Question

The coefficient of static friction between the box and the train's floor is 0.2. The maximum acceleration of the train in which a box lying on its floor will remain stationary is
(Take g=10 $m{s}^{-2}$)

• A. 2$m{s}^{-2}$
• B. 4$m{s}^{-2}$
• C. 6 $m{s}^{-2}$
• D. 8 $m{s}^{-2}$

Answer 1

Answer: (A)

2$m{s}^{-2}$

As acceleration of the box is due to static friction,
$\therefore ma=f_s\le {\mu }_{s}N={\mu }_{s}mg$
$a\le {\mu }_{s}g$
$\therefore a_{max}=\mu g=0.2\times 10m{s}^{-2}$=2$m{s}^{-2}$