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# What is the coefficient of friction between the surface and the box?

A 44.3kg box is sliding down a 40.0 degrees incline. If the acceleration of the box is 0.250m/s^2, what is the coefficient of friction between the surface and the box?

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- billrussell42Lv 74 weeks agoFavourite answer
F = ma = 44.3 kg x 0.25 m/s² = 177.2 N

weight = 44.3 kg x 9.8 N/kg = 434.14

component of that parallel to surface is 434.14 sin 40 = 279 N, this causes a force down the slope. This force minus friction causes acceleration.

component of that normal to surface is 434.14 cos 40 = 332.6 N, this causes a force of friction.

net force down the slope = 279 – µ332.6 = 177.2

µ332.6 = 101.8

µ = 0.306

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