An object with a mass of 12 kg slides 9 meters across a horizontal surface with a coefficient of kinetic friction of 0.22. For the duration of the slide an external force, F1, with a magnitude of 43 Newtons acts on the mass in the positive y direction while another force, F2, is angled at 33 degrees above the x axis. The net work done over the distance of 9 meters, 414 Joules. What is the magnitude of F2 in Newtons?

To find the magnitude of force F2, we will use the work-energy principle, which states that the net work done on an object is equal to the change in its kinetic energy.

First, let's calculate the work done by force F1. We know that F1 is acting in the positive y direction, so the force and displacement are in the same direction. Therefore, the work done by F1 is simply the magnitude of the force multiplied by the displacement:

Work_F1 = F1 * d = 43 N * 9 m = 387 J

Next, we can calculate the work done by the frictional force. The work done by friction can be found using the equation:

Work_friction = μ * m * g * d

where μ is the coefficient of kinetic friction, m is the mass of the object, g is the acceleration due to gravity, and d is the displacement.

Given that the coefficient of kinetic friction μ is 0.22, the mass m is 12 kg, the acceleration due to gravity g is approximately 9.8 m/s^2, and the displacement d is 9 m, we can substitute these values into the equation:

Work_friction = 0.22 * 12 kg * 9.8 m/s^2 * 9 m = 23.1656 J

Now, we can find the net work done by subtracting the work done by friction from the total work:

Net work = Work_F1 + Work_friction = 387 J + 23.1656 J = 410.1656 J

However, the given net work is 414 J. The discrepancy could be due to rounding errors or inaccuracies in the given values. Nonetheless, we will proceed with the calculated value of the net work.

Finally, we can calculate the work done by force F2. Since the net work is the sum of the work done by all forces, we can write:

Net work = Work_F2 + Work_F1 + Work_friction

Rearranging the equation, we find:

Work_F2 = Net work - (Work_F1 + Work_friction)

Substituting the values, we get:

Work_F2 = 414 J - (387 J + 23.1656 J) = 3.8344 J

The magnitude of force F2 is equal to the work done by force F2 divided by the displacement:

F2 = Work_F2 / d = 3.8344 J / 9 m ≈ 0.426 J/m ≈ 0.426 N

Therefore, the magnitude of force F2 is approximately 0.426 Newtons.