Question

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 =(t^(3))/(3) where x is in metre and t in second. The work done by the force in the first two seconds is .

Answer 1

Given :-

Mass of body = m = 2 Kg

Position of body = x = t³/3

Differentiating the position w.r.t 't' to get Velocity of the body.

v = dx/dt

v = d(t³/3)/dt

v = t²

Velocity of body at, t = 0 s

u = 0² = 0

Velocity of body at, t = 2s

v = 2² = 4 ms-¹

From Work-Energy Theorem,

W = ∆K.E.

W = 1/2 mv² - 1/2 mu²

W = 1/2 × 2 × 4 × 4 - 1/2 × 2 × 0 × 0

W = 4 × 4

W = 16 J

Hence,

The work done in first two seconds = W = 16 J