A monkey of mass `20kg` is holding a vertical rope. The rope will not break when a mass of `25kg` is suspended from it but will break it the mass exeeds `25kg` . What is the maximum acceleration with which the monkey can climb up along the rope? `(g=10m//s^(2))` .

Options

A. `25m//s^(2)`
B. `2.5m//s^(2)`
C. `5m//s^(2)`
D. `10m//s^(2)`

Answer

To solve the problem step by step, we will analyze the forces acting on the monkey and use Newton's second law of motion.

### Step 1: Identify the forces acting on the monkey
The forces acting on the monkey are:
1. The gravitational force (weight) acting downward: \( W_m = m_m \cdot g \)
2. The tension \( T \) in the rope acting upward.

Where:
- \( m_m = 20 \, \text{kg} \) (mass of the monkey)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 2: Calculate the weight of the monkey
Using the formula for weight:
\[ W_m = m_m \cdot g = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Determine the maximum tension in the rope
The problem states that the rope can hold a maximum mass of \( 25 \, \text{kg} \) before breaking. Therefore, the maximum tension \( T_{\text{max}} \) in the rope is:
\[ T_{\text{max}} = m_{\text{max}} \cdot g = 25 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 250 \, \text{N} \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force acting on the monkey when it climbs up with acceleration \( a \) is given by:
\[ T - W_m = m_m \cdot a \]
Substituting the known values:
\[ T - 200 \, \text{N} = 20 \, \text{kg} \cdot a \]

### Step 5: Substitute the maximum tension into the equation
Using the maximum tension calculated:
\[ 250 \, \text{N} - 200 \, \text{N} = 20 \, \text{kg} \cdot a \]
This simplifies to:
\[ 50 \, \text{N} = 20 \, \text{kg} \cdot a \]

### Step 6: Solve for the maximum acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{50 \, \text{N}}{20 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration with which the monkey can climb up along the rope is:
\[ \boxed{2.5 \, \text{m/s}^2} \]