A 27.0-kg block is initially at rest on a horizontal surface. A horizontal force of 70.0 N is required to set the block in motion, after which a horizontal force of 64.0 N is required to keep the block moving with constant speed. Find (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the block and the surface.

(a) The block has a weight of (27.0 kg) g, and the normal force of the surface pushing upward on the block has the same magnitude, so that static friction exerts a maximum force of

µ (27.0 kg) g = 70.0 N

where µ is the coefficient of static friction. Solving for µ gives

µ = (70.0 N) / ((27.0 kg) g) ≈ 0.265

(b) As it's moving, the block still has the same weight and thus feels the same normal force, (27.0 kg) g. In order to move at a constant speed, kinetic friction must exert the same force as the push, so

µ (27.0 kg) g = 64.0 N

where µ is now the coefficient of kinetic friction. Solve for µ :

µ = (64.0 N) / ((27.0 kg) g) ≈ 0.242

the answer is a. hope this .

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