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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A spring is stretched 60 cm, and it takes a force of 40 Newtons to hold it there. How much work did it take to stretch the spring to this point?

$\begin{array}{1 1}\text{12 Joules} \\\text{10 Joules}\\\text{15 Joules}\\\text{25 Joules}\end{array} $

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Answer : 12 Joules
We know that F = 40 N
when $x = 60 cm = 0.60$ m.
Since $F = kx$
$k = F/x = (40 N)/(0.60 m) = 67 N/m$
Now, work = $(\large\frac{1}{2})$$kx^2 $
$\Rightarrow \large\frac{1}{2}$$(67N/m)(0.60m)^2$
$\Rightarrow 12$ Joules
answered Sep 2, 2014 by sreemathi.v

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