Question

If the potential energy of a spring when stretched through a distance ‘a ‘ is 25J , then what is the amount of wok done on the same spring so ad to stretch it by an additional distance ‘5a‘.

Answer 1

$$\begin{gathered} Potentialenergyofthespring=\frac{1}{2}k{x}^2 \\ 25=\frac{1}{2}k{a}^2 \\ newpotentialenergy=\frac{1}{2}k{\left(5a+a\right)}^{02} \\ \frac{\frac{1}{2}k{\left(5a+a\right)}^{02}}{\frac{1}{2}k{a}^2}=\frac{U}{25} \\ \frac{U}{25}=36 \\ U=36\times 25 \\ changeinpotentialenergy=workdone \\ W=36\times 25-25 \\ W=35\times 25=875J \end{gathered}$$