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

2 m s^{- 2}

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4 m s^{- 2}

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6 m s^{- 2}

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8 m s^{- 2}

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Solution

The correct option is A $2 m s^{- 2}$
As acceleration of the box is due to static friction,
$∴ m a = f_{s} \leq μ_{s} N = μ_{s} m g$
$a \leq μ_{s} g$
$∴ a_{m a x} = μ_{s} g = 0.2 \times 10 m s^{- 2} = 2 m s^{- 2}$

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The coefficient of static friction between the box and the train’s floor is 0.2. The maximum acceleration of the train upto which a box lying on its floor will remain stationary with respect to train is: (Take g=10ms−2)

The correct option is A 2ms−2As acceleration of the box is due to static friction, ∴ma=fs≤μsN=μsmg a≤μsg ∴amax=μsg=0.2×10ms−2=2ms−2

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

The correct option is A $2 m s^{- 2}$
As acceleration of the box is due to static friction,
$∴ m a = f_{s} \leq μ_{s} N = μ_{s} m g$
$a \leq μ_{s} g$
$∴ a_{m a x} = μ_{s} g = 0.2 \times 10 m s^{- 2} = 2 m s^{- 2}$