A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs `1,000N` exerts a force of `450N` on the chair downwards while pulling the rope on the other side. If the chair weighs `250N` then the acceleration of the chair is .

Options

A. `0.45m//s^(2)`
B. `0`
C. `2m//s^(2)`
D. `9//25m//s^(2)`

Explanation

To solve the problem, we need to analyze the forces acting on the system consisting of the man and the chair.

### Step 1: Identify the Forces
1. The weight of the man (W_m) = 1000 N (downward).
2. The force exerted by the man on the chair (F_m) = 450 N (downward).
3. The weight of the chair (W_c) = 250 N (downward).
4. The tension in the rope (T) acts upward on the chair.

### Step 2: Write the Net Force Equation
The net force acting on the chair can be expressed as:
\[ F_{\text{net}} = T - (W_c + F_m) \]
Where:
- \( W_c + F_m \) is the total downward force acting on the chair.

### Step 3: Calculate the Total Downward Force
Calculate the total downward force on the chair:
\[ W_c + F_m = 250 \, \text{N} + 450 \, \text{N} = 700 \, \text{N} \]

### Step 4: Apply Newton's Second Law
According to Newton's second law, the net force is also equal to the mass of the chair multiplied by its acceleration (a):
\[ F_{\text{net}} = m_c \cdot a \]
Where \( m_c \) is the mass of the chair. The weight of the chair is given by:
\[ W_c = m_c \cdot g \]
Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). Thus, we can find the mass of the chair:
\[ m_c = \frac{W_c}{g} = \frac{250 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 25.5 \, \text{kg} \]

### Step 5: Substitute and Solve for Acceleration
Now, we can substitute the values into the net force equation:
\[ T - 700 \, \text{N} = m_c \cdot a \]
To find the tension (T), we need to consider the forces acting on the man. The man is pulling the rope with a force of 450 N, which means the tension in the rope is:
\[ T = 450 \, \text{N} \]

Now substitute T into the net force equation:
\[ 450 \, \text{N} - 700 \, \text{N} = 25.5 \, \text{kg} \cdot a \]
\[ -250 \, \text{N} = 25.5 \, \text{kg} \cdot a \]
Now, solve for acceleration (a):
\[ a = \frac{-250 \, \text{N}}{25.5 \, \text{kg}} \approx -9.8 \, \text{m/s}^2 \]

### Step 6: Interpret the Result
The negative sign indicates that the chair is accelerating downward.

### Final Answer
The acceleration of the chair is approximately \( 9.8 \, \text{m/s}^2 \) downward.

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