$A$ simple pendulum bob of mass $m$ is raised to a maximum height $h$. Find the work done by the gravitational force as the pendulum moves from one extreme end to the other.

The kinetic energy acquired by a body of mass $m$ traveling a distance $s$,starting from rest under the action of a constant force,is directly proportional to:

$A$ force of $4 \,N$ acts on a $10 \,kg$ body initially at rest. Let $W_1$ be the work done by the force during $0 \leq t \leq 1 \,s$. Likewise, $W_2$ is the work done by the force during $1 \,s \leq t \leq 2 \,s$, where $t$ is time in seconds. The ratio $\frac{W_2}{W_1}$ is

$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec{F} = (3\hat{i} + 4\hat{j})$ through two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

$A$ body,acted upon by a force of $50 \, N$,is displaced through a distance of $10 \, m$ in a direction making an angle of $60^\circ$ with the force. The work done by the force is .......... $J$.

$A$ particle moves in the $x-y$ plane under the action of a force,$F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$,where $K$ is a constant. The work done by the force when the particle moves from $(0, a)$ to $(a, 0)$ along a circular path of radius $a$ about the origin is: