$A$ pendulum bob on a $2\, m$ string is displaced $60^\circ$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path? (Take $g = 9.8\, m/s^2$)

$A$ body falls freely under gravity. Its speed is $v$ when it has lost an amount $U$ of the gravitational potential energy. Then its mass is

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

$A$ bullet of mass $10 \; g$ leaves a rifle at an initial velocity of $1000 \; m/s$ and strikes the earth at the same level with a velocity of $500 \; m/s$. The work done in joule $(J)$ in overcoming the resistance of air will be:

$A$ body of mass $2 \,kg$ moving in $X-Y$ plane has a potential energy given by $U = (6x + 8y) \,J$. The body is at rest at the point $(3, 2) \,m$. The work done by the body to reach another position after $2 \,s$ is: (in $\,J$)

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 $x = \frac{t^2}{3}$,where $x$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is .......... $J$.