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The two forces of interest are the car's weight (W = 9000) acting vertically down and the pushing force, P, acting parallel to, and up, the hill. If the car drops a vertical distance h and an inclined distance s then the works are; Work done by gravity = Wh (positive because they are both in same direction) and Work done by pusher = -Ps (negative because they are opposite). Now I don't know what your 5 meters is, h or s. Assuming negligible friction, the net work done by all forces, Wh - Ps = change in objects kinetic energy. If the change in kinetic energy is negligible or zero then you get , Ps = Wh. Now if h=5 meters then you can calculate Wh = 9000x5 = 45000 joules., and then Ps = 45000 joules. If s is the 5 meters and not h ,then I don't know how you can calculate a number without more information. So if all the assumptions are right; negligible friction, h = 5 , and negligible KE change then; (1) work done by gravity = 45000 j ; work done by pusher = - 45000 j ; net work done = 0

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