A vertical spring with force constant k is fi | vedclass

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Question

$\begin{array}{l}
\,\,\,\,When\,a\,mass\,falls\,on\,a\,spring\,from\
a\,height\,h\,the\,work\,done\,by\,the\,loss\,of\
gravitational\,potential\,

Vedclass · Accepted Answer

$mg\left( {h + d} \right) - \frac{1}{2}k{d^2}$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,When\,a\,mass\,falls\,on\,a\,spring\,from\
a\,height\,h\,the\,work\,done\,by\,the\,loss\,of\
gravitational\,potential\,energy\,of\,the\
mass\,is\,stored\,as\,the\,potential\,energy\
of\,the\,spring.\
One\,can\,write\,mg\left( {h + d} ight) = \frac{1}{2}k{d^2}\
mg\left( {h + d} ight) = \frac{1}{2}k{x^2} = \frac{1}{2}k{d^2}
\end{array}$

$\begin{array}{l}
The\,two\,energies\,are\,equal.\
If\,work\,done\,is\,initial\,P.E - final\
P.E.,\,it\,is\,zero;\
Work\,done\,is\,totally\,converted\,(a{m{ssuming}}\
{m{there}}\,{m{is}}\,{m{no}}\,{m{loss}}{m{.)}}\,The\,work\,done\,in\,\
compression\,or\,\exp ansion\,is\,always\
positive\,as\,it\,is\, \propto \,{x^2}.\,The\,answer\,\exp ected\
is\
mg\,\left( {h + d} ight) - \frac{1}{2}k{d^2}\,or,\,\,\frac{1}{2}k{d^2} - mg\left( {h + d} ight)
\end{array}$

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

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