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Q.

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ  as shown in figure. The coefficient of kinetic friction is μk. Then, the block’s acceleration ‘a’ is given by:  (g is acceleration due to gravity)

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a

Fmcosθ−μKg−Fmsinθ

b

−Fmcosθ−μKg−Fmsinθ

c

Fmcosθ−μKg+Fmsinθ

d

Fmcosθ+μKg−Fmsinθ

answer is A.

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Detailed Solution

In vertical,  Fsinθ+N =mg ⇒N=mg-FsinθIn horizontal,  ma=Fcosθ -fk ⇒ma=Fcosθ -μkN ⇒a=Fcosθ−μkmg-Fsinθm
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