Posted by **umesh** on Sunday, December 30, 2012 at 7:58am.

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.

- dynamics -
**Henry**, Sunday, December 30, 2012 at 11:57am
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.

- dynamics -
**Henry**, Sunday, December 30, 2012 at 12:55pm
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.

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