Recent content by Syndrix

Ah, Thank you so much for your help. This will help me a lot with my studies.

So if Force friction is the only force, then; F(f) = F(net) F(f) = ma F(f) = a M 3678 = a 750 a = 4.9m/s

Wouldn't the breaks act as an applied force?

Homework Statement The coefficient of kinetic friction between rubber tires and wet pavement is 0.50. The brakes are applied to a 750kg car traveling 30.0 m/s and the car skids to a stop. What would be the magnitude and direction of the acceleration on the car? u = 0.50 m = 750kg V(i) =...