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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Suppose 2 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 42 cm.Then how far beyond its natural length will a force of 30 N keep the spring stretched?

$\begin{array}{1 1}\large\frac{27}{10}\\\large\frac{29}{10}\\27\\\large\frac{17}{10}\end{array} $

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Answer : $\large\frac{27}{10}$
From Hooke's law : $F=kx$ where $x$ is the distance the sprin is stretched from its natural length,and k is the (usually unknown) spring constant.
The spring's length is 30cm,so it is stretched 12cm from its natural length,we can use
$W=\int dW$
$\;\;\;\;=\int Fdx$
$2J=\int_0^{.12m} kxdx$
$\quad\;=\large\frac{k}{2}$$x^2\bigg|_0^{.12m}$
$\quad\;=\large\frac{k}{2}$$(0.0144m^2)$
$k =\large\frac{4Nm}{0.0144m^2}$
$\;\;\;=\large\frac{100}{9}$N/m
Once we have $k$,we can calculate the distance,since we know that $F=30N$ :
$F=kx$
$30=\large\frac{100}{9}$$x$
$x=\large\frac{27}{10}$
answered Sep 2, 2014 by sreemathi.v
 

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