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To solve this problem, we need to find the net work done on the cargo canister by the three forces during a displacement of 3.50 m. The forces are acting on the canister at given angles, and the formula for work is:

\[
W = F_{\text{net}} \cdot d \cdot \cos(\theta)
\]

Where:
- \( F_{\text{net}} \) is the net force in the direction of the displacement,
- \( d \) is the displacement (3.50 m), and
- \( \theta \) is the angle between the force and the displacement direction.

### Steps:
1. **Decompose the forces**: Since the forces are acting at angles, we need to find the components of each force in the horizontal (\(x\)) and vertical (\(y\)) directions.

   - Force \(F_1 = 2.80 \, \text{N}\) is along the negative x-axis, so:
     \[
     F_{1x} = -2.80 \, \text{N}, \quad F_{1y} = 0 \, \text{N}
     \]

   - Force \(F_2 = 3.70 \, \text{N}\) is at an angle \(\theta_2 = 51.0^\circ\) from the negative x-axis:
     \[
     F_{2x} = -3.70 \cdot \cos(51.0^\circ) \approx -2.33 \, \text{N}, \quad F_{2y} = 3.70 \cdot \sin(51.0^\circ) \approx 2.88 \, \text{N}
     \]

   - Force \(F_3 = 10.0 \, \text{N}\) is at an angle \(\theta_3 = 33.0^\circ\) from the positive x-axis:
     \[
     F_{3x} = 10.0 \cdot \cos(33.0^\circ) \approx 8.38 \, \text{N}, \quad F_{3y} = 10.0 \cdot \sin(33.0^\circ) \approx 5.45 \, \text{N}
     \]

2. **Calculate the net force in each direction**:
   \[
   F_{\text{net},x} = F_{1x} + F_{2x} + F_{3x} = -2.80 - 2.33 + 8.38 \approx 3.25 \, \text{N}
   \]
   \[
   F_{\text{net},y} = F_{1y} + F_{2y} + F_{3y} = 0 + 2.88 + 5.45 \approx 8.33 \, \text{N}
   \]

3. **Find the magnitude of the net force**:
   \[
   F_{\text{net}} = \sqrt{(F_{\text{net},x})^2 + (F_{\text{net},y})^2} = \sqrt{(3.25)^2 + (8.33)^2} \approx 8.92 \, \text{N}
   \]

4. **Calculate the angle of the net force**:
   \[
   \theta_{\text{net}} = \tan^{-1} \left( \frac{F_{\text{net},y}}{F_{\text{net},x}} \right) = \tan^{-1} \left( \frac{8.33}{3.25} \right) \approx 68.1^\circ
   \]

5. **Find the work done**: Since the displacement is along the x-axis, we only need the horizontal component of the net force:
   \[
   W = F_{\text{net},x} \cdot d = 3.25 \cdot 3.50 \approx 11.38 \, \text{J}
   \]

Thus, the net work done on the canister by the three forces is approximately **11.38 joules**.

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the significance of decomposing forces into components?
2. How do angles affect the calculation of work done by forces?
3. What would happen if the displacement was in a different direction?
4. How would you calculate the work done by individual forces?
5. Can the net work be negative in any scenario?

**Tip**: Always remember to decompose vectors into components before applying them to different directions of motion!

The figure here 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 = 2.80 N, F2 = 3.70 N, and F3 = 10.0 N, and the indicated angles are θ2 = 51.0° and θ3 = 33.0°. What is the net work done on the canister by the three forces during the first 3.50 m of displacement?

Learn how to calculate the net work done by three forces acting at different angles on a cargo canister moving across a frictionless floor. This problem covers key concepts in physics, including vector decomposition, trigonometry, and the work formula, making it suitable for students in Grades 11-12.

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