$A$ small sphere of mass $m$ and radius $r$ slides down the smooth surface of a large hemispherical bowl of radius $R$. If the sphere starts sliding from rest, the total kinetic energy of the sphere at the lowest point $A$ of the bowl will be [given, moment of inertia of sphere $= \frac{2}{5} mr^2$].

$A$ block of mass '$m$' (as shown in the figure) moving with kinetic energy $E$ compresses a spring through a distance $25\,cm$ when its speed is halved. The value of the spring constant of the used spring will be $nE\;N\,m^{-1}$ for $n=$

$A$ child is swinging on a swing. The minimum and maximum heights of the swing from the earth's surface are $0.75\,m$ and $2\,m$ respectively. The maximum velocity of this swing is ............. $m/s$.

Consider an elliptically shaped rail $PQ$ in the vertical plane with $OP = 3 \ m$ and $OQ = 4 \ m$. $A$ block of mass $1 \ kg$ is pulled along the rail from $P$ to $Q$ with a force of $18 \ N$,which is always parallel to the line $PQ$ (see the figure). Assuming no frictional losses,the kinetic energy of the block when it reaches $Q$ is $(n \times 10) \ J$. The value of $n$ is (take acceleration due to gravity $g = 10 \ m/s^2$):

$A$ bullet when fired at a target with a velocity of $100\,m/s$ penetrates $1\,m$ into it. If the bullet is fired at a similar target with a thickness of $0.5\,m$,then it will emerge from it with a velocity of:

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