Homework Help: Physics

Posted by Mike Joseph on Saturday, June 16, 2012 at 5:10pm.

A 13.0 -kg box is released on a 33 degree incline and accelerates down the incline at 0.20 m/s^2

Find the friction force impeding its motion?

Can someone please take me through this problem step by step? Thank you so very much

Physics - Elena, Sunday, June 17, 2012 at 7:51am
ma=mgsinα - F(fr)
F(fr) =m(gsinα –a)
Physics - Henry, Sunday, June 17, 2012 at 10:07pm
Wb = m*g = 13kg * 9.8N/kg = 127.4 N. =
Wt. of box.

Fb = 127.4N @ 33 Deg.
Fp = 127.4*s.n33 = 69.4 N. = Force parallel to the incline.
Fv = 127.4*cos33 = 106.8 N. = Force perpendicular to plane.

Fn = Fp - Fk = m*a
69.4 - Fk = 13*0.20 = 2.6
-Fk = 2.6 - 69.4 = -66.8
Fk = 66.8 = Force of kinetic energy.

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