$A$ body of mass $m$ slides down a smooth hemispherical surface of radius $r$. What will be its velocity at the bottom of the surface?

$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

The work-energy theorem is valid in the presence of:

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

$A$ 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:

$A$ force acts on a $2\,kg$ object so that its position is given as a function of time as $x = 3t^2 + 5$. What is the work done by this force in the first $5\,seconds$? ................ $J$