A monkey of 25 kg is holding a vertical rope. The rope does not break if a body of mass less than 30 kg is suspended with the rope. What will be the maximum acceleration with which the monkey can climb up along the rope ? (Take `g=10 ms^(-2)`)

Options

A. `2 ms^(-2)`
B. `2.5 ms^(-2)`
C. `3 ms^(-2)`
D. `4 ms^(-2)`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the forces acting on the monkey
The monkey has a mass of 25 kg and is climbing up a rope. The forces acting on the monkey are:
- The gravitational force acting downwards, which is \( W_m = m_m \cdot g \)
- The tension in the rope acting upwards, which we will denote as \( T \)

### Step 2: Calculate the gravitational force on the monkey
Using the given value of \( g = 10 \, \text{m/s}^2 \):
\[ W_m = m_m \cdot g = 25 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 250 \, \text{N} \]

### Step 3: Determine the maximum tension in the rope
The problem states that the rope can support a mass of less than 30 kg. Therefore, the maximum tension \( T_{max} \) in the rope is:
\[ T_{max} = m_{max} \cdot g = 30 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 300 \, \text{N} \]

### Step 4: Write the equation of motion for the monkey
When the monkey climbs up with an acceleration \( A \), the net force acting on the monkey can be expressed as:
\[ T - W_m = m_m \cdot A \]
Substituting the values we have:
\[ T - 250 \, \text{N} = 25 \, \text{kg} \cdot A \]

### Step 5: Substitute the maximum tension into the equation
To find the maximum acceleration, we will use the maximum tension:
\[ 300 \, \text{N} - 250 \, \text{N} = 25 \, \text{kg} \cdot A \]
This simplifies to:
\[ 50 \, \text{N} = 25 \, \text{kg} \cdot A \]

### Step 6: Solve for acceleration \( A \)
Now, we can solve for \( A \):
\[ A = \frac{50 \, \text{N}}{25 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration with which the monkey can climb up along the rope is:
\[ \boxed{2 \, \text{m/s}^2} \]