One end of massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum vertical force that the clamp can bear is 360 N. With what value of minimum safe acceleration ( in `ms^(-2)`) can a man of 60 kg climp down the rops ?

Options

A. `10`
B. `5`
C. `8`
D. `4`

Explanation

To solve the problem step by step, we will analyze the forces acting on the man climbing down the rope and apply Newton's second law of motion.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The weight of the man (W) acting downwards: \( W = mg \)
- The tension (T) in the rope acting upwards.

2. **Write Down the Given Values:**
- Mass of the man, \( m = 60 \, \text{kg} \)
- Maximum tension the clamp can bear, \( T_{\text{max}} = 360 \, \text{N} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

3. **Calculate the Weight of the Man:**
\[ W = mg = 60 \, \text{kg} \times 10 \, \text{m/s}^2 = 600 \, \text{N} \]

4. **Apply Newton's Second Law:**
- According to Newton's second law, the net force acting on the man is equal to the mass of the man multiplied by his acceleration:
\[ W - T = ma \]
where \( a \) is the acceleration of the man downwards.

5. **Substitute the Known Values:**
- Rearranging the equation gives:
\[ 600 \, \text{N} - T = 60 \, \text{kg} \times a \]
- We know that the maximum tension \( T \) can be \( 360 \, \text{N} \):
\[ 600 \, \text{N} - 360 \, \text{N} = 60 \, \text{kg} \times a \]

6. **Calculate the Net Force:**
\[ 240 \, \text{N} = 60 \, \text{kg} \times a \]

7. **Solve for Acceleration \( a \):**
\[ a = \frac{240 \, \text{N}}{60 \, \text{kg}} = 4 \, \text{m/s}^2 \]

### Final Answer:
The minimum safe acceleration with which the man can climb down the rope is \( 4 \, \text{m/s}^2 \).

---