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You want to move a 41-kg bookcase to a different place in the living room. If you push with a force of 65 n and the bookcase accelerates at 0. 12 m/s2, what is the coefficient of kinetic friction between the bookcase and the carpet?.

Sagot :

LammettHashidn

The net force on the bookcase in the direction of its motion is

∑ F = 65 N - F[friction] = (41 kg) (0.12 m/s²)

so that friction has a magnitude of

F[friction] = 65 N - (41 kg) (0.12 m/s²) = 60.08 N

The net force on the bookcase perpendicular to the surface is

∑ F = F[normal] - (41 kg) g = 0

so that

F[normal] = (41 kg) g = 401.8 N

If µ is the coefficient of kinetic friction, then

F[friction] = µ F[normal]

so that

µ = F[friction] / F[normal] = (60.08 N) / (401.8 N) ≈ 0.15

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