To calculate the work done, we can apply the following principle:

The work done by a force over a distance is given by the area under the force vs. distance graph. In this case, the graph consists of two sections: a triangle and a rectangle.

1. Triangle:

   • Base = 4 meters (from 0 to 4 m)
   • Height = 30 N (the force reaches a maximum of 30 N)

   The area of the triangle is given by: $\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \, \text{m} \times 30 \, \text{N} = 60 \, \text{J}$

2. Rectangle:

   • Length = 2.5 meters (from 4 m to 6.5 m)
   • Height = 30 N (force is constant at 30 N)

   The area of the rectangle is: $\text{Area of rectangle} = \text{Length} \times \text{Height} = 2.5 \, \text{m} \times 30 \, \text{N} = 75 \, \text{J}$

Finally, the total work done is the sum of the areas of the triangle and rectangle: $\text{Total work} = 60 \, \text{J} + 75 \, \text{J} = 135 \, \text{J}$

Thus, the work done in moving the object over a distance of 6.5 meters is 135 joules.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How is work calculated in cases where force varies non-linearly?
2. What is the physical significance of the area under a force-distance graph?
3. How would the work change if the object were moved an additional 1 meter?
4. How is work related to energy in the context of Newtonian mechanics?
5. How would you calculate the work done if the force decreases instead of remaining constant?

Tip: The work done by a constant force over a distance is simply the product of the force and the distance, but when the force varies, integrating or calculating the area under the graph is necessary.