A person holds a bucket of weight 60N . He walks 7m along the horizontal path and then climbs up a vertical distance of 5m. The work done by the man is

Options

A. 300 J
B. 420 J
C. 720 J
D. None of these

Answer

To solve the problem of calculating the work done by the man holding a bucket while walking and climbing, we can break it down into steps:

### Step-by-Step Solution:

1. **Identify the Forces and Displacements**:
- The weight of the bucket is given as \( W = 60 \, \text{N} \).
- The man walks horizontally for a distance of \( d_1 = 7 \, \text{m} \).
- He then climbs vertically for a height of \( h = 5 \, \text{m} \).

2. **Calculate Work Done While Walking Horizontally**:
- When the man walks horizontally, the force exerted by the bucket (weight) acts downward while the displacement is horizontal.
- The angle \( \theta \) between the force and displacement is \( 90^\circ \).
- The formula for work done \( W \) is given by:
\[ W = F \cdot d \cdot \cos(\theta) \]
- Since \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \).
- Therefore, the work done while walking horizontally is:
\[ W_{AB} = 60 \, \text{N} \cdot 7 \, \text{m} \cdot \cos(90^\circ) = 0 \, \text{J} \]

3. **Calculate Work Done While Climbing Vertically**:
- When climbing, the man exerts an upward force equal to the weight of the bucket against gravity.
- The displacement is vertical, and the angle between the force and displacement is \( 0^\circ \) (since both are in the same direction).
- The work done while climbing is:
\[ W_{BC} = F \cdot h \cdot \cos(0^\circ) = 60 \, \text{N} \cdot 5 \, \text{m} \cdot 1 = 300 \, \text{J} \]

4. **Total Work Done**:
- The total work done by the man is the sum of the work done in both segments:
\[ W_{\text{total}} = W_{AB} + W_{BC} = 0 \, \text{J} + 300 \, \text{J} = 300 \, \text{J} \]

### Final Answer:
The total work done by the man is \( \boxed{300 \, \text{J}} \).