A 3.1 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 19.5◦ with the horizontal. An average frictional force of 4.4×103 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 3.1 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 19.5◦ with the horizontal. An average frictional force of 4.4×103 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

Wc = m*g = 3100kg * 9.8N/kg = 30,380 N.=
Wt. of car.

Fp = 30,380*sin19.5 = 10,141 N. = Force parallel to incline.
Fv = 30,380*cos19.5 = 28,637 N. = Force
perpendicular to incline.

Fk=4400 N. = Force of kinetic friction.

Fp-Fk = m*a
a=(Fp-Fk)/m
a = (10,141-4400)/3100=1.852 m/s^2.

L = (V^2-Vo^2)/2a=(4.9^2-0)/3.704=6.48 m