Question

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

• A. $6.9\times {10}^{-2}J$
• B. $12.2\times {10}^{-2}J$
• C. $8.1\times {10}^{-2}J$
• D. $12.2\times {10}^{-1}J$

Answer 1

The correct option is C $8.1\times {10}^{-2}J$

Given,

$F=-5x-16x^3$

$-kx=-5x-16x^3$

$k=5+16x^2$

work done,

$=\frac{1}{2}(k_1{x}_1^2+k_2{x}_2^2)$

upon solving the above equation, we get,

work done $=8.7\times {10}^{-2}J$