99 Percentile Qs Bank for JEE MainPhysicsWork, Power and Energy
One end of a spring of natural length 2 m and spring constant k = 100 N/m is fixed at the ground and the other is fitted with a smooth ring of mass 1 kg which is allowed to slide on a horizontal rod fixed at a height 2 m (diagram). Initially, the spring makes an angle of 37 o with the vertical when the system is released from rest. Find the speed (in m/s) of the ring when the spring becomes vertical. ( Take sin 37 o
Correct answer
5
Step-by-step solution
tan 3 7 ∘ = l h , 3 4 = l h , l = 3 h 4 x = extension in spring = h 2 + l 2 - h = 9 h 2 + 1 6 h 2 1 6 - h = h 4 1 2 kx 2 = 1 2 mv 2 [By work energy theorem] ⇒ K m h 4 2 = v 2 ⇒ v = h 4 k m = 5 m/s