Posted by Tina on Monday, June 27, 2011 at 11:45pm.
Question: A car is on a ramp 100m from the bottom of the ramp (angle of incline=30 degrees):
a. If the potential energy of the car is 80902 J, how much does the car weigh?
b. What is the KE half way down the ramp?
c. What speed is the car moving at the bottom of the ramp?
Please indicate which equation you use, what variable corresponds to each number, and how you got the number that goes with the variable.
Thanks!

Physics  Henry, Wednesday, June 29, 2011 at 4:11pm
a. h = 100sin30 = 50m.
PE = mgh = 80,902J,
m*9.8*50 = 80,902,
m = 165.1kg.
F = mg = 165.1kg * 9.8N/kg = 1618N =
Wt. of car in Newtons.
Wt. = 165.1kg / 0.454kg/lb = 364lbs =
Weight of car in lbs.
b. h = (100/2)sin30 = 25m.
V^2 = Vo^2 + 2gd,
V^2 = 0 + 2*9.8*25 = 490,
V = 22.1m/s.
KE = 0.5mV^2 = 0.5 * 165.1 * (22.1)^2 =
40,318J.
c. V^2 = Vo^2 + 2gd,
V^2 = 0 + 2*9.8*50 = 980,
V = 31.3m/s.
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