JEE Main Physics question
A body of mass 1 kg is moving in a vertical circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest points is
Answer: (A)
Solution
(KE)_L - (KE)_H = \tfrac12 m(v_L^2-v_H^2) = \tfrac12 m(4gr) = 2mgr = 2\times1\times9.8\times1 \approx 20 J.