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

Options

A. `300 N-m`
B. `420 N-m`
C. `720 N-m`
D. `210 N-m`

Explanation

To find the work done by the man holding the bucket, we need to consider the work done in both horizontal and vertical movements.

### Step-by-Step Solution:

1. **Identify the Forces**: The weight of the bucket is given as 60 N. This is the force that the man is working against when he lifts the bucket vertically.

2. **Calculate Work Done in Vertical Movement**:
- The formula for work done (W) is given by:
\[ W = F \times d \times \cos(\theta) \]
- In the vertical movement, the force (F) is the weight of the bucket (60 N), the distance (d) is the vertical distance climbed (5 m), and the angle (θ) is 0 degrees (since the force and displacement are in the same direction).
- Therefore, the work done in lifting the bucket is:
\[ W_{\text{vertical}} = 60 \, \text{N} \times 5 \, \text{m} \times \cos(0) = 60 \times 5 \times 1 = 300 \, \text{J} \]

3. **Calculate Work Done in Horizontal Movement**:
- When the man walks horizontally, he is not doing any work against gravity (the vertical component of the weight does not change). The force exerted horizontally does not contribute to work done against the weight of the bucket.
- Therefore, the work done in the horizontal movement is:
\[ W_{\text{horizontal}} = 0 \, \text{J} \]

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

### Final Answer:
The total work done by the man is **300 J**.