a = (V-Vo)/t = (0.37-0)/12 = 0.0308
(Fx-Fk) = M*a
(127*Cos40-Fk) = 35 * 0.0308
97.3 - Fk = 1.079
-Fk = 1.079 - 97.3 = -96.2
Fk = 96.2 N. = Force of kinetic friction.
A 35kg lawnmower starts from rest. Its pushed with a force of 127N at an angle 40 degrees below horizontal. After 12s it reaches speed of 0.37m/s.
Find acceleration and force of friction.
Attempts: Well I figured out Fnormal=343 my multiplying mass by gravity and assuming as it didn't move vertically the Fn=Fg. I also did 127cos40degrees to get 97.288 as the length on one side. I don't know how to get Ffriction without mue or how to get acceleration. I tried vf=d/t to get d=4.44 and was thinking i could sub that into kinematic equation d=vfxt - a/2 x t^2 and got .12 m/s/s. Pretty sure that I'm. Help's Appreciated.
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