Question

When a spring is stretched by a distance $x,$ it exerts a force given by
$F=(-5x-16x^3)N$
The work done, when the spring is stretched from $0.1m$ to $0.2m$ 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
$F=(-5x-16x^3)N$
or $F=(-5+16x^2)x$ ...(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
$k=5+16x^2$
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, $x_1=0.1mandx=0.2m$
So, $W=\frac{1}{2}[5+16{(0.2)}^2]{(0.2)}^2$
$-\frac{1}{2}[5+16{(0.1)}^2]{(0.1)}^2$
$=2.82\times 4\times {10}^{-2}-2.58\times {10}^{-2}$
$=8.7\times {10}^{-2}J$