A 1200-kg car is being driven up a 7.76° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 496 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 204 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 197 kJ? in doing this problem i got the work of the force F to be 204*Force of F, the work of the gravity to be 3.24E5, the work of the normal force to be 0, and the work of the kinetic friction to be 1.01E5. The problem says that the net work is 197000 J so i added all those forces and made them equal to that and then i solved for F, but i got a negative force (-1117.65), does that make sense?