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A 1. 5 kg block is pulled across a horizontal surface that has a coefficient of kinetic friction of 0. 60. A. What is the force of friction exerted by the surface on the block?.

Sagot :

LammettHashidn

Perpendicular to the surface, the net force on the block is

∑ F = F[normal] - (1.5 kg) g = 0

so that the normal force exerted by the surface on the block is

F[normal] = (1.5 kg) g = 14.7 N

Then the magnitude of kinetic friciton felt by the block is

0.60 (14.7 N) = 8.82 N ≈ 8.8 N

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