V^2 = Vo^2 + 2ad.
a = (V^2-Vo^2)/2d.
a = (0-36)/10 = -3.6 m/s^2.
Fk = m*a = 1500 * -3.6 = -5400 N. Acting
in opposite direction of motion.
the 1500 kg automobile is travelling up the 20 degree inclined at a speed of 6 m/s , if the driver wishes to stop his car in a distance of 5m , determine the frictional force at the pavement which must be supplied by the rear wheels.
2 answers
CORRECTION:
PE = KE.
h = 5*sin20 = 1.71 m.
mg*h-Fk*d = 0.5m*V^2.
1500*9.8*1.71-Fk*5 = 0.5*1500*6^2
25,137-5Fk = 27,000
-5Fk = 27000-25,137 = 1863
Fk = -373 N. = Force of kinetic friction.
PE = KE.
h = 5*sin20 = 1.71 m.
mg*h-Fk*d = 0.5m*V^2.
1500*9.8*1.71-Fk*5 = 0.5*1500*6^2
25,137-5Fk = 27,000
-5Fk = 27000-25,137 = 1863
Fk = -373 N. = Force of kinetic friction.
3 answers