A 1540-kg car is being driven up a 6.39 ° hill. The frictional force is directed

Question

A 1540-kg car is being driven up a 6.39 ° 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 227 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 126 kJ?

Explanation / Answer

Since "work done by ALL forces" = the work done by the NET force (NF):
NF x 227 = 126,000 J
NF = 126,000/227 = 555.066 N

NF = F - [(mg sin 6.39°) + 495]
555.066 = F - (1540(9.81)(sin 6.39) + 495) {where 6.39° = hill angle}
555.066 = F - (1681.385 + 495) = F - 2176.385
F = 2731.45 N ANS*

the normal force is a REACTION force to the car weight component that acts perpendicular to the road, it is used to calculate the friction force which is already given, so is not used in these calculations.

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