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Question: The coefficient of static friction between the box and the train’s floor is 0.2. The maximum acceler...

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=10ms−2g = 10ms^{- 2}) :

A

2ms−22ms^{- 2}

B

4ms−24ms^{- 2}

C

16ms−216ms^{- 2}

D

18ms−218ms^{- 2}

Answer

2ms−22ms^{- 2}

Explanation

Solution

As acceleration of the box is due to static friction,

}{a \leq \mu_{s}g }$$$\therefore{a{s^{- 2}}^{- 2}}_{\max}$