A body of mass 2 kg is moving in the xy -plane such that its position vector as a function of time is given by r (t) = 2t^2 i + t^3 j (where r is in meters and t is in seconds). The work done on the body during the time interval from t = 1 s to t = 2 s is

Step-by-step solution

The velocity vector is the time derivative of the position vector: v (t) = d r dt = 4t i + 3t^2 j At t = 1 s, the initial velocity vector is: v ₁ = 4(1) i + 3(1)^2 j = 4 i + 3 j m/s The square of its magnitude is: v₁^2 = 4^2 + 3^2 = 16 + 9 = 25 m ^2 /s ^2 At t = 2 s, the final velocity vector is: v ₂ = 4(2) i + 3(2)^2 j = 8 i + 12 j m/s The square of its magnitude is: v₂^2 = 8^2 + 12^2 = 64 + 144 = 208 m ^2 /s ^2 Using the work-energy theorem, the total work done is equal to the change in kinetic energy: W = KE = 1