(3096) Problem 1: A car ofmass m = 1030 kg is traveling down a -14 degree inclin
ID: 1796106 • Letter: #
Question
(3096) Problem 1: A car ofmass m = 1030 kg is traveling down a -14 degree incline. When the car's speed is vo 13 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is uy = 0.45. 0 ©theexpertta.com 33% Part (a) Write an expression for the magnitude of the force of kinetic friction, 33% Part b) Write an expression for the magnitude of the change in the cars height, h, along the y-direction, assuming it travels a distance L down the incline A 33% Part (c) Calculate the distance the car travels down the hill L in meters until it comes to a stop at the endExplanation / Answer
a) fk = mu_k*m*g*cos(theta) = 0.45*1030*9.81*cos(14) = 4411.87 N
b) sin(theta) = h/L
h = L*sin(theta) = L*sin(14)
C) Work done by the Net force = change in kinetic energy
Wg+ wf = 0-0.5*m*vo^2
(m*g*sin(14) - 4411.87)*L = -0.5*1030*13^2
((1030*9.81*sin(14))-4411.87)*L = -0.5*1030*13^2
L = 44.23 m
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