Physics - NLM Question with Solution | TestHub
PhysicsNLMSimple Application of NLMMedium2 minQB
PhysicsMediumsingle choice
Passage / Comprehension
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.
What is the initial acceleration of the block of mass ?
Options:
Answer:
A
Solution:
Initially the spring is undeformed, so it exerts no force. For block m: F - T = ma. For the hanging block m/4 (which rises as m moves toward F): T - (m/4)g = (m/4)a. Adding: mg/2 - mg/4 = (5m/4) a, so a = g/5. Answer: (A).
Stream:JEESubject:PhysicsTopic:NLMSubtopic:Simple Application of NLM
⏱ 2mℹ️ Source: QB
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