Question

A car of mass m = 1030 kg is traveling down a θ = 13-degree incline. When the car's speed is v0 = 14 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
a. Write an expression for the magnitude of the force of kinetic friction
b. Write an expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline.
c. Calculate the distance the car travels down the hill 1 in meters until it comes to a stop at the end

Answer 1

Answer:

a. $F_f = \mu mg cos\theta$

b. h = Lsinθ

c. 22.78 m

Explanation:

a. The kinetic friction is the product of kinetic coefficient and normal force N, which is the gravity force in the direction normal to the incline

$F_f = \mu N = \mu mg cos\theta$

b. As the car travels a distance L down the incline of θ degrees, vertically speaking it would have traveled a distance of:

h = Lsinθ

As we can treat L and h in a right triangle where L is the hypotenuse and h is a side length in opposite of incline angle θ

c. Let g = 9.81 m/s2. the acceleration caused by kinetic friction according to Newton's 2nd law is

$a = F_f/m = \mu g cos\theta = 0.45*9.81*cos13^o = 4.3 m/s^2$

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the car when it stops, $v_0$ = 14m/s is the initial velocity of the car when it starts braking, a = -4.3 m/s2 is the deceleration of the car, and $\Delta s$ is the distance traveled, which we care looking for:

$0^2 - 14^2 = 2(-4.3)\Delta s$

$\Delta s = 14^2 / (2*4.3) = 22.78 m$