A 1230-kg car is being driven up a 5.56 ° hill. The frictional force is directed

Question

A 1230-kg car is being driven up a 5.56 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 495 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 252 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 220 kJ?

Explanation / Answer

Gravity force down hill = mgsin(theta) = 1230*9.8*sin(5.56) = 1167.89 N

Work done against gravity = 1167.89*252 = 294308.28 = 294.308 kJ

Work done against friction = 495*252 = 124740 = 124.74 kJ

Total minimum work to be delivered = 419.048 kJ

Now work done by F

F*d = (419.048 + 220)*10^3

F = 639.048*10^3 / 252 = 2535.9 N

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