To determine the work done on the UFO by its propulsion system from 0 m to 12 m, we need to calculate the area under the force-position graph, as work is given by the integral of force over distance.

Step-by-Step Solution

The graph is divided into different segments. Let’s find the area for each segment individually:

1. From 0 m to 4 m: This is a rectangle.

   • Force (height) = 10 N
   • Distance (width) = 4 m
   • Area = $10 \, \text{N} \times 4 \, \text{m} = 40 \, \text{J}$

2. From 4 m to 8 m: This is a trapezoid.

   • Top side (Force at 8 m) = 5 N
   • Bottom side (Force at 4 m) = 10 N
   • Height (distance) = 4 m
   • Area = $\frac{1}{2} \times (10 + 5) \, \text{N} \times 4 \, \text{m} = 30 \, \text{J}$

3. From 8 m to 12 m: This is a rectangle.

   • Force = 5 N
   • Distance = 4 m
   • Area = $5 \, \text{N} \times 4 \, \text{m} = 20 \, \text{J}$

Adding these areas gives the total work done from 0 m to 12 m:

$40 \, \text{J} + 30 \, \text{J} + 20 \, \text{J} = 90 \, \text{J}$

Answer

The total work done on the UFO over the interval from 0 m to 12 m is 90 J, but this option does not appear in the provided choices. The closest answer, though incorrect, would be C. 75 J.

Would you like more details on this calculation or have any questions?

Related Questions

1. How do we calculate work done when force varies with position?
2. What is the significance of the area under a force-position graph?
3. How would the work done change if the force were constant?
4. What would happen to the work done if the force acted in the opposite direction?
5. How can we interpret negative work in the context of physics?

Tip

For piecewise linear graphs, calculating work as the sum of areas under each segment simplifies the process.