Determine the maximum acceleration in $m/s^2$ of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is $0.15.$
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Answer since the acceleration of the box is due to the static friction.
$m a=f_{s} \leq \mu_{s} N=\mu_{s} m g$
ie. $a \leq \mu_{s} g$
$\therefore a_{\max }=\mu_{ s } g=0.15 \times 10 m s ^{-2}$
$=1.5 m s ^{-2}$
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