$A$ spring of spring constant $200 \, Nm^{-1}$ is initially stretched by $10 \, cm$ from the unstretched position. The work to be done to stretch the spring further by another $10 \, cm$ is (in $J$)

In the figure,the ball $A$ is released from rest when the spring is at its natural (unstretched) length. For the block $B$ of mass $M$ to leave contact with the ground at some stage,the minimum mass of $A$ must be

$A$ spring is stretched by a distance $s$,and the potential energy stored in it is $10 \ J$. What is the work done to stretch the spring by an additional distance $s$ (in $J$)?

An elastic spring of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is

$A$ ring of mass $m$ is attached to a horizontal spring of spring constant $k$ and natural length $l_0$. The other end of the spring is fixed,and the ring can slide on a smooth horizontal rod as shown. Now,the ring is shifted to position $B$ and released. The speed of the ring when the spring attains its natural length is:

$A$ mass of $1 \ kg$ falls from a height of $1 \ m$ and lands on a massless platform supported by a spring having spring constant $15 \ N \ m^{-1}$ as shown in the figure. The maximum compression of the spring is. (acceleration due to gravity $= 10 \ m \ s^{-2}$)