Physics - W.E.P. Question with Solution | TestHub
A constant horizontal force F = mg/2 starts acting on the block of mass m from the position shown. The spring is undeformed in the position shown and has a natural length L, while the blocks are initially stationary. The spring constant is k. The floor is frictionless. The mass of the hanging block is m/4, while the pulley is massless and frictionless. Assume that the block never leaves contact with the surface.
The maximum displacement of the heavier block is found to be . Based on this information, find the spring constant :-
Options:
Answer:
Solution:
The spring is vertical with natural length L, anchored directly above the block's initial position. When the block shifts by x = sqrt(3) L, the spring length becomes sqrt(L^2 + 3L^2) = 2L, i.e. the stretch is L. Energy from start to maximum displacement (both blocks at rest at the ends): F x = (m/4) g x + (1/2) k L^2. So (mg/2 - mg/4)(sqrt(3) L) = (1/2) k L^2, giving k = (sqrt(3)/2)(mg/L). Answer: (B).
