The code calculates the total work done against gravity by a set of forces applied to lift an object, using the formula for work based on the dot product of force and displacement vectors. Let’s analyze the setup:

1. Force Values and Angles: The forces applied are:

   • Force 1: $1830 \, \text{N}$ at $120^\circ$
   • Force 2: $1295 \, \text{N}$ at $45^\circ$
   • Weight Force: $2500 \, \text{N}$ at $270^\circ$ (pointing downward)

2. Displacement Vector: The displacement vector is $(5, 90^\circ)$, indicating 5 meters straight up (in the positive y-direction).

3. Code Functionality: For each force:

   • The code converts the magnitude and angle to $x$ and $y$ components.
   • It does the same for the displacement.
   • Then, it calculates the dot product of each force vector with the displacement vector to compute the work done by each force.
   • It sums up the work values to obtain the total work.

Let's break down the work done by each force:

• For each force, calculate the $x$ and $y$ components using $\text{magnitude} \times \cos(\text{angle})$ and $\text{magnitude} \times \sin(\text{angle})$.
• The displacement vector components are $(0, 5)$ because it’s purely vertical.

4. Work Calculation: Work is given by the dot product: $W = F_x \cdot d_x + F_y \cdot d_y$ where $d_x = 0$ and $d_y = 5$, so the work simplifies to $W = F_y \times 5$.

5. Option Analysis:

   • The correct work calculation yields Option C: $1.25 \times 10^4$ Joules as the result.

Would you like more details on each calculation step?

Related Questions

1. How is the dot product used to compute work in physics?
2. What is the significance of angles in force vector decomposition?
3. Why is only the vertical component relevant in this example?
4. How does gravitational force influence work in lifting scenarios?
5. What changes if the displacement direction changes?

Tip

In physics, always confirm the direction of vectors when calculating work to ensure correct component usage.