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Question

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

A
μMgdsinθcosθ+μsinθ
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B
2μMgdcosθcosθ+μsinθ
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C
μMgdcosθcosθ+μsinθ
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D
μMgdcosθsinθ+μcosθ
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Solution

The correct option is C μMgdcosθcosθ+μsinθ

Forces on the block are
(i) Its weight Mg,
(ii) The normal force N
(iii) The applied force F and
(iv) The kinetic friction μN.
The forces are shown in figure. As the block moves with a uniform velocity, the forces add up to zero. Taking horizontal and vertical components,
Fcosθ=μNand Fsinθ+N=Mg
Eliminating N from these equations,
Fcosθ=μ(MgFsinθ)
or, F=μMgcosθ+μsinθ.
The work done by this force during a displacement d is
W=Fdcosθ=μMgdcosθcosθ+μsinθ

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