$A$ small block of mass $m$ slides along a smooth track as shown in the figure. If it starts from rest at point $P$,what is the velocity of the block at point $Q$?

$A$ particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$ will be:

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

The total work done on a particle is equal to the change in its kinetic energy. This is applicable:

$A$ 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

For a body of mass $1 \ kg$,the $U-x$ graph is shown in the figure. If the body is released from rest at $x = 2 \ m$,then its speed when it crosses $x = 5 \ m$ is: