A 4.0 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 19.6◦ with the horizontal. An average frictional force of 4.0×103 N impedes the car’s motion so that the car’s speed at the bottom of the driveway is 4.9 m/s.

Question

A 4.0 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 19.6◦ with the horizontal. An average frictional force of 4.0×103 N impedes the car’s motion so that the car’s speed at the bottom of the driveway is 4.9 m/s.
The acceleration of gravity is 9.81 m/s2 . What is the length of the driveway?

Henry · Answer

Fc = m*g = 4000kg * 9.8N/kg = 39,200 N. = Force of car.

Fp = 39200*sin19.6 = 13,150 N. = Force
parallel to the slope.
Fn = 39200*cos19.6 = 36,929 N. = Normal
Force = Force perpendicular to the slope

Fk = 4,000 N. = Force of kinetic friction.

a=(Fp-Fk)/m
a = (13150-4000)/4000=2.28m/s^2

L=(V^2-Vo^2)/2a = (4.9^2-0)/4.56=5.27 m.