A car of mass m = 1070 kg is traveling down a theta = 11 degree incline. When th

Question

A car of mass m = 1070 kg is traveling down a theta = 11 degree incline. When the car's speed is v_0 = 12 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is mu_k = 0.45. Write an expression for the magnitude of the force of kinetic friction, f_k. f_k = _______________ 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. Calculate the distance the car travels down the hill Lin meters until it comes to a stop at the end.

Explanation / Answer

a) Sum of the forces is perpendicular to the incline (which is away from the plane is positive) = 0
0 = N - m*g*cos()
N = m*g*cos()

We can define friction by
friction =k*N
friction = k*m*g*cos() ----------------> ANS A

b) Magnitude of change in cars height

h = L*sin() -------------------> ANS B

c) Sum of the forces which is parallel to the incline (down the incline is positive) = -m*a <--- acceleration is retarding the car down and acts up the incline, hence negative

-m*a = m*g*sin() - friction

With friction substituted in
-m*a = m*g*sin() - k*m*g*cos()

m terms cancel out
-a = gx(sin() - kxcos())

=9.8x(sin(11) - 0.45xcos(11))

substitute numbers
a = 2.45 m/s2

We can Use kinematics to find the distance
vf^2 = vi^2 + 2*a*L

Solve for L
L = (vf^2 - vi^2) / (2*a)

substitute numbers
L = (0 - (12 m/s)^2) / (2*2.45 m/s^2) = 29.3 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.