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A 2100 kg car starts from rest at the top of a driveway 5.0 m long that is sloped at 20.0 degrees with the horizontal. If an average friction force of 4000 N impedes the motion, what is the speed of the car at the bottom of the driveway?

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ian7555

Answer:

h = 5*sin20 = 1.71 m.

PE = mgh = 2100*9.8*1.71 = 35,192 J.

KE = PE-Fk*d.

0.5mV^2 = 35192 - 4000*5

1050V^2 = 15,192

V^2 = 14.47

V = 3.80 m/s.

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