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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

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MechEngineer

Answer:

a. [tex]F_f = \mu mg cos\theta[/tex]

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

[tex]F_f = \mu N = \mu mg cos\theta[/tex]

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

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

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

[tex]v^2 - v_0^2 = 2a\Delta s[/tex]

where v = 0 m/s is the final velocity of the car when it stops, [tex]v_0[/tex] = 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 [tex]\Delta s[/tex] is the distance traveled, which we care looking for:

[tex]0^2 - 14^2 = 2(-4.3)\Delta s[/tex]

[tex]\Delta s = 14^2 / (2*4.3) = 22.78 m[/tex]

(a) the kinetic friction is given by [tex]f=\mu mgcos\theta[/tex]

(b) The magnitude of height is h = Lcosθ

(c) The car travels down the hill for 22.78m until it stops.

Inclined Plane:

(a) The kinetic friction acts opposite to the motion of the car and it is proportional to the normal reaction force as shown below:

[tex]f=\mu mgcos\theta[/tex]

where μ = 0.45 is the coefficient of kinetic friction

m is the mass of the car = 1030kg

and θ is the angle of incline = 13°

(b) If the car slides down the incline for a distance L then the magnitude of change of height h is given by the following expression:

h = Lcosθ

(c) when the breaks of the car lock, the speed of the car is u = 14 m/s, only frictional force will be acting at this point. the acceleration is given by:

[tex]a=-\frac{f}{m} =\mu gcos\theta\\\\a=-0.45\times9.8\times cos13\\\\a=-4.3m/s^2[/tex]

from the third equation of motion:

[tex]v^2=u^2+2as\\\\0=14^2-2\times2.3s\\\\s=22.78m[/tex]

Learn more about laws of motion:

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