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A block of mass m slides down an inclined plane which makes an angle $θ$ with the horizontal. The coefficient of friction between the block and the plane is $μ$ . The force exerted by the block on the plane is

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$\frac{μ mgcos θ}{\sqrt{μ^{2} + 1}}$

$m g \cos θ$

$μ mgcos θ$

$(\sqrt{μ^{2} + 1}) m g \cos θ$

answer is B.

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Detailed Solution

$f_{k} = μ m g \cos θ, N = m g \cos θ,$

$R = \sqrt{f_{k}^{2} + N^{2}} = (\sqrt{μ^{2} + 1}) m g \cos θ$

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