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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A spring whose natural length is 15 cm exerts a force of 45 N when stretched to a length of 20 cm.Find the work done in stretching the spring from a length of 20 cm to a length of 25 cm

$\begin{array}{1 1}3.375J\\2.375J\\4.375J\\3.415J\end{array} $

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Answer : 3.375J
The change of length is 5cm =.05m.Since $F=kx$ we have
$45=k(.05)$
$k=\large\frac{4500}{5}$
$\;\;\;=900$ N/m
The work done is
$W=900\int\limits_{0.05}^{.1} xdx$
$\;\;\;\;=\large\frac{900x^2}{2}\bigg]_{0.05}^{0.1}$
$\;\;\;\;=450[.1^2-.05^2]$
$\;\;\;\;=3.375J$

 

answered Sep 2, 2014 by sreemathi.v
edited Sep 2, 2014 by balaji.thirumalai
 

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