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A 3.5 × 103 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 16.9◦ with the horizontal. An a...Asked by Artemis
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?
The acceleration of gravity is 9.81 m/s2 . What is the length of the driveway?
Answers
Answered by
Henry
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.
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.