A 15.0 kg box is released on a 32 degree incline and accelerates down the incline at 0.30m/s sqaured ? find the friction force impeding its motion. What is the coefficient of kinectic friction?

Answer 1

Fb = 15kg * 9.8 = 147N @ 32deg = Force of box.

Fp = 147*sin32 = 77.9N = Force parallel to the plane.

Fv = 147*cos32 = 124.7N = Force perpendicular to the plane.

Fp - uFv = ma,
77.9 - 124.7u = 15 * 0.30,
77.9 - 124.7u = 4.5,
-124.7u = 4.5 - 77.9 = - 73.4,
u = -73.4 /- 124.7 = 0.59 = coefficient of friction.

Ff=uFv = 0.59 * 124.7N = 73.6 = Force of friction.

Answer 2

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