A time dependent force $$F = 6t$$   acts on a particle of mass $$1 \,kg.$$  If the particle starts from rest, the work done by the force during the first $$1 \,second$$   will be-

Solution :
$$\eqalign{ & {\text{Using, }}F = ma = m\frac{{dV}}{{dt}} \cr & 6t = 1\,.\,\frac{{dV}}{{dt}}\,\,\,\,\left[ {\because \,\,\,m = 1{\text{ kg given}}} \right] \cr & \int\limits_0^v {dV} = \int {6t} \,dt \cr & V = 6\left[ {\frac{{{t^2}}}{2}} \right]_0^1 = 3\,m{s^{ - 1}}\,\,\,\,\,\left[ {\because \,\,\,t = 1{\text{ sec given}}} \right] \cr} $$
From work-energy theorem,
$$W = \Delta KE = \frac{1}{2}m\left( {{V^2} - {u^2}} \right) = \frac{1}{2} \times 1 \times 9 = 4.5\,J$$