A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

Options

A. Moves it over an inclined plane
B. Moves it over a horizontal surface
C. Lifts it vertically upwards
D. None of these

Answer

To solve the problem of determining when the maximum amount of work is done by a man carrying a load on his head through a distance of 5 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Work Done**:
Work done (W) is defined as the product of the force (F) applied in the direction of displacement (S) and the distance moved. Mathematically, it is given by:
\[ W = F \cdot S \]
where \( F \) is the component of the force in the direction of the displacement.

2. **Identifying the Force**:
When the man carries the load, the force acting on the load is the normal force (N) exerted by his head to support the load. The weight of the load acts downwards and is equal to \( Mg \) (where \( M \) is the mass of the load and \( g \) is the acceleration due to gravity).

3. **Considering the Angle of Movement**:
The man can carry the load at different angles (θ) with respect to the vertical. The displacement vector \( S \) can be broken down into its components:
- Horizontal component: \( S \cos \theta \)
- Vertical component: \( S \sin \theta \)

4. **Calculating Work Done**:
The work done by the man can be expressed as:
\[ W = F \cdot S = N \cdot S \sin \theta \]
Here, \( S = 5 \, m \) (the distance moved).

5. **Maximizing Work**:
To find the maximum work done, we need to maximize \( \sin \theta \). The value of \( \sin \theta \) reaches its maximum when \( \theta = 90^\circ \), which means the force is acting vertically upward.

6. **Conclusion**:
Therefore, the maximum amount of work is done when the man lifts the load vertically upward through the distance of 5 m.

### Final Answer:
The maximum amount of work is done when he lifts it vertically upward.

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