A 1210-kg car is being driven up a 7.18 ° hill. The frictional force is directed

Question

A 1210-kg car is being driven up a 7.18 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 490 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 223 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 201 kJ? Please provide steps to solve

Explanation / Answer

The net force is non zero along the direction of the slope since the car is moving there.

so, 201000 = F(223) - 490(223) - 1210(9.8)sin7.18o (223)

=> F = 2873.44 N = 2.873 kN.

net work done perpendicular to the slope is zero since the car is not accelerating vertically upwards.

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