Question
Determine the maximum acceleration 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.

Answer

Since the acceleration of the box is due to the static friction,
$
\begin{aligned}
\quad m a & =f_s \leq \mu_s N=\mu_s m g \\
\text { i.e. } \quad a & \leq \mu_s g \\
\therefore a_{\max } & =\mu_s g=0.15 \times 10 m s ^{-2} \\
& =1.5 m s ^{-2}
\end{aligned}
$

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