A block of mass `2kg` is placed on the surface of trolley of mass `20kg` which is on a smooth surface.The coefficient of friction between the block and the surface of the trolley is `0.25`.If a horizontal force of `2 N` acts on the block, the acceleration of the system in `ms^(-2)` is `(g=10ms^(-2))`

Options

A. `1.8`
B. `1.0`
C. `0.9`
D. `0.09`

Explanation

To solve the problem, we need to analyze the forces acting on the block and the trolley, and then calculate the acceleration of the entire system.

### Step-by-Step Solution:

1. **Identify the Masses:**
- Mass of the block (m1) = 2 kg
- Mass of the trolley (m2) = 20 kg

2. **Calculate the Maximum Friction Force:**
- The coefficient of friction (μ) between the block and the trolley = 0.25
- The normal force (N) acting on the block is equal to its weight, which is given by:
\[ N = m1 \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]
- The maximum friction force (f_max) that can act on the block is:
\[ f_{\text{max}} = \mu \cdot N = 0.25 \cdot 20 \, \text{N} = 5 \, \text{N} \]

3. **Determine the Applied Force:**
- A horizontal force (F) of 2 N is applied to the block.

4. **Analyze the Motion:**
- Since the applied force (2 N) is less than the maximum friction force (5 N), the block will not slip off the trolley. Instead, both the block and the trolley will move together.

5. **Calculate the Total Mass of the System:**
- The total mass (M) of the system (block + trolley) is:
\[ M = m1 + m2 = 2 \, \text{kg} + 20 \, \text{kg} = 22 \, \text{kg} \]

6. **Calculate the Acceleration of the System:**
- The acceleration (a) of the system can be calculated using Newton's second law:
\[ a = \frac{F}{M} = \frac{2 \, \text{N}}{22 \, \text{kg}} = \frac{1}{11} \, \text{m/s}^2 \approx 0.0909 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the system is approximately \(0.0909 \, \text{m/s}^2\).

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