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A block of mass m is pulled slowly by a minimum constant force on a horizontal surface through a distance x. The coefficient of kinetic friction is $μ$ . Find the work done by force F.

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$\frac{μ m g x}{1 + μ^{2}}$

$\frac{μ m g x}{1 - μ^{2}}$

$\frac{μ m g x}{\sqrt{1 + μ^{2}}}$

$\frac{μ m g x}{2 + μ^{2}}$

answer is A.

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Detailed Solution

The work done by the force F is

$W = F_{\min} x \cos θ$

$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}, \sin θ \frac{μ}{\sqrt{1 + μ^{2}}}$

F is minimum when $\tan θ = μ$

$F_{\min} = m g \sin θ$

Using equation (i) and (iii),

$W = (m g \sin θ) \times x \cos θ$

$= m g . \frac{μ}{\sqrt{1 + μ^{2}}} \times \frac{1}{\sqrt{1 + μ^{2}}} = \frac{μ m g x}{1 + μ^{2}}$

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