Posted by Mike Joseph on .
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
ma=mgsinα - F(fr)
F(fr) =m(gsinα –a)
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.