$A$ knife of mass $m$ is at a height $x$ from a large wooden block. The knife is allowed to fall freely,strikes the block and comes to rest after penetrating distance $y$. The work done by the wooden block to stop the knife is ..............

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying a retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?

$A$ small block slides down from rest at point $A$ on the surface of a smooth cylinder,as shown. At point $B$,the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ until the mass comes to rest,as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $A$,the net work done on the block is

The adjoining diagram shows the velocity versus time plot for a particle. The work done by the force on the particle is positive from