A man of mass `65kg` is standing stationary with respect to a conveyor belt which is accelerating with `1m//s^(2)`. if `mu_(s)` is 0.2 the net force on the man and the maximum acceleration of the belt so that the man is stationary relative to the belt are `(g =10 m//s^(2))` .

Options

A. `zero, 2m//s^(2)`
B. `65N,2m//s^(2)`
C. `zero,1m//s^(2)`
D. `65N,1m//s^(2)`

Answer

To solve the problem step by step, we will analyze the forces acting on the man standing on the conveyor belt and calculate the required values.

### Step 1: Identify the Given Data
- Mass of the man, \( m = 65 \, \text{kg} \)
- Acceleration of the conveyor belt, \( a_b = 1 \, \text{m/s}^2 \)
- Coefficient of static friction, \( \mu_s = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Maximum Static Friction Force
The maximum static friction force (\( F_{f} \)) that can act on the man is given by the formula:
\[ F_{f} = \mu_s \cdot N \]
where \( N \) is the normal force. Since the man is standing on a horizontal surface, the normal force is equal to the weight of the man:
\[ N = m \cdot g = 65 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 650 \, \text{N} \]
Now, substituting the values:
\[ F_{f} = 0.2 \cdot 650 \, \text{N} = 130 \, \text{N} \]

### Step 3: Determine the Net Force on the Man
The net force (\( F_{net} \)) acting on the man due to the acceleration of the conveyor belt can be calculated using Newton's second law:
\[ F_{net} = m \cdot a_b = 65 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 65 \, \text{N} \]

### Step 4: Compare the Forces
For the man to remain stationary with respect to the conveyor belt, the maximum static friction force must be greater than or equal to the net force acting on him:
\[ F_{f} \geq F_{net} \]
Substituting the values we found:
\[ 130 \, \text{N} \geq 65 \, \text{N} \]
This condition is satisfied, meaning the man can remain stationary relative to the belt.

### Step 5: Calculate the Maximum Acceleration of the Belt
To find the maximum acceleration of the belt (\( a_{max} \)) that allows the man to remain stationary relative to it, we set the maximum static friction force equal to the force required to accelerate the man:
\[ F_{f} = m \cdot a_{max} \]
Substituting the known values:
\[ 130 \, \text{N} = 65 \, \text{kg} \cdot a_{max} \]
Solving for \( a_{max} \):
\[ a_{max} = \frac{130 \, \text{N}}{65 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Final Answers
- The net force on the man is \( 65 \, \text{N} \).
- The maximum acceleration of the belt so that the man is stationary relative to the belt is \( 2 \, \text{m/s}^2 \).