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The question asks about the work done by a boy holding a 40-N weight at arm's length for 10 seconds while his arm is 1.5 meters above the ground.

To solve this, recall the formula for work:

\[
W = F \cdot d \cdot \cos(\theta)
\]

Where:
- \( W \) is the work done,
- \( F \) is the force applied (40 N),
- \( d \) is the displacement of the object,
- \( \theta \) is the angle between the force and the displacement direction.

In this scenario:
- The boy holds the weight at a fixed height, meaning there is **no vertical displacement** of the weight (i.e., \( d = 0 \)). 
- Since the weight is not moving, no work is being done because work requires displacement.

Thus, the work done by the boy is:

\[
W = 0 \, \text{Joules}
\]

The correct answer is **0 J**.

Would you like further clarification or details on the concept of work? Here are some related questions:

1. How is work affected if the object is in motion?
2. What is the role of the angle \(\theta\) in the work formula?
3. How does time factor into work calculations?
4. Why doesn't the height above the ground matter in this case?
5. Can holding a weight cause energy expenditure if no work is done?

**Tip:** Work is only done when a force causes displacement. Holding an object in place does not result in work, even though energy may be expended to maintain the position.

A boy holds a 40-N weight at arm's length for 10 seconds. His arm is 1.5 m above the ground. The work done by the force of the boy on the weight while he is holding it is:

Learn how to calculate the work done by a 40-N force holding a weight at arm's length for 10 seconds with no displacement. This problem involves the work-energy theorem and key physics concepts like force and displacement. Suitable for students in Grades 9-12.

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