Maximum acceleration of the train in which a 50 kg box lying on its floor will ramain stationary is (Given: Co-efficient of static friction between the box and the train's floor is `0.3 and g =10ms^(-2))`

Options

A. `5.0ms^(-2)`
B. `3.0 ms^(-2)`
C. `1.5 ms^(-2)`
D. `15 ms^(-2)`

Answer

To solve the problem of finding the maximum acceleration of the train such that a 50 kg box remains stationary on its floor, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Data**:
- Mass of the box, \( m = 50 \, \text{kg} \)
- Coefficient of static friction, \( \mu = 0.3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Normal Force**:
- The normal force \( N \) acting on the box is equal to its weight, which can be calculated using:
\[ N = m \cdot g = 50 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 500 \, \text{N} \]

3. **Calculate the Maximum Frictional Force**:
- The maximum static frictional force \( F_{\text{friction}} \) that can act on the box is given by:
\[ F_{\text{friction}} = \mu \cdot N = 0.3 \cdot 500 \, \text{N} = 150 \, \text{N} \]

4. **Set Up the Equation for Pseudo Force**:
- When the train accelerates, a pseudo force \( F_{\text{pseudo}} \) acts on the box in the opposite direction of the train's acceleration. This force can be expressed as:
\[ F_{\text{pseudo}} = m \cdot a \]
- Here, \( a \) is the acceleration of the train.

5. **Equate the Pseudo Force to the Maximum Frictional Force**:
- For the box to remain stationary relative to the train, the pseudo force must be equal to the maximum static frictional force:
\[ m \cdot a = F_{\text{friction}} \]
- Substituting the values we have:
\[ 50 \, \text{kg} \cdot a = 150 \, \text{N} \]

6. **Solve for the Maximum Acceleration \( a \)**:
- Rearranging the equation gives:
\[ a = \frac{150 \, \text{N}}{50 \, \text{kg}} = 3 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration of the train in which the 50 kg box will remain stationary is:
\[ \boxed{3 \, \text{m/s}^2} \]