Work done on a certain spring when it is stretched to $1 \text{ mm}$ from its mean position is $10 \text{ J}$. The amount of work that must be done on the spring to stretch it further by $1 \text{ mm}$ is: (in $\text{ J}$)

$A$ body of mass $0.1 \ kg$ moving with a velocity of $10 \ m/s$ hits a spring (fixed at the other end) of force constant $1000 \ N/m$ and comes to rest after compressing the spring. The compression of the spring is .............. $m$.

The force constants of two springs are $K_1$ and $K_2$. Both are stretched until their elastic potential energies are equal. If the stretching forces applied are $F_1$ and $F_2$,then the ratio $F_1:F_2$ is:

If the potential energy of a spring is $V$ on stretching it by $2 \, cm$,then its potential energy when it is stretched by $10 \, cm$ will be

$A$ spring has a natural length $l$ with one end fixed to the ceiling. The other end is fitted with a smooth ring which can slide on a horizontal rod fixed at distance $l$ below the ceiling. Initially, the spring makes an angle of $60^{\circ}$ with the vertical, when the system is released from rest. Find the angle of the spring with the vertical, when the velocity of the ring reaches half of the maximum velocity, which the ring can attain during the motion.

As per the given figure,two blocks each of mass $250\,g$ are connected to a spring of spring constant $2\,N/m$. If both are given velocity $V$ in opposite directions,then the maximum elongation of the spring is: