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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

Potential energy of certain spring when stretched through a distance of $x=10 J$. The amount of work that must be done on this spring to stretch it through an additional distance $x$ will be

\[(a)\;30\;J \quad (b)\;20\;J \quad (c)\;10\;J \quad (d)\;40\;J\]

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1 Answer

Given P.E to stretch throgh 'x' is
$\large\frac{1}{2}$$ k x^2=10\;J$
To stretch through additonal distance 'x'
$\large\frac{1}{2} $$k[(2s)^2-s^2]=3 \times \large\frac{1}{2} $$\times ks^2$
$\qquad=3 \times 10 J$
$\qquad=30\;J$
Hence a is the correct answer.
answered Jul 22, 2013 by meena.p
edited Jun 6, 2014 by lmohan717
 

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