A 1100-kg car is being driven up a 4.0° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f = 518 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 Fn directed perpendicular to the road surface. The length of the road up the hill is 310 m. What should be the magnitude of F , so that the net work done by all the forces acting on the car is +140 kJ?

To solve this problem, we need to find the force F that is applied to the car by the road.

First, we need to calculate the work done by each force. The work done by a force is given by the formula: work = force * distance * cos(angle), where the angle is the angle between the force and the direction of motion.

1. The work done by the frictional force is: work_friction = force_friction * distance * cos(180°) = -518 N * 310 m * cos(180°) = -160,780 J (we use a negative sign because the frictional force is opposite to the motion of the car).

2. The work done by the weight force is: work_weight = force_weight * distance * cos(180°) = m * g * distance * cos(180°), where m is the mass of the car and g is the acceleration due to gravity. We can calculate the force_weight as: force_weight = m * g = 1100 kg * 9.8 m/s². Substituting the values, we get: work_weight = 1100 kg * 9.8 m/s² * 310 m * cos(180°) = -3,233,240 J (again, we use a negative sign because the weight force is opposite to the motion of the car).

3. The work done by the normal force is zero since it is perpendicular to the displacement of the car (cos(90°) = 0).

4. The work done by the force applied by the road (F) is: work_F = F * distance * cos(0°) = F * 310 m * cos(0°) = 310F J.

Now, the net work done by all the forces is given as +140 kJ, which is +140,000 J. So we can write the equation: 310F J - 160,780 J - 3,233,240 J = 140,000 J.

Simplifying the equation, we get: 310F J = 303,040 J.

Finally, we can solve for F by dividing both sides of the equation by 310: F = 303,040 J / 310 m ≈ 979.48 N.

Therefore, the magnitude of the force F that should be applied to the car by the road is approximately 979.48 Newtons.