Question bank/JEE Main/Physics
JEE MainPhysicsModerateNewton's Laws of Motion
Initially spring is in its natural length and blocks A and B are
at rest. Find maximum value of constant force F that can be applied on B
such that block A remains at rest. (g = 10 ms-2).
Show answer and solution
Value of for A and B are 2N and 2N.
Work energy theorem on B gives
Fx - 2x - (k/2) x^2 = 0
F = 2 + kx/2....................(i)
For equilibrium of A, kx =
So, F = 2 + 1 = 3N
Work energy theorem on B gives
Fx - 2x - (k/2) x^2 = 0
F = 2 + kx/2....................(i)
For equilibrium of A, kx =
So, F = 2 + 1 = 3N
Source: Grass-G / P-16 Physics paper 25 Nov First.docx
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