In the system shown in the figure the mass m moves in a circular arc of angular amplitude 60°. Mass 4m is stationary. Then

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the minimum value of coefficient of friction between the mass 4m and the surface of the table is 0.5

the work done by gravitational force on the block m is positive when it moves from A to B

the power delivered by the tension when m moves from A to B is zero

the kinetic energy of m in position B equals the work done by gravitational force on the block when it moves from position A to B

answer is A.

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Detailed Solution

Tension in the string will be maximum when mass m is at point B, soT=mg+mv2ℓBy Law of Conservation of Energy,12mv2=mgh=mgℓ(1−cos⁡θ)⇒v2=2gℓ1−cos⁡60∘⇒T=mg+2mgℓℓ1−12⇒T=2mgFor mass 4m to be stationary, f≥T⇒ μ(4mg)≥2mg⇒ μ≥0.5Hence, (1) is correctWhen m moves from A to B it has vertical displacement downwards, i.e., along the direction of gravitational forceSo, (2) is correct.Tension is always perpendicular to the velocity, so power delivered by the tension is zero. So, (3) is correct.According to Work Energy Theorem,Wtotal =ΔKSince work done by tension is zero therefore work done only by the gravitational force. So, (4) is also correct.

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