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### Given:
- Mass of the box, \( m = 2.20 \, \text{kg} \)
- Distance, \( d = 4.00 \, \text{m} \)
- Applied force, \( F = 16.0 \, \text{N} \)
- Angle, \( \theta = 30.0^\circ \)
- The surface is frictionless.

### (a) Work done by the applied force:

The formula for work done by a force is:
\[
W = F \cdot d \cdot \cos(\theta)
\]
where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.

Substituting the given values:
\[
W = 16.0 \, \text{N} \cdot 4.00 \, \text{m} \cdot \cos(30^\circ)
\]
\[
W = 16.0 \cdot 4.00 \cdot 0.866 = 55.424 \, \text{J}
\]
So, the work done by the applied force is approximately **55.42 J**.

### (b) Work done by the normal force:

The normal force acts perpendicular to the direction of motion. Since work is only done when a force has a component along the direction of displacement, and the normal force is perpendicular to the displacement, the work done by the normal force is **0 J**.

### (c) Work done by the gravitational force:

The gravitational force also acts vertically (downward), which is perpendicular to the horizontal displacement. Thus, similar to the normal force, the work done by the gravitational force is **0 J**.

### (d) Work done by the net force:

The net force is the combination of all the forces acting on the box. However, since the only force doing work along the direction of motion is the horizontal component of the applied force, the work done by the net force is equal to the work done by the applied force, which is **55.42 J**.

### Summary of Answers:
- (a) \( W = 55.42 \, \text{J} \)
- (b) \( W = 0 \, \text{J} \)
- (c) \( W = 0 \, \text{J} \)
- (d) \( W = 55.42 \, \text{J} \)

Would you like more details or have any questions about these calculations?

Here are 5 related questions you might find useful:
1. How would the work change if friction were present?
2. What if the force were applied vertically upward instead?
3. How would you calculate the net force if the surface had friction?
4. What is the relationship between work and energy in this system?
5. How does the angle affect the amount of work done by the applied force?

**Tip:** Always ensure the direction of forces and displacement align properly when calculating work, as perpendicular forces do no work!

A dock worker pushes a 2.20 kg box a distance of 4.00 m along the floor by a constant force of magnitude 16.0 N directed at an angle of 30.0° below the horizontal. Assume the floor is frictionless. Determine the work done by the applied force, the normal force, the gravitational force, and the net force.

Learn how to calculate the work done on a 2.20 kg box moved 4.00 meters by a 16.0 N force at an angle of 30°. Detailed step-by-step solution, including the work done by the applied force, normal force, gravitational force, and the net force. Suitable for high school physics students.

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