Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = \frac{K}{v}$,where $K$ is a constant. The work done by this force in time $t$ is

$A$ body of mass $m$ moves in a circle of radius $r$ with a constant speed $V$. $A$ force of $mV^2/r$ acts towards the center of the circle. Find the work done by this force when the body covers half the circumference of the circle.

$A$ force $\overrightarrow{F} = 5\hat{i} + 6\hat{j} + 4\hat{k}$ acting on a body produces a displacement $\overrightarrow{S} = 6\hat{i} - 5\hat{k}$. The work done by the force is: (in $J$)

The figure shows three forces applied to a trunk that moves leftward by $3\, m$ over a smooth floor. The force magnitudes are $F_1 = 5\, N$,$F_2 = 9\, N$,and $F_3 = 3\, N$. The net work done on the trunk by the three forces is ............... $J$.

$A$ 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

$A$ block of mass $2 \,kg$ is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at $30^{\circ}$. Then the work done by the tension in the rope in pulling it a distance $4 \,m$ along the inclined plane in joule is (acceleration due to gravity $= 10 \,ms^{-2}$)