8\ \text{m},\quad T=12\ \text{N}

    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=1.8−1.5=0.3 mx=1.8-1.5=0.3\ \text{m}

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

  3. Rearrange Hooke's law to make λ\lambda the subject

    λ=Tlx\lambda=\frac{Tl}{x}

    Multiply through by ll and divide by the extension.

  4. State the final answer

    answer=60 N\text{answer}=60\ \text{N}

    This is the value asked for in the question.

Answer
60 N60\ \text{N}
Question 5
2 markseasy
A light elastic spring of natural length 0.4 m0.4\ \text{m} has one end attached to a fixed point AA. The spring is stretched until its total length is 0.55 m0.55\ \text{m}, and in this position the tension in it is 9 N9\ \text{N}. Find the modulus of elasticity of the spring.

Worked solution

  1. Write down the information given in the question

    l=0.4 m,length=0.55 m,T=9 Nl=0.4\ \text{m},\quad \text{length}=0.55\ \text{m},\quad T=9\ \text{N}

    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=0.55−0.4=0.15 mx=0.55-0.4=0.15\ \text{m}

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

  3. State the final answer

    answer=24 N\text{answer}=24\ \text{N}

    This is the value asked for in the question.

Answer
24 N24\ \text{N}

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