JEE Main Physics question
Under the action of a force, a 2 kg body moves such that its position x as a function of time t 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
Answer: (B)
Solution
v = dx/dt = t². At t=0, v=0; at t=2 s, v=4 ms⁻¹. By the work-energy theorem, W = ΔKE = ½(2)(4²−0²) = 16 J.