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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 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
W=μMg dcosθcosθ+μsinθ
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B
W=μMg dcosθsinθ+μcosθ
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C
W=μMg dsinθsinθ+μcosθ
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D
W=μMg dsinθsinθ+μcosθ
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Solution

The correct option is A W=μMg dcosθcosθ+μsinθ


Resolving forces in vertical direction

N=MgFsinθ(1)
Resolving forces in horizontal direction:
Fcosθ=fk=μN(2)
From eq. (1) and (2)
Fcosθ=μ[MgFsinθ]
Fcosθ+μFsinθ=μMg
F=μMgcosθ+μsinθ
As work done W=Fdcosθ
substituting the value of F in above equation we get,
W=μMg dcosθcosθ+μsinθ

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