Question

A spring stores $\text{1J}$ of energy for a compression of $\text{1mm}$. The additional work to be done to compress it further by $\text{1mm}$ is
$\begin{aligned} & \text{A}\text{. 1J} \\\ & \text{B}\text{. 2J} \\\ & \text{C}\text{. 3J} \\\ & \text{D}\text{. 4J} \\\ & \text{E}\text{. 0}\text{.5J} \\\ \end{aligned}$

Answer 1

Hint: When spring gets compressed, a restoring force is produced. Use a work energy formula which states work done is equal to kinetic energy change. Work done depends on displacement and restoring force. Value of the spring constant is not given. So calculate this value by using first condition and then use this value in second case to calculate final work done. Additional work can be calculated by subtracting the value of work.

Complete step by step solution:
In the question it is given that a spring stores $\text{1J}$ of energy to compress $\text{1mm}$ of spring. Now there is additional work we have to do to compress by $\text{1mm}$ again. We need to find additional work done to compress $\text{1mm}$ of spring.

Aim: Find additional work done to compress $\text{1mm}$ of spring.

When spring is compressed a potential energy is stored in it and later it gets converted in kinetic energy. We know that work done is nothing but energy to do work.

Find energy for 1st case:

${{\text{W}}_{1}}=1J,x=1\times {{10}^{-3}}m$
So work done is equal to energy.
Mathematically,
${{W}_{1}}=\dfrac{1}{2}k{{x}^{2}}$
Where,

$\begin{aligned} & {{W}_{1}}=\text{work done to compress 1mm} \\\ & x=\text{ spring constant} \\\ & k=\text{displacement} \\\ \end{aligned}$
Put the value in the above equation given in question.
We get,

& 1J=\dfrac{1}{2}k{{\left( 1\times {{10}^{-3}} \right)}^{2}} \\\ & k=2\times {{10}^{6}} \\\ \end{aligned}$$ From this we get the value of k. Find energy for 2nd case: ${{\text{W}}_{1}}=?,x=1\times {{10}^{-3}}m,k=2\times {{10}^{6}}$ So work done is equal to energy. Mathematically, ${{W}_{2}}=\dfrac{1}{2}k{{x}^{2}}$ Put value in the above equation. We get, ${{W}_{2}}=\dfrac{1}{2}(2\times {{10}^{6}}){{(1\times {{10}^{-3}}\times 1\times {{10}^{-3}})}^{2}}=4J$ Now the question is how to get additional work. In the first case we calculated work for 1mm and in second we calculated final work for 2mm i.e. 1mm plus 1mm. So additional work done can be calculated by subtracting initial from final. So we get, $$\text{Additional work = }{{\text{W}}_{2}}-\text{ }{{\text{W}}_{1}}=4-1=3J$$ Hence the answer is 3 joule. Answer- (C) Note: We know that when spring is compressed or relaxed, a restoring force is stored in it which is directly proportional to displacement. The proportionality constant in restoring force is known as spring constant denoted by k. additional work done can be calculated by subtracting initial from final. Since work is equal to energy therefore the unit of work is joule (J).