Question

A passenger is travelling inside a train moving at $40m{s}^{-1}$. His suitcase is kept on the berth. The driver of train applies breaks such that the speed of the train decreases at a constant rate to $20m{s}^{-1}$ in $4s$. What should be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not to slide during retardation of the train? (Take $g=10m/s^2$)

• A. $0.3$
• B. $0.5$
• C. $0.1$
• D. $0.2$

Answer 1

The correct option is B $0.5$

Let us assume that the train was initially moving in the rightward direction.
So, the below figure shows the FBD of the suitcase when there is retardation in the train.
Given, initial speed of the train $v_i=40m/s$
final speed of the train $v_f=20m/s$ and time take $t=4s$

$\therefore$ Retardation of train is given by
$a=\frac{v_i-v_f}{t}=\frac{40-20}{4}=\frac{20}{4}=5m{s}^{-2}$

Now, analyse the suitcase in the frame of train
From the FBD we have
$N=mg$
Since, the suitcase should not slide, the minimun coefficient of friction will be if the friction is limiting
So, the limiting friction is given by
$f_{\text{max}}=\mu N=\mu mg$
Hence, from the condition of equilibrium in the horizontal direction
$\Rightarrow f_{\text{max}}=ma\Rightarrow \mu mg=ma$
$\Rightarrow \mu =\frac{a}{g}=\frac{5}{10}=0.5$