$A$ body of mass $0.15 \ kg$ moving with a velocity of $15 \ ms^{-1}$ comes to rest when it hits a spring that is fixed at the other end. If the force constant of the spring is $1500 \ Nm^{-1}$,then the compression in the spring is: (in $m$)

When a spring is stretched by $10 \ cm$, the potential energy stored is $E$. When the spring is stretched by $10 \ cm$ more, the potential energy stored in the spring becomes (in $E$)

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 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 $J$)

$A$ spring of spring constant $5 \times 10^3 \, N/m$ is stretched initially by $5 \, cm$ from the unstretched position. Then the work required to stretch it further by another $5 \, cm$ is .............. $J$.

$A$ block of mass $m$ is pushed against a spring with spring constant $k$ which is attached to a wall. The block slides on a frictionless table as shown in the figure. The natural length of the spring is $\ell_0$ and it is compressed to half of its natural length when the block is released. What will be the final velocity of the block?