To solve the problem of calculating the work done by the person holding a bucket while walking and climbing, we can break the solution down into clear steps:
### Step 1: Understand the Forces Involved
The person is holding a bucket with a weight of 60 N. The work done by the person can be calculated based on the vertical and horizontal movements.
### Step 2: Calculate Work Done in Horizontal Motion
When the person walks horizontally, the force exerted (which is vertical due to holding the bucket) does not contribute to work done in the horizontal direction. The formula for work done is:
\[ W = F \cdot d \cdot \cos(\theta) \]
where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.
In this case, since the force (weight of the bucket) is vertical and the distance walked is horizontal:
- \( F = 60 \, \text{N} \)
- \( d = 7 \, \text{m} \)
- \( \theta = 90^\circ \) (since the force is perpendicular to the direction of motion)
Thus, the work done in the horizontal direction is:
\[ W_H = 60 \, \text{N} \cdot 7 \, \text{m} \cdot \cos(90^\circ) = 0 \, \text{J} \]
### Step 3: Calculate Work Done in Vertical Motion
Next, we need to calculate the work done while climbing vertically. The person climbs a vertical distance of 5 m while holding the bucket.
Using the same formula for work:
- \( F = 60 \, \text{N} \)
- \( d = 5 \, \text{m} \)
- \( \theta = 0^\circ \) (since the force and the direction of motion are in the same direction)
Thus, the work done in the vertical direction is:
\[ W_V = 60 \, \text{N} \cdot 5 \, \text{m} \cdot \cos(0^\circ) = 60 \, \text{N} \cdot 5 \, \text{m} = 300 \, \text{J} \]
### Step 4: Calculate Total Work Done
The total work done by the person is the sum of the work done in the horizontal and vertical directions:
\[ W_{\text{net}} = W_H + W_V = 0 \, \text{J} + 300 \, \text{J} = 300 \, \text{J} \]
### Final Answer
The total work done by the man is **300 Joules**.
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