When a $1.0\,kg$ mass hangs attached to a spring of length $50\,cm$,the spring stretches by $2\,cm$. The mass is pulled down until the length of the spring becomes $60\,cm$. What is the amount of elastic energy stored in the spring in this condition,if $g = 10\,m/s^2$? (in $Joule$)

$A$ spring of force constant $k$ is cut into two pieces such that one piece is three times the length of the other. The longer piece will have a force constant of

$A$ long spring is stretched by $2 \ cm$ and its potential energy is $U$. If the spring is stretched by $10 \ cm$,its potential energy will be

Two masses of $10 \, kg$ and $20 \, kg$ respectively are connected by a massless spring as shown in the figure. $A$ force of $200 \, N$ acts on the $20 \, kg$ mass. At the instant shown,the $10 \, kg$ mass has an acceleration of $12 \, m/s^2$ towards the right. The acceleration of the $20 \, kg$ mass at this instant is ........ $m/s^2$.

The length of a spring is $l$ and its force constant is $k$. When a weight $W$ is suspended from it,its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them,then the extension will be

To the free end of a spring hanging from a rigid support,a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly,the amount of maximum stretching is: (force constant,$k$)