$A$ particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

$A$ 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

$A$ 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? $(g = 9.8 \, m/s^2)$

$A$ body moves from rest with a constant acceleration. Which one of the following graphs represents the variation of its kinetic energy $K$ with the distance travelled $x$?

Explain the work-energy theorem.

$A$ particle of mass $m \ kg$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to