As shown in the figure,a block of mass $2 \ kg$ is released from point $P$. It slides $0.5 \ m$ on the inclined plane before hitting a spring. The spring constant is $4000 \ N/m$. The coefficient of friction between the block and the inclined plane is $0.3$. The compression in the spring will be ............... $mm$.

$A$ ball is falling freely from a height. When it reaches $10 ~m$ height from the ground, its velocity is $v_0$. It collides with the ground, loses $50 \%$ of its energy, and rises back to a height of $10 ~m$. Then the velocity $v_0$ is (in $~m / s$)

$A$ spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \ N \ m^{-1}$ and the mass of the block is $2.0 \ kg$. Ignore the mass of the spring. Initially,the spring is in an unstretched condition. Another block of mass $1.0 \ kg$ moving with a speed of $2.0 \ m \ s^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \ kg$ block does not hit the wall. The distance,in metres,between the two blocks when the spring returns to its unstretched position for the first time after the collision is . . . . .

$A$ machine which is $70 \%$ efficient raises a $10 \,kg$ body through a certain distance and spends $100 \,J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

$A$ small disc of mass $m = 1 \,g$ slides down a smooth hill of height $h = 10 \,cm$ from rest and gets onto a plank of mass $M = 100 \,g$ as shown in the figure. Due to friction between the disc and the plank, the disc slows down and moves as one piece with the plank. The work done by the frictional force is approximately (Use $g = 10 \,m/s^2$): (in $\,J$)

$A$ raindrop of mass $1.00 \, g$ falling from a height of $1 \, km$ hits the ground with a speed of $50 \, m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$? If not,why?
Take $g = 10 \, m s^{-2}$.