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A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d.$ The net work done in the process is
$mg\left( {h + d} \right) - \frac{1}{2}k{d^2}$
$\;mg\left( {h - d} \right) - \frac{1}{2}k{d^2}$
$\;mg\left( {h - d} \right) + \frac{1}{2}k{d^2}$
$\;mg\left( {h + d} \right) + \frac{1}{2}k{d^2}$
Solution
$\begin{array}{l}
\,\,\,\,When\,a\,mass\,falls\,on\,a\,spring\,from\\
a\,height\,h\,the\,work\,done\,by\,the\,loss\,of\\
gravitational\,potential\,energy\,of\,the\\
mass\,is\,stored\,as\,the\,potential\,energy\\
of\,the\,spring.\\
One\,can\,write\,mg\left( {h + d} \right) = \frac{1}{2}k{d^2}\\
mg\left( {h + d} \right) = \frac{1}{2}k{x^2} = \frac{1}{2}k{d^2}
\end{array}$
$\begin{array}{l}
The\,two\,energies\,are\,equal.\\
If\,work\,done\,is\,initial\,P.E – final\\
P.E.,\,it\,is\,zero;\\
Work\,done\,is\,totally\,converted\,(a{\rm{ssuming}}\\
{\rm{there}}\,{\rm{is}}\,{\rm{no}}\,{\rm{loss}}{\rm{.)}}\,The\,work\,done\,in\,\\
compression\,or\,\exp ansion\,is\,always\\
positive\,as\,it\,is\, \propto \,{x^2}.\,The\,answer\,\exp ected\\
is\\
mg\,\left( {h + d} \right) – \frac{1}{2}k{d^2}\,or,\,\,\frac{1}{2}k{d^2} – mg\left( {h + d} \right)
\end{array}$
