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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} $
jeemain
physics
class11
work-power-energy
ch6
work-done-by-a-spring
hookes-law
easy
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asked
Sep 2, 2014
by
sreemathi.v
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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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