The coefficient of static friction between the box and the train's floor is 0.2. The maximum acceleration of the train in which a box lying on its floor will remain stationary is `("Take g"=10ms^(-2))`

Options

A. `2"ms"^(-2)`
B. `4"ms"^(-2)`
C. `6"ms"^(-2)`
D. `8"ms"^(-2)`

Explanation

To solve the problem, we need to find the maximum acceleration of the train such that the box remains stationary relative to the train. We will use the concept of static friction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Coefficient of static friction (μ_s) = 0.2
- Acceleration due to gravity (g) = 10 m/s²

2. **Understand the Forces Acting on the Box:**
- The weight of the box (W) acts downward and is given by:
\[ W = mg \]
- The normal force (N) acting on the box is equal to its weight when the box is resting on the train's floor:
\[ N = mg \]

3. **Determine the Maximum Static Friction Force:**
- The maximum static friction force (F_friction) that can act on the box is given by:
\[ F_{friction} = \mu_s \cdot N = \mu_s \cdot mg \]
- Substituting the value of N:
\[ F_{friction} = \mu_s \cdot mg = 0.2 \cdot mg \]

4. **Relate the Forces to the Acceleration of the Train:**
- For the box to remain stationary relative to the train, the maximum static friction force must be equal to the force required to accelerate the box. This force is given by:
\[ F = ma \]
- Setting the two forces equal gives:
\[ ma \leq \mu_s \cdot mg \]

5. **Simplify the Equation:**
- We can cancel the mass (m) from both sides (assuming m ≠ 0):
\[ a \leq \mu_s \cdot g \]

6. **Substitute the Values:**
- Now substituting the values of μ_s and g:
\[ a \leq 0.2 \cdot 10 \]
- This simplifies to:
\[ a \leq 2 \, \text{m/s}^2 \]

7. **Conclusion:**
- The maximum acceleration of the train in which the box will remain stationary is:
\[ a = 2 \, \text{m/s}^2 \]

### Final Answer:
The maximum acceleration of the train is **2 m/s²**.