Question

When a spring is stretched by a distance $x$, it exerts a force, given by $F=\left(-5 x-16 x^{3}\right) N$ 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. $ 8.7\times 10^{-1}J $
• D. $ 12.2\times {{10}^{-1}}J $

Answer 1

Answer: (A)

Work done is equal to difference in potential energies for two different positions of spring. Given, $F=-5 x-16 x^{3}$
$=-\left(5+16 x^{2}\right) x$
$=-k x$
where $k\left(=5+16 x^{2}\right)$ is force constant of spring. Therefore, work done in stretching the spring from position $x_{1}$ to position $x_{2}$ is
$W=\frac{1}{2} \,k_{2}\, x_{2}^{2}-\frac{1}{2} \,k_{1} \,x_{1}^{2}$
we have, $x_{1}=0.1 \,m$ and $x_{2}=0.2\, m$,
$\therefore W=\frac{1}{2}\left[5+16(0.2)^{2}\right](0.2)^{2}$
$-\frac{1}{2}\left[5+16(0.1)^{2}\right](0.1)^{2}$
$=2.82 \times 4 \times 10^{-2}-2.58 \times 10^{-2}$
$=8.7 \times 10^{-2} \,J$