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A block of mass $m$ is pulled along a horizontal surface by applying a force at an angle $θ$ with the horizontal. If the block travels with a uniform velocity and has a displacement $d$ and the coefficient of friction is $µ$ , then the work done by the applied force is

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$\frac{μ m g d}{cosθ + sinθ}$

$\frac{μ m g d cosθ}{cosθ + μ sinθ}$

$\frac{μ m g d sinθ}{cosθ + μ sinθ}$

$\frac{μ m g d cosθ}{cosθ - μ sinθ}$

answer is B.

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Detailed Solution

$N = m g - F \sin θ$

Block moves with uniform velocity. Hence, $net force = 0$

$o r, F \cos θ = μ N = μ (m g - F \sin θ) ∴ F = \frac{μ m g}{cosθ + μ sinθ} W = F s \cos θ = \frac{μ m g d cosθ}{cosθ + μ sinθ}$

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