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  4. Maximum acceleration of the train in which a 50 kg box lying on its floor will remain stationary (Given: Co-efficient of static friction between the box and the train’s floor is 0.3 and g = 10 ms-2)

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Q. Maximum acceleration of the train in which a $50\, kg$ box lying on its floor will remain stationary (Given : Co-efficient of static friction between the box and the train’s floor is $0.3$ and $g = 10 \,ms^{-2}$)

KCETKCET 2016Laws of Motion

Solution:

Key Idea Motion is possible when $m a \geq \mu m g$.
The maximum static frictional force $=\mu_{s} m g$ should be equal to force $F=m a$
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$\therefore F=m a=\mu_{s} m g $
$ \Rightarrow a=\mu_{s} g $
Here, $ \mu_{s}=0.3 . g =10 \,ms ^{-2} $
$ a=(0.3) \times 10 $
$ \Rightarrow a=3\, ms ^{-2}$