# physics

A 20.0 kg box slides down a 12.0 m long incline at an angle of 3.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is

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1. 593j

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2. its wrong!!!!!!

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3. M*g = 20 * 9.8 = 196 N. = Wt. of box.

Fp = 196*sin3 = 10.26 N. = Force parallel to the incline.

Fn = 196*Cos3 = 195.7 N. = Force perpendicular to the incline.

Fk = u*Fn = 0.1 * 195.7 = 19.57 N. = Force of kinetic friction.

a = (Fap-Fp-Fk)/M = (50-10.26-19.57)/20 =1.01 m/s^2

V^2 = Vo^2 + 2a*d = 0 + 2.02*12 = 24.24
V = 4.92 m/s.

KE = 0.5*20*4.92^2 = 242 J.

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