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 in Newtons, we need to calculate the net work done on the mass and then solve for F2.
The net work done on an object is equal to the sum of the work done by each individual force acting on the object. In this case, we have two forces acting on the mass: F1 in the positive y direction and F2 at an angle of 33 degrees above the x-axis.
The work done by a force is given by the formula:
Work = Force * Distance * cos(θ)
Where:
Force is the magnitude of the force in Newtons,
Distance is the distance over which the force acts,
θ is the angle between the force and the direction of motion.
In this problem, the net work done on the object is given as 414 Joules. Let's calculate the individual work done by each force:
Work1 = F1 * 9m * cos(90°) --> Since the angle is 90 degrees between the y-axis and direction of motion, the cosine of 90° is zero. Hence, the work done by F1 is zero.
Work2 = F2 * 9m * cos(33°) --> Here, we have the magnitude of F2 and the angle of 33 degrees above the x-axis. So we need to calculate the cosine of 33°.
cos(33°) = 0.8387
Now we can solve for F2:
414 J = 0 + F2 * 9m * 0.8387
F2 = 414 J / (9m * 0.8387)
F2 ≈ 54.56 N
Therefore, the magnitude of force F2 is approximately 54.56 Newtons.