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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.

Sagot :

LammettHash

(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

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