solved!!!!! Work and Energy, friction problem 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? m = 1200 kg Ɵ = 5 degrees s = 290 m friction = f = 524 N W = 150,000 J Find the force forward = F So the net forces acting on the car are mg, Fn, and friction. W = FscosƟ In the solution to this problem, they say that for the above equation, cosƟ = cos5....Why? The net forces in the y direction cancel each other out. And the only forces left are the forward force F, mgsinƟ, and frictional force f The Ɵ is supposed to be the angle between the force and the displacement. However, if forward forward force, mgsinƟ, and frictional force are the only net forces acting on the car, and they all lie along the x-axis, then why isn't Ɵ = 0?