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Question

# A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. Coefficient of friction between block and surface is μ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.

A
μMgdcosθcosθ+μsinθ
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B
μMgdsinθcosθ+μsinθ
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C
Mgdcosθcosθ+μsinθ
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D
Mgdsinθcosθ+μsinθ
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Solution

## The correct option is A μMgdcosθcosθ+μsinθThe forces acting on the block are shown in the figure, As the block moves with uniform velocity the resultant force on it is zero. ∴Fcosθ=μN .........(1) Applying equation of equilibrium in vertical direction Fsinθ+N=Mg ........(2) Eliminating N from equations (1) and (2), Fcosθ=μ(Mg−Fsinθ) Fcosθ+μFsinθ=μMg F=μMgcosθ+μsinθ Work done by this force during a horizontal displacement d will be W=(Fcosθ)d=μMgdcosθcosθ+μsinθ Therefore, option (a) is correct.

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