A passengee is traveling a train moving at `40 ms^(-1)` Hit suitcase is kept on the berth ,The drive of train applies breaks such that the speed of the train decreases at a constant rate to `20 ms^(-1)` in `5s` What should be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not to slide during and retardation of the train ?

Options

A. `0.3`
B. `0.5`
C. `0.1`
D. `0.2`

Explanation

To solve the problem step by step, we need to determine the minimum coefficient of friction required to prevent the suitcase from sliding on the berth during the train's deceleration.

### Step 1: Calculate the Retardation of the Train
The train's speed decreases from \(40 \, \text{m/s}\) to \(20 \, \text{m/s}\) in \(5 \, \text{s}\). We can calculate the retardation (deceleration) using the formula:

\[ a = \frac{v_f - v_i}{t} \]

Where:
- \(v_f = 20 \, \text{m/s}\) (final velocity)
- \(v_i = 40 \, \text{m/s}\) (initial velocity)
- \(t = 5 \, \text{s}\) (time)

Substituting the values:

\[ a = \frac{20 \, \text{m/s} - 40 \, \text{m/s}}{5 \, \text{s}} = \frac{-20 \, \text{m/s}}{5 \, \text{s}} = -4 \, \text{m/s}^2 \]

The negative sign indicates that this is a deceleration.

### Step 2: Understand the Forces Acting on the Suitcase
For the suitcase to not slide, the frictional force must be sufficient to provide the necessary acceleration (retardation in this case) to the suitcase. The frictional force \(F_f\) can be expressed as:

\[ F_f = \mu m g \]

Where:
- \(\mu\) is the coefficient of friction
- \(m\) is the mass of the suitcase
- \(g\) is the acceleration due to gravity (\(g \approx 10 \, \text{m/s}^2\))

### Step 3: Set Up the Equation for Forces
The suitcase must experience a backward acceleration of \(4 \, \text{m/s}^2\) (the same as the train's retardation). The net force required to achieve this acceleration is given by:

\[ F = ma = m \cdot 4 \]

### Step 4: Equate the Frictional Force to the Required Force
For the suitcase not to slide, the frictional force must be equal to or greater than the force required to accelerate the suitcase:

\[ \mu mg \geq ma \]

Cancelling \(m\) from both sides (assuming \(m \neq 0\)) gives:

\[ \mu g \geq a \]

### Step 5: Solve for the Coefficient of Friction
Substituting \(g = 10 \, \text{m/s}^2\) and \(a = 4 \, \text{m/s}^2\):

\[ \mu \cdot 10 \geq 4 \]

Dividing both sides by \(10\):

\[ \mu \geq \frac{4}{10} = 0.4 \]

### Conclusion
The minimum coefficient of friction required to prevent the suitcase from sliding is:

\[ \mu \geq 0.4 \]