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?
The net work done on the canister by the three forces during the first 4.00 m of displacement is 48.00 J. This is calculated by multiplying the magnitude of each force by the displacement (4.00 m) and then adding the three values together: (2.00 N x 4.00 m) + (3.00 N x 4.00 m) + (13.00 N x 4.00 m) = 48.00 J.
To find the net work done on the canister by the three forces, we need to calculate the work done by each individual force and then sum them up.
The formula for work is given by:
Work = Force x Distance x cos(theta)
where theta is the angle between the force vector and the displacement vector.
Given that the canister moves horizontally, there is no vertical displacement (θ = 0°). Therefore, the cosine of 0° is 1.
For the first force, F1 = 2.00 N, since it is acting horizontally, the angle between the force and displacement vectors is also 0°. Therefore, the work done by F1 is:
Work1 = F1 x Distance x cos(0°) = 2.00 N x 4.00 m x 1 = 8.00 J
For the second force, F2 = 3.00 N, the angle between the force and displacement vectors is also 0°. Therefore, the work done by F2 is:
Work2 = F2 x Distance x cos(0°) = 3.00 N x 4.00 m x 1 = 12.00 J
For the third force, F3 = 13.00 N, the angle between the force and displacement vectors is also 0°. Therefore, the work done by F3 is:
Work3 = F3 x Distance x cos(0°) = 13.00 N x 4.00 m x 1 = 52.00 J
Finally, to find the net work done on the canister, we sum up the work done by each force:
Net Work = Work1 + Work2 + Work3 = 8.00 J + 12.00 J + 52.00 J = 72.00 J
Therefore, the net work done on the canister by the three forces during the first 4.00 m of displacement is 72.00 Joules.
To find the net work done on the canister, we need to calculate the work done by each force and then add them together.
The work done by a force is given by the equation:
Work = Force × Displacement × cos(θ)
Where:
- Force is the magnitude of the force,
- Displacement is the magnitude of the displacement,
- θ is the angle between the force and the direction of displacement.
In this case, the angle between each force and the direction of displacement is not given, but since the forces are acting horizontally and the displacement is also horizontal, the angle between the forces and the displacement is 0°. Therefore, cos(θ) = 1.
Let's calculate the work done by each force:
Work1 = F1 × 4.00 m × cos(0°) = 2.00 N × 4.00 m × 1 = 8.00 J
Work2 = F2 × 4.00 m × cos(0°) = 3.00 N × 4.00 m × 1 = 12.00 J
Work3 = F3 × 4.00 m × cos(0°) = 13.00 N × 4.00 m × 1 = 52.00 J
Now we can find the net work done by adding the three individual works together:
Net Work = Work1 + Work2 + Work3
= 8.00 J + 12.00 J + 52.00 J
= 72.00 J
Therefore, the net work done on the canister during the first 4.00 m of displacement is 72.00 Joules.