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A wedge of mass m is pushed with a speed v₀ on a rough horizontal plane. The angle of friction between the wedge and horizontal plane is $ϕ$ . The angle of inclination $θ$ of the pendulum is

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90∘−ϕ

tan−1⁡mMtan⁡ϕ

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detailed solution

Correct option is A

Since frictional force acts on the block, its deceleration is μg. So in the frame of wedge, a pseudo force μmg acts on the block and it is inclined at angle θ=ϕ with vertical.

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A wedge of mass m is pushed with a speed v0 on a rough horizontal plane. The angle of friction between the wedge and horizontal plane is ϕ. The angle of inclination θ of the pendulum is

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A wedge of mass m is pushed with a speed v₀ on a rough horizontal plane. The angle of friction between the wedge and horizontal plane is $ϕ$ . The angle of inclination $θ$ of the pendulum is