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 .
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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
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