A 3000 kg car is parked at the top of a 6.9 m long driveway that is sloped at 16 degrees with the horizontal. The parking brake fails. If an average friction force of 4100 N impedes the motion, find the speed of the car at the bottom of the driveway. The acceleration of gravity is 9.8 m/s^2. Answer in units of m/s.
2007-10-21 14:28:27 · 1 answers · asked by PhysicsDumbass 1 in Science & Mathematics ➔ Physics
A 16 degree slope that's 6.9 m long rises 6.9sin(16)=1.9 meters. (Don't you wish you'd paid attention in trig?) So the total potential energy of the car is 3000*9.8*1.9 = 55860 J. Since it travelles 6.9 m with 4100 N acting on it, the frictional work was 6.9*4100=28290 J and there are 55860-28290=27570 J of kinetic energy remaining so the velocity must be √(2*27570/3000) = 4.29 m/s
Doug
2007-10-21 14:54:15 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋