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Q.

A block of mass m, 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 the surface is µ. The force required to slide the block with uniform velocity on the floor is

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a

μmg(sin⁡θ−μcos⁡θ)

b

(sin⁡θ−μcos⁡θ)μmg

c

μmg(sin⁡θ+μcos⁡θ)

d

(sin⁡θ+μcos⁡θ)μmg

answer is C.

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Detailed Solution

See fig.For vertical equilibriumR=Fcos⁡θ+mg …(i)For horizontal motion,Fsin⁡θ−μR=maFsin⁡θ=μR ...(ii)Substituting the value of R from eq. (i) in eq. (ii), we get Fsin⁡θ=μ[Fcos⁡θ+mg]or Fsin⁡θ+μFcos⁡θ=μmg∴ F=μmg(sin⁡θ+μcos⁡θ).

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