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in Physics by (70.8k points)

As shown in the accompanying figure, a force F is exerted at an angle of θ. The block of weight mg is initially moving the right with speed v. The coefficient of friction between the rough floor and the block is µ. The frictional force acting on the block is:

(A) µmg to the left. 

(B) µmg to the right. 

(C) µmg – F sinθ to the left. 

(D) µ(mg – F cosθ) to the right. 

(E) µ(mg + F sinθ) to the left. 

1 Answer

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by (65.3k points)
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Best answer

Correct option (E) The reaction exists only after the action force is removed.

Friction opposes the motion of the block and therefore points to the left. The normal force is found from ΣFy = 0 = FN – mg – Fsinθ and the force of friction Ff = µFN

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