Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

$A$ body of mass $6 \,kg$ is acted upon by a force which causes a displacement in it given by $x = \frac{t^2}{4} \,m$, where $t$ is the time in seconds. The work done by the force in $2 \,seconds$ is: (in $\,J$)

$A$ simple pendulum is released from point $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at point $B$ is:

$A$ particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface,then the speed of the particle is

The mechanical energy of a freely falling body is......

$A$ 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 $= \frac{2}{5} mr^2$].