Q. A passenger is travelling in a train moving at 72 $kmh^{- 1}.$ His suitcase is kept on the berth. The driver of the train applies brakes such that the speed of the train decreases at a constant rate of 36 $kmh^{- 1}$ in 5 s. What should be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not the slide during retardation of the train?
A) $\frac{10}{49}$
B) $\frac{10}{98}$
C) $\frac{28}{49}$
D) $\frac{30}{98}$
Explanation
(a) Retardation of train = $\frac{36\ kmh^{- 1}}{5\ s}$
$= \frac{35 \times \frac{5}{18}ms^{- 1}}{5s} = 2\ ms^{- 2}$
It acts in the backward direction
Fictitious force on suitcase = 2$m$ N,
Where $m$ is the mass of suitcase
It acts in the forward direction
Due to this force, the suitcase has a tendency to slide forward. If suitcase is not to slide, then $2m =$ force $f$ of friction
or $2m = \mu mg$ or $\mu = \frac{2}{g} = \frac{2}{9.8} = \frac{20}{98} = \frac{10}{49}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.