The distance $x$ moved by a body of mass $0.5 kg$ by a force varies with time $t$ as

                                       $x=3t^2+4t+5$

where $x$ is expressed in meter and t in second. What is the work done by the force in the first $2 seconds$? 

Velocity (v) $=\frac{dx}{dt}=\frac{d}{dt}\left(3t^2+4t+5\right)=6t+4.$

Acceleration is a$=\frac{dv}{dt}=\frac{d}{dt}(6t+4)=6m{s}^{-2}.$

Therefore, applied force is$F= ma= 0.5\times 6= 3 N.$ Now $t = 2s$, the distance moved is $x = 3\times {\left(2\right)}^2 + 4\times 2 + 5 = 25 m$ $(\therefore u = 4m{s}^{-1} and x=5 m at t=0)$.

$\therefore$ Work done $W = Fx = 3 x 25 = 75 J$.