A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down `(g=10(m)/(s^2)`)

Options

A. `1(m)/(s^2)`
B. `2.5(m)/(s^2)`
C. `10(m)/(s^2)`
D. `5(m)/(s^2)`

Explanation

To solve the problem, we need to analyze the forces acting on the fireman as he slides down the pole.

### Step-by-Step Solution:

1. **Identify the Forces:**
- The weight of the fireman (W) acting downwards:
\[ W = m \cdot g = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \]
- The normal force (R) with which he presses the pole is given as 600 N.
- The frictional force (f) opposing the motion is given by:
\[ f = \mu \cdot R = 0.5 \cdot 600 \, \text{N} = 300 \, \text{N} \]

2. **Apply Newton's Second Law:**
- According to Newton's second law, the net force (F_net) acting on the fireman can be expressed as:
\[ F_{\text{net}} = W - f \]
- Substituting the values:
\[ F_{\text{net}} = 600 \, \text{N} - 300 \, \text{N} = 300 \, \text{N} \]

3. **Calculate the Acceleration:**
- Using Newton's second law, \( F_{\text{net}} = m \cdot a \), we can find the acceleration (a):
\[ 300 \, \text{N} = 60 \, \text{kg} \cdot a \]
- Rearranging for acceleration:
\[ a = \frac{300 \, \text{N}}{60 \, \text{kg}} = 5 \, \text{m/s}^2 \]

4. **Conclusion:**
- The acceleration of the fireman as he slides down the pole is \( 5 \, \text{m/s}^2 \).

### Final Answer:
The acceleration with which the fireman slides down is \( 5 \, \text{m/s}^2 \).