A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation s=1/3t², where t is in seconds. Work done by the force in 2 seconds is

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(4) 8/3 J. Given: m=3 kg and s=1/3t², with the force constant; find the work over the first 2 s. Idea: differentiate twice for the acceleration, get the force, and multiply by the distance covered. v=(ds)/(dt)=2/3t and a=(d²s)/(dt²)=2/3 m s⁻², indeed constant. F=ma=3×2/3=2 N. Displacement in 2 s: s=1/3(2)²=4/3 m. W=Fs=2×4/3=8/3 J. Check by kinetic energy: v(2)=4/3 m s⁻¹, so 1/2(3)(4/3)²=8/3 J.

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation s=1/3t², where t is in seconds. Work done by the force in 2 seconds is

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