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 1

Answer:

The maximum acceleration of the  = 1.47 ms⁻²

Explanation:

As per the question,

Its given that the acceleration of the box lying on the floor is due to the static friction between box and the train's floor as the box lying is always in stationary condition.

Therefore,

Force applied on box is always less than static frictional force.

Since,

Static frictional force is given by

$f_{s}=\mu_{s}R$

Where,

R = mg

Now,

Force applied = ma

Therefore,

$ma\leq \mu_{s}mg$

$a\leq \mu_{s}g$

$a\leq 0.15\times 9.8\\a\leq 1.47ms^{-2}$

Therefore,

The maximum acceleration of the train in which a box lying on its floor will remain stationary = 1.47 ms⁻²