JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Laws of Motion

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A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is

θ

. The magnitude of the contact force will be :

(A)Mg✓ Correct
(B) $Mg cos θ$
(C) $Mg sin θ + Mg cos θ$
(D) $Mg sin θ 1 + μ$

Answer: (A)

Solution

Since the block moves with constant velocity, it is in equilibrium. The normal force

N = M g cos θ

and friction force

f = M g sin θ

. The contact force (resultant of N and f) has magnitude

R = N^{2} + f^{2} = (M g cos θ)^{2} + (M g sin θ)^{2} = M g cos^{2} θ + sin^{2} θ = M g

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option

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