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 N directed perpendicular to the road surface. The length of the road up the hill is 310 m. What should be the magnitude of , so that the net work done by all the forces acting on the car is +140 kJ?
To find the magnitude of the force F that needs to be applied to the car, we need to calculate the net work done by all the forces acting on the car.
The net work done is equal to the change in kinetic energy of the car. Since the car is being driven up a hill, we need to consider the change in gravitational potential energy as well.
The work done by the force F is given by the equation:
Work = Force * Distance * cos(θ)
Where:
- Force is the magnitude of the force F
- Distance is the length of the road up the hill (310 m)
- θ is the angle between the force F and the direction of motion (0°, since the force is propelling the car forward)
The work done by the frictional force f is equal to the magnitude of the force times the distance:
Work_f = f * Distance
The work done against gravity is given by:
Work_g = weight * Distance * sin(θ)
Where:
- weight is the weight of the car, given by mass * gravitational acceleration (1100 kg * 9.8 m/s^2)
- θ is the angle of the hill (4.0°)
The net work done is the sum of all the individual work done:
Net work = Work + Work_f + Work_g
Given that the net work is +140 kJ, we can convert it to joules:
Net work = 140,000 J
Now we can substitute the known values and solve for the force F:
140,000 J = (F * 310 m * cos(0°)) + (518 N * 310 m) + ((1100 kg * 9.8 m/s^2) * 310 m * sin(4.0°))
Simplifying this equation will give us the magnitude of the force F.