Q: Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x = t³/3, where x is in metre and t in second. The work done by the force in the first two seconds is

(a) 1600 J

(b) 160 J

(c) 16 J

(d) 1.6 J

Ans: (c)

Sol: x = t³/3

Differentiating with respect to time ,

$ \displaystyle \frac{dx}{dt} = \frac{d}{dt}(t^3/3) $

v = t^{2}

At , t=2 , v= 2×2 = 4 m/s

Work done W = ΔK.E

$ \displaystyle W = \frac{1}{2}mv^2 – \frac{1}{2}mu^2 $

$ \displaystyle W = \frac{1}{2}2\times (4^2) – 0 $ = 16 J