Frictional Forces of a steep hill and a car

Question

You live at the top of a steep (a slope of 15^° above the horizontal) hill and must park your 2200 kg car on the street at night.

a) you unwisely leave your car out of gear one night and your handbrake fails. assuming no significant frictional forces are acting on the car, how quickly will is accelerate down the hill?

b) the increase in insurance premiums due to the results of your mistake mean that you cannot afford to fix you hand-break properly. you resolve to always leave your car in gear when parked on a slope. if the rolling frictional force caused by leaving the drive train connected to the wheels is 5000N at what rate will your car accelerate down the hill if the hand break fails again?

Answer 1

a) Weight of the car is mg = 2200 kg x 10 m/s² = 22000 N. Find the component of the weight that acts parallel to the slope: Parallel F = weight x sin 15 degrees = 5694 N. This is the force that accelerate the car down the hill, so a = F/m = 5694/2200 = 2.6 m/s²

b) With friction = 5000 N acting parallel up the hill, we need to find the net force down the slope. We already know the parallel force acting down the slope = 5694 N, so F_net = 5694 - 5000 = 694 N. Again, a = F_net/m = 694/2200 = 0.32 m/s²