$A$ 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 $P$ slides on a frictionless semi-circular wire $(ACB)$. It is at point $S$ at $t = 0$, and at this instant, the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal wire $AB$ with speed $v$. Friction between the beads and the respective wires may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach point $B$. Then the relation between $t_P$ and $t_Q$ is:

$A$ uniform flexible chain of mass $m$ and length $2l$ hangs in equilibrium over a smooth horizontal pin of negligible diameter. One end of the chain is given a small vertical displacement so that the chain slips over the pin. The speed of the chain when it leaves the pin is

$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

$A$ truck accelerates from speed $v$ to $2v$. The work done during this process is:

$A$ body of mass $5 \ kg$ is placed on a horizontal surface with a coefficient of friction $\mu = 0.2$. It is pulled by a horizontal force of $25 \ N$ for a distance of $10 \ m$. The kinetic energy gained by the body is ..... $J$. (Take $g = 10 \ m/s^2$)