A body of mass 60 kg is holding a vertical rope . The rope can break when a mass of 75 kg is suspended from it. The maximum acceleration with which the boy can climb the rope without breaking it is `: ( g = 10 m//s^(2))`

Options

A. `2.5 m//s^(2)`
B. `5.0 m//s^(2)`
C. ` 7.5 m//s^(2)`
D. `22.5m//s^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the maximum tension in the rope
The maximum tension (T_max) that the rope can withstand is given by the weight of the mass that can break it. This mass is 75 kg. We can calculate the maximum tension using the formula:
\[ T_{\text{max}} = m \cdot g \]
where \( m = 75 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).

Calculating:
\[ T_{\text{max}} = 75 \, \text{kg} \times 10 \, \text{m/s}^2 = 750 \, \text{N} \]

### Step 2: Write the equation for the forces acting on the body
When the boy of mass 60 kg climbs the rope with an upward acceleration \( a \), the forces acting on him are:
- Weight acting downwards: \( W = m \cdot g = 60 \, \text{kg} \times 10 \, \text{m/s}^2 = 600 \, \text{N} \)
- Tension in the rope acting upwards: \( T \)

According to Newton's second law, the net force acting on the boy can be expressed as:
\[ T - W = m \cdot a \]
Substituting the values:
\[ T - 600 = 60 \cdot a \]

### Step 3: Substitute the maximum tension into the equation
We know that the maximum tension \( T \) can be at most \( 750 \, \text{N} \). Therefore, we can substitute \( T \) with \( T_{\text{max}} \):
\[ 750 - 600 = 60 \cdot a \]

### Step 4: Solve for the maximum acceleration \( a \)
Rearranging the equation gives:
\[ 150 = 60 \cdot a \]
Now, divide both sides by 60:
\[ a = \frac{150}{60} = 2.5 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration with which the boy can climb the rope without breaking it is \( 2.5 \, \text{m/s}^2 \).

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