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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A block of mass $m$ initially at rest is dropped from a height $ h$ on to a spring of force constant $k$. The maximum compression in spring is $x$ ,then

\[(a)\;mgh=\frac{1}{2} kx^2 \quad (b)\;mg(h+x)=\frac{1}{2}kx^2 \quad (c)\;mgh=\frac{1}{2}k(x+h)^2 \quad (d)\;mg(h+x)=\frac{1}{2}k(x+h)^2\]

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Answer: $mg(h+x)=\frac{1}{2}kx^2$
The loss in potential energy of the mass =elastic potential energy stored in spring .
Let x be compression in spring
The mass falls through height (h+x)
$\Rightarrow mg(h+x)=\frac{1}{2}kx^2$
Hence b is the correct answer.
answered Jul 18, 2013 by meena.p
edited Aug 27, 2014 by balaji.thirumalai

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