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Q.

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 :

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a

$Mg \sin θ \sqrt{1 + μ}$

b

Mg

c

$\sqrt{Mg \sin θ + Mg \cos θ}$

d

$Mg \cos θ$

answer is A.

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Detailed Solution

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Question Image

$N = Mg \cos θ$

Since it is moving with a constant velocity

$f = Mg \sin θ$

$\begin{array}{l} R = \sqrt{N^{2} + f^{2}} \\ R = Mg \end{array}$

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