**12. A 1200-kg car is being driven up a 5.0° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f = 524 N. A force  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  and the normal force  directed perpendicular to the road surface. The length of the road up the hill is 290 m. What should be the magnitude of , so that the net work done by all the forces acting on the car is + 150 kJ?

The forces N and mg cos(q) act perpendicular to the displacement, so their work done is zero!

 

The minus sign is due to force acting in opposite direction of displacement.

 

 

 

 

 

Solving for F

 

 

 

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This page last updated on January 11, 2020