A person holds a bucket of weight 80 N. He walks 8 m along the horizontal and then climbs up a vertical distance of 5 m. The work done by the person is

Options

A. 640 J
B. 400 J
C. 720 J
D. zero

Explanation

To solve the problem of the work done by a person holding a bucket while walking and climbing, we can break it down into two parts: the horizontal movement and the vertical movement.

### Step-by-Step Solution:

1. **Identify the Forces and Displacements**:
- The weight of the bucket (force) is given as \( F = 80 \, \text{N} \).
- The horizontal distance walked is \( d_1 = 8 \, \text{m} \).
- The vertical distance climbed is \( d_2 = 5 \, \text{m} \).

2. **Calculate Work Done During Horizontal Movement (A to B)**:
- When the person walks horizontally, the force (weight of the bucket) acts downward while the displacement is horizontal.
- The angle \( \theta \) between the force and displacement is \( 90^\circ \).
- The formula for work done is:
\[ W = F \cdot d \cdot \cos(\theta) \]
- Substituting the values:
\[ W_{AB} = 80 \, \text{N} \cdot 8 \, \text{m} \cdot \cos(90^\circ) \]
- Since \( \cos(90^\circ) = 0 \):
\[ W_{AB} = 80 \cdot 8 \cdot 0 = 0 \, \text{J} \]

3. **Calculate Work Done During Vertical Movement (B to C)**:
- When the person climbs vertically, the force (weight of the bucket) acts downward while the displacement is upward.
- The angle \( \theta \) between the force and displacement is \( 180^\circ \).
- Using the work done formula again:
\[ W_{BC} = F \cdot d \cdot \cos(\theta) \]
- Substituting the values:
\[ W_{BC} = 80 \, \text{N} \cdot 5 \, \text{m} \cdot \cos(180^\circ) \]
- Since \( \cos(180^\circ) = -1 \):
\[ W_{BC} = 80 \cdot 5 \cdot (-1) = -400 \, \text{J} \]

4. **Total Work Done**:
- The total work done by the person is the sum of the work done in both movements:
\[ W_{\text{total}} = W_{AB} + W_{BC} = 0 \, \text{J} + (-400 \, \text{J}) = -400 \, \text{J} \]

### Final Answer:
The total work done by the person is \( -400 \, \text{J} \).

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