$A$ body of mass $0.5\, kg$ travels on a straight-line path with velocity $v = (3x^2 + 4)\, m/s$. The net work done by the force during its displacement from $x = 0$ to $x = 2\, m$ is $.......J$.

$A$ particle is released from a height $H$. At a certain height $h$,its kinetic energy is two times its potential energy. The height $h$ and the speed $v$ of the particle at that instant are:

$A$ 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:

$A$ 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:

$A$ car moving with a velocity of $10 \, m/s$ can be stopped by the application of a constant force $F$ in a distance of $20 \, m$. If the velocity of the car is $30 \, m/s$,it can be stopped by this same force in what distance (in $, m$)?

The pulley and spring are massless,and friction is absent everywhere. $A$ $5\, kg$ block is released from rest. Find the speed of the $5\, kg$ block when the $2\, kg$ block leaves contact with the ground. (Take the force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2$.)