$A$ force acts on a $3 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in metres and $t$ is in seconds. The work done during the first $4 \,s$ is ..... $mJ$.

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

What is the difference between work done on a body by a force and work done by a body?

Three different objects of masses $m_1, m_2$ and $m_3$ are allowed to fall from rest and from the same point $O$ along three different frictionless paths. The speeds of the three objects,on reaching the ground,will be in the ratio of

$A$ bullet moving with a speed of $100 \ m/s$ can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be