Let's analyze the problem step by step.

Problem Summary:

• A 321-kg crate hangs from a rope of length 11.4 m.
• You apply a horizontal force to move the crate 3.8 m to the right.
• The applied force required was 1108.4 N.
• We're interested in calculating:
  • Work done by the weight of the crate.
  • Work done by the force applied on the crate.

To solve for the work done by the weight of the crate:

1. Weight of the Crate: The weight of the crate $W$ is given by: $W = mg$ where $m = 321 \, \text{kg}$ is the mass of the crate, and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

2. Vertical Displacement: In this scenario, the force of gravity acts vertically downward. The crate, however, is only displaced horizontally by $d = 3.8 \, \text{m}$. There is no vertical movement of the crate, so the vertical displacement ($h$) is zero. Since work is defined as $W = F \cdot d \cdot \cos\theta$, and in this case, $\theta = 90^\circ$ between the weight (vertical) and displacement (horizontal):

   $\cos(90^\circ) = 0$

3. Work Done by Weight: Therefore, the work done by the weight of the crate is: $W_{\text{weight}} = F_{\text{gravity}} \times d \times \cos(90^\circ) = 0$

Conclusion: The work done by the weight of the crate is 0 J.

Would you like me to help calculate the work done by the applied force or discuss any other details?

Related Questions:

1. How is the work done calculated when the force is not aligned with displacement?
2. What is the difference between conservative and non-conservative forces in terms of work done?
3. How does gravitational potential energy change if the crate moves vertically?
4. What happens to the work done if the crate were moved diagonally instead of horizontally?
5. Can you explain how tension in the rope changes as the crate is displaced horizontally?

Tip: In problems involving forces, always remember to decompose forces into components that are parallel and perpendicular to the direction of movement to correctly determine the work done.