The figure below shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 2.00 N, F2 = 3.00 N, and F3 = 13.00 N. What is the net work done on the canister by the three forces during the first 4.00 m of displacement?
You may need to provide a figure for anyone to solve this question.
To find the net work done on the canister by the three forces during the first 4.00 m of displacement, we need to calculate the work done individually by each force and then add them together.
Work (W) is calculated using the formula: W = F * d * cos(theta), where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and the direction of displacement.
Let's calculate the work done by each force:
1. Force F1 = 2.00 N:
The magnitude of the force is given as 2.00 N, and since the floor is frictionless, the angle between the force and the displacement is 0 degrees (cos(0) = 1).
So, the work done by F1 = F1 * d * cos(theta) = 2.00 N * 4.00 m * 1 = 8.00 J.
2. Force F2 = 3.00 N:
Similar to the previous force, the magnitude is 3.00 N, and the angle between the force and the displacement is also 0 degrees (cos(0) = 1).
Therefore, the work done by F2 = F2 * d * cos(theta) = 3.00 N * 4.00 m * 1 = 12.00 J.
3. Force F3 = 13.00 N:
The magnitude is 13.00 N; however, we don't have the angle between the force and the displacement. Without this angle, we cannot calculate the work accurately.
Since we do not have the angle between F3 and the displacement, we cannot determine the work done by F3 accurately. Therefore, we cannot determine the net work done on the canister by the three forces during the first 4.00 m of displacement without knowing the angle of F3.