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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

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?

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

1 Answer

Answer : 525J
$100 =\int_0^{0.5} kxdx$
$\quad\;\;\;= \large\frac{1}{8}$$k$
$\Rightarrow k=800$
$W=\int\limits_{0.5}^{1.25} kxdx$
$\;\;\;\;=(400x^2)\bigg|_{0.5}^{1.25}$
$\;\;\;\;=525J$
answered Sep 2, 2014 by sreemathi.v
 

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