Q.
A 0.098 kg block slides down a frictionless track as shown. The vertical component of the velocity of block at $A$ is
A) $\sqrt{g}$
B) $2\sqrt{g}$
C) $3\sqrt{g}$
D) $4\sqrt{g}$
Explanation
(a) $\frac{1}{2}mv^{2} = mg(3 - 1) = 2mg$ or $v = \sqrt{4}g = 2\sqrt{g}$
Vertical component at $A = 2\sqrt{g}\sin{30{^\circ}} = \sqrt{g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.