It takes 100J of work to stretch a spring 0.5m from its equilibrium position.How much work is needed to stretch it an additional 0.75m?

Question

$\begin{array}{1 1}525J\\625J\\725J\\325J\end{array} $

Answer 1

Answer : 525J

$100 =\int_0^{0.5} kxdx$

$\quad\;\;\;= \large\frac{1}{8}$$k$

$\Rightarrow k=800$

$W=\int\limits_{0.5}^{1.25} kxdx$

$\;\;\;\;=(400x^2)\bigg|_{0.5}^{1.25}$

$\;\;\;\;=525J$