A monkey of mass 30 kg climbs a rope which can withstand a maximum tension of 360 N. The maximum acceleration which this rope can tolerate for the climbing of monkey is:
`(g=10m//s^(2))`

Options

A. `2 m//s^(2)`
B. `3 m//s^(2)`
C. `4 m//s^(2)`
D. `5 m//s^(2)`

Explanation

To find the maximum acceleration that the rope can tolerate while the monkey climbs, we can follow these steps:

### Step 1: Identify the forces acting on the monkey
When the monkey climbs the rope, two forces are acting on it:
1. The gravitational force (weight) acting downwards, which is given by \( W = mg \).
2. The tension \( T \) in the rope acting upwards.

### Step 2: Write down the equation of motion
According to Newton's second law, the net force acting on the monkey is equal to the mass of the monkey multiplied by its acceleration (\( a \)):
\[ F_{\text{net}} = T - mg = ma \]

### Step 3: Rearrange the equation
From the equation above, we can rearrange it to express the tension in terms of mass, gravitational force, and acceleration:
\[ T = mg + ma \]

### Step 4: Substitute known values
We know:
- The mass of the monkey \( m = 30 \, \text{kg} \)
- The maximum tension \( T = 360 \, \text{N} \)
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Now, we can calculate the weight of the monkey:
\[ mg = 30 \, \text{kg} \times 10 \, \text{m/s}^2 = 300 \, \text{N} \]

### Step 5: Substitute into the tension equation
Now substitute \( T \) and \( mg \) into the rearranged equation:
\[ 360 \, \text{N} = 300 \, \text{N} + 30 \, \text{kg} \cdot a \]

### Step 6: Solve for acceleration \( a \)
Now, isolate \( a \):
\[ 360 \, \text{N} - 300 \, \text{N} = 30 \, \text{kg} \cdot a \]
\[ 60 \, \text{N} = 30 \, \text{kg} \cdot a \]
\[ a = \frac{60 \, \text{N}}{30 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration that the rope can tolerate for the climbing of the monkey is \( 2 \, \text{m/s}^2 \).

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