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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
+1 vote

A spring has a natural length of 14 meters.If a force of 5N is required to keep the spring stretched 2 meters.How much work is done in stretching the spring from its natural length to a length of 18 meters.

$\begin{array}{1 1}20Joules\\25Joules\\40Joules\\34Joules\end{array} $

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1 Answer

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Answer : 20 Joules
We must first find the spring constant $k$.
By Hooke's law,$F(x) =kx$
Thus,$F(2) =k(2)=5$ and so we can calculate k as follows
$k=\large\frac{F(2)}{2}$
$\;\;\;=\large\frac{5N}{2m}$
$\;\;\;=2.5N/m$
Therefore,$F(x)=2.5x$ and so the work done by the force is
Work done = $\int\limits_0^4 2.5xdx$
$\qquad\qquad\;=\big[\large\frac{5}{4}$$x^2\big]_0^4$
$\qquad\qquad\;=20$ Joules
answered Sep 1, 2014 by sreemathi.v
 

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