Answer :
To determine the amount of work done by the construction worker, we must consider the physics definition of work. Work is calculated as the product of the force exerted in the direction of the motion and the distance over which that force is exerted. The formula for work is:
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
Where:
- \( W \) is the work done,
- \( F \) is the force,
- \( d \) is the distance,
- \( \theta \) is the angle between the force and the direction of the motion.
In this problem:
- The load has a mass of 40 kilograms.
- The worker walks a distance of 50 meters.
- The worker is carrying the load over his head and walking horizontally.
The crucial aspect to consider is the direction of the force and the movement:
1. The force exerted by the load (its weight) is directed downwards due to gravity.
2. The worker's movement is horizontal.
Since the force is vertical (downward due to gravity) and the movement is horizontal, the angle \(\theta\) between the force and the direction of the motion is 90 degrees. The cosine of 90 degrees is 0 ( \(\cos(90^\circ) = 0\)).
Substituting these values into the formula for work:
[tex]\[ W = F \cdot d \cdot \cos(90^\circ) \][/tex]
[tex]\[ W = F \cdot d \cdot 0 \][/tex]
[tex]\[ W = 0 \][/tex]
Therefore, no work is done in this scenario because the force exerted by the load is perpendicular to the direction of the worker's movement.
The correct answer is:
A. 0 joules
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
Where:
- \( W \) is the work done,
- \( F \) is the force,
- \( d \) is the distance,
- \( \theta \) is the angle between the force and the direction of the motion.
In this problem:
- The load has a mass of 40 kilograms.
- The worker walks a distance of 50 meters.
- The worker is carrying the load over his head and walking horizontally.
The crucial aspect to consider is the direction of the force and the movement:
1. The force exerted by the load (its weight) is directed downwards due to gravity.
2. The worker's movement is horizontal.
Since the force is vertical (downward due to gravity) and the movement is horizontal, the angle \(\theta\) between the force and the direction of the motion is 90 degrees. The cosine of 90 degrees is 0 ( \(\cos(90^\circ) = 0\)).
Substituting these values into the formula for work:
[tex]\[ W = F \cdot d \cdot \cos(90^\circ) \][/tex]
[tex]\[ W = F \cdot d \cdot 0 \][/tex]
[tex]\[ W = 0 \][/tex]
Therefore, no work is done in this scenario because the force exerted by the load is perpendicular to the direction of the worker's movement.
The correct answer is:
A. 0 joules
