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Question

A block placed on a horizontal surface is being pushed by a force F making an angle θ with the vertical. The coefficient of friction between block and surface is μ. The force required to slide the block with uniform velocity on the floor is

A
μmg(sinθμcosθ)
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B
(sinθμcosθ)μmg
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C
μmg
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D
None of these
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Solution

The correct option is A μmg(sinθμcosθ)
The given condition can be reframed as shown below in the figure

For vertical equilibrium of the block
N=Fcosθ+mg......(1)
As block will move with uniform velocity, the acceleration a will be zero.
Hence, for horizontal motion
FsinθμN=0
Fsinθ=μN....(2)

From Eqs. (1) and (2) we get
Fsinθ=μ(Fcosθ+mg)
i.e., F=μmg(sinθμcosθ)

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