Question

When a spring is stretched by a distance $ x, $ it exerts a force given by
$ F=(-5x-16{{x}^{3}})N $
The work done, when the spring is stretched from $0.1\, m$ to $0.2\, m$ is :

• A. $ 8.7\times {{10}^{-2}}J $
• B. $ 12.2\times {{10}^{-2}}J $
• C. $ 8.1\times {{10}^{-1}}J $
• D. $ 12.2\times {{10}^{-1}}J $

Answer 1

Answer: (A)

Spring force is given by
$ F=(-5x-16{{x}^{3}})N $
or $ F=(-5+16{{x}^{2}})x $ ...(i)
but we are familiar with linear restoring force,
$ F=-kx $ ...(ii) where $k$ is force constant of spring.
Comparing Eqs. (i) and (ii), we have
$ k=5+16{{x}^{2}} $
Therefore, work done in stretching the spring from position
$ {{x}_{1}} $ to $ {{x}_{2}} $ is,
$ W=\frac{1}{2}{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2} $ ..(iv)
Given, $ {{x}_{1}}\ =0.1m\,and\,x=0.2m $
So, $ W=\frac{1}{2}[5+16{{(0.2)}^{2}}]{{(0.2)}^{2}} $
$ -\frac{1}{2}[5+16{{(0.1)}^{2}}]{{(0.1)}^{2}} $
$=2.82\times 4\times {{10}^{-2}}-2.58\times {{10}^{-2}} $
$=8.7\times {{10}^{-2}}J $