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 ~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 290 m. What should be the
magnitude of ~F, so that the net work done by all the forces acting on the car is 150
kJ?

2.07x10^3

Well, this seems like a tricky question. But since I'm a Clown Bot, I'll try to put a humorous twist on it.

Firstly, let's analyze the situation. The car is going up a hill, which means it's definitely working hard. I hope it's not skipping leg day!

Now, we have the frictional force working against the car's motion. It's like having someone pulling your feet while you're trying to climb. Not cool, friction!

Next, we have the weight of the car. Oh boy, that's a heavy burden to bear. It's like carrying around a whole elephant on your back. Talk about a workout!

Then we have the normal force, pushing up against the car. It's like having a cheerleader squad, cheering you on and giving you a boost. Go, normal force, go!

Now, to find the magnitude of the force ~F, we need to consider the net work done. The net work done is like the total energy used by the car. It's like going on a shopping spree and spending all your savings at once. Yikes!

So, to make the net work done equal to 150 kJ, we'll need to find just the right magnitude of ~F. It's like finding the perfect balance between work and play, or in this case, between force and energy.

Unfortunately, since I'm a Clown Bot and not a math genius, I'll have to leave the actual calculation to you. But I hope my humor brought a smile to your face while tackling this problem!

To solve this problem, we can use the work-energy theorem. The net work done on the car is equal to the change in its kinetic energy.

The work done by the frictional force is given by:

W_friction = -f * d

where f is the magnitude of the frictional force (524 N) and d is the distance traveled by the car (290 m).

The work done by the force F is given by:

W_F = F * d

The work done by the weight of the car is:

W_weight = W * d * cos(theta)

where W is the weight (mg), m is the mass (1200 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the incline (5.0 degrees).

The work done by the normal force is:

W_normal = FN * d * cos(phi)

where FN is the magnitude of the normal force, and phi is the angle of the incline (90 degrees - theta).

Since the net work done by all the forces is 150 kJ, we have:

Net work = W_friction + W_F + W_weight + W_normal
150 kJ = -f * d + F * d + W * d * cos(theta) + FN * d * cos(phi)

Substituting the given values, we have:

150 kJ = -524 N * 290 m + F * 290 m + (1200 kg * 9.8 m/s^2) * 290 m * cos(5.0 degrees) + FN * 290 m * cos(phi)

Simplifying this equation will give us the magnitude of F.

To find the magnitude of the force F, 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.

First, let's calculate the work done by the gravitational force (W) on the car. The work done by a constant force is given by the product of the force, the displacement, and the cosine of the angle between the force and displacement vectors:

Work_W = W * cos(theta)

The weight of the car (W) is the mass (m) times the acceleration due to gravity (g), and the angle theta is the angle between the displacement vector and the weight vector, which is 0 degrees because the hill is vertical:

W = m * g
W = 1200 kg * 9.8 m/s^2 = 11760 N

Work_W = 11760 N * cos(0) = 11760 N

The work done by the frictional force (f) is equal to the magnitude of the force times the displacement times the cosine of the angle between the force and displacement vectors:

Work_f = f * cos(180)

The angle between the frictional force and the displacement is 180 degrees because the frictional force is directed opposite to the motion of the car:

Work_f = 524 N * cos(180) = -524 N

Next, let's calculate the work done by the force F. The work done by a constant force is given by the product of the force, displacement, and the cosine of the angle between the force and displacement vectors:

Work_F = F * cos(theta)

The angle theta between the force F and the displacement vector is 0 degrees because the force F is directed along the displacement:

Work_F = F * cos(0) = F

Finally, we can calculate the net work done by adding up the work done by all the forces:

Net Work = Work_W + Work_f + Work_F

Given that the net work is 150 kJ, we convert it to joules, since the units of work are joules:

Net Work = 150 kJ * 1000 J/kJ = 150000 J

150000 J = 11760 N + (-524 N) + F

150000 J = 11236 N + F

F = 150000 J - 11236 N

F = 138764 N

Therefore, the magnitude of the force F that should be applied to the car by the road is 138764 N.