$A$ body of mass $3\, kg$ is under a constant force which causes a displacement $s$ in metres in it,given by the relation $s = \frac{1}{3}t^2$,where $t$ is in seconds. Work done by the force in $2$ seconds is

When an object is dragged $10 \, m$ along a road,a frictional force of $200 \, N$ acts in the direction opposite to its motion. What is the work done by the frictional force on the object (in $, J$)?

$A$ body moves a distance of $10 \ m$ along a straight line under the action of a force of $5 \ N$. If the work done in this process is $25 \ J$,what is the angle between the force and the displacement in degrees?

$A$ body of mass $2 \ kg$ is moving along the $x$-direction such that its displacement as a function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma \ m$, where $\alpha = 1 \ m/s^2$, $\beta = 1 \ m/s$, and $\gamma = 1 \ m$. The work done on the body during the time interval $t = 2 \ s$ to $t = 3 \ s$ is . . . . . . $J$.

In the figure shown,all the surfaces are frictionless,and the mass of the block $m$ is $1\,kg$. The block and wedge are held initially at rest. Now,the wedge is given a horizontal acceleration of $5\,m/s^2$ by applying a force on the wedge,such that the block does not slip on the wedge. Then,the work done by the normal force in the ground frame on the block in $2\,sec$ is ............... $J$.

$A$ horizontal force of $10 \ N$ is applied on a block of mass $1.5 \ kg$ which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of $6 \ s$ from the beginning of the motion is (Acceleration due to gravity $= 10 \ m/s^2$; the coefficient of kinetic friction between the block and the surface is $0.2$). (in $J$)