$A$ bullet moving with a speed of $100 \ m/s$ can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be

$A$ 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 $n=$

$A$ body is thrown vertically upwards and reaches a maximum height $h$. What is the ratio of its kinetic energy to its potential energy at a height of $\frac{3h}{4}$?

$A$ block of mass $m$ is sliding down an inclined plane with constant speed. At a certain instant $t_0$,its height above the ground is $h$. The coefficient of kinetic friction between the block and the plane is $\mu$. If the block reaches the ground at a later instant $t_g$,then the energy dissipated by friction in the time interval $(t_g - t_0)$ is

$A$ bead of mass $\frac{1}{2} \, kg$ starts from rest from point $A$ to move in a vertical plane along a smooth fixed quarter ring of radius $5 \, m$,under the action of a constant horizontal force $F = 5 \, N$ as shown. The speed of the bead as it reaches point $B$ is ................ $m/s$. [Take $g = 10 \, m/s^2$]

$A$ mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is