Posted by Anonymous on Thursday, October 17, 2013 at 5:26pm.
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

physics  Henry, Saturday, October 19, 2013 at 10:01pm
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.
FpFk = m*a
a=(FpFk)/m
a = (10,1414400)/3100=1.852 m/s^2.
L = (V^2Vo^2)/2a=(4.9^20)/3.704=6.48 m
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