A 10 kg block kept on an inclined plane is pulled by a string applying 200 N force. A 10 N force is also applied on 10 kg block as shown in figure. Find :
acceleration of 10 kg block.

Explanation

To solve the problem, we need to analyze the forces acting on the 10 kg block on the inclined plane. Let's break it down step by step.

### Step 1: Identify the Forces Acting on the Block
1. **Weight of the Block (W)**: The weight of the block is given by \( W = m \cdot g = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \).
2. **Force Applied via String (T)**: The tension in the string pulling the block up the incline is \( T = 200 \, \text{N} \).
3. **Additional Force (F)**: There is also a downward force of \( F = 10 \, \text{N} \) acting on the block.
4. **Normal Force (N)**: The normal force acts perpendicular to the inclined surface.
5. **Component of Weight Along the Incline**: The component of the weight acting down the incline can be calculated as \( W_{\text{parallel}} = W \cdot \sin(\theta) \), where \( \theta = 30^\circ \).

### Step 2: Calculate the Components of the Weight
- The component of the weight acting down the incline is:
\[ W_{\text{parallel}} = 100 \cdot \sin(30^\circ) = 100 \cdot \frac{1}{2} = 50 \, \text{N} \]

### Step 3: Set Up the Equation of Motion
- The net force acting along the incline can be expressed as:
\[ F_{\text{net}} = T - W_{\text{parallel}} - F \]
- Substituting the known values:
\[ F_{\text{net}} = 200 \, \text{N} - 50 \, \text{N} - 10 \, \text{N} = 140 \, \text{N} \]

### Step 4: Apply Newton's Second Law
- According to Newton's second law, \( F_{\text{net}} = m \cdot a \):
\[ 140 \, \text{N} = 10 \, \text{kg} \cdot a \]
- Solving for acceleration \( a \):
\[ a = \frac{140 \, \text{N}}{10 \, \text{kg}} = 14 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the 10 kg block is \( \mathbf{14 \, m/s^2} \).

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