Q. 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 point is
A) 20 J
B) 10 J
C) 4$\sqrt{5}$ J
D) 10$\sqrt{5}$ J
Explanation
(a) $(KE)_{L} - (KE)_{H} = \frac{1}{2}m(v_{L}^{2} - v_{H}^{2})$
$= \frac{1}{2}m(5\ gr - gr) = 2mgr$
$= 2 \times 1 \times 10 \times 1$=20 J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.