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 find the maximum acceleration with which the boy can climb the rope without breaking it, we can follow these steps:

### Step 1: Identify the forces acting on the boy
The forces acting on the boy (mass \( m = 60 \, \text{kg} \)) are:
- The gravitational force acting downward, which is \( F_g = m \cdot g = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \).
- The tension \( T \) in the rope acting upward.

### Step 2: Determine the maximum tension in the rope
The rope can break when a mass of \( 75 \, \text{kg} \) is suspended from it. The maximum tension \( T_{\text{max}} \) that the rope can withstand is given by the weight of this mass:
\[ T_{\text{max}} = 75 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 750 \, \text{N} \]

### Step 3: Apply Newton's second law
When the boy climbs the rope with an acceleration \( a \), the net force acting on him can be expressed using Newton's second law:
\[ T - F_g = m \cdot a \]
Substituting the values we have:
\[ T - 600 \, \text{N} = 60 \, \text{kg} \cdot a \]

### Step 4: Substitute the maximum tension into the equation
We know that the maximum tension \( T \) cannot exceed \( 750 \, \text{N} \). Therefore, we set \( T = T_{\text{max}} \):
\[ 750 \, \text{N} - 600 \, \text{N} = 60 \, \text{kg} \cdot a \]
\[ 150 \, \text{N} = 60 \, \text{kg} \cdot a \]

### Step 5: Solve for the acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{150 \, \text{N}}{60 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]

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

---