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 find the workdone by the applied force

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\frac{μ mgdcos θ}{(\cos θ + μ \sin θ)}

\frac{μ mgd}{(\cos θ + μ \sin θ)}

\frac{μ mgdsin θ}{(\cos θ + μ \sin θ)}

\frac{μ mgdcos θ}{(\cos θ - μ \sin θ)}

answer is B.

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Detailed Solution

$F_{net} = F \cos θ - f = 0$

$F \cos θ = f = μ N$

$Fcos θ = μ (mg - F \sin θ)$

$F = \frac{μ mg}{\cos θ + μ \sin θ}$

$∴ W = F \cos θ d$

$= \frac{μ mg \cos θ d}{\cos θ + μ \sin θ}$

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