Jim is driving a 2268-kg pickup truck at 30.0 m/s and releases his foot from the

Question

Jim is driving a 2268-kg pickup truck at 30.0 m/s and releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 800 N .

a)Determine the stopping distance of the truck.

b)Determine the average acceleration while stopping of the truck.

c) Determine the coefficient of kinetic friction between tire and surface.

d)Would a heavier car travel farther before stopping or stop sooner if the average friction force were the same?

Explanation / Answer

a) KEi = ½mv² = ½ * 2268kg * (30m/s)² = 1020600 J
work done = KE = F * d = 1020600 J = 800N * d
distance d = 1275.75 m

b) Easiest is to use Torricelli's:
v² = u² + 2as
0² = (30m/s)² + 2 * a * 1275.75  
a = -0.353 m/s²

c) µk = F/mg = 800/22226.4 0.036

d) heavier car stop sooner if the average friction force were the same.

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