Asked by Mike Joseph
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
Find the friction force impeding its motion?
Can someone please take me through this problem step by step? Thank you so very much
Answers
Answered by
Elena
ma=mgsinα - F(fr)
F(fr) =m(gsinα –a)
F(fr) =m(gsinα –a)
Answered by
Henry
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.
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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