Asked by Michael
The coefficiant of friction between rubber tires and wet pavement is 0.5. Breaks are alllied to a 750kg car, intitally travelling at 30 m/s and the car skids to a stop. What is the magnitude and direction of the acceleration. Ik the magnitude and i kinda get how to do the question, but i dont understand why we have to use force of friction as Fnet
Answers
Answered by
Henry
M*g = 750 * 9.8 = 7350 N. = Wt. of car = Normal force(Fn).
Fp = 7350*sin 0o = 0 = Force parallel to the plane.
Fk = u*Fn = 0.5 * 7350 = 3675 N. = Force of kinetic friction.
Fp-Fk = M*a.
0-3675 = 750a, a = -4.9 m/s^2.
Or
u = a/g.
0.5 = a/-9.8, a = -4.9 m/s^2.
Fp = 7350*sin 0o = 0 = Force parallel to the plane.
Fk = u*Fn = 0.5 * 7350 = 3675 N. = Force of kinetic friction.
Fp-Fk = M*a.
0-3675 = 750a, a = -4.9 m/s^2.
Or
u = a/g.
0.5 = a/-9.8, a = -4.9 m/s^2.
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