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)

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

Tardigrade Admin · Accepted Answer

As acceleration of the box is due to static friction, ∴ ma=fs le μs N=μs mg a le μs g ∴ amax =μs g =0.2×10 m s-2 =2 m s-2

Q. 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 $g = 10 m s^{- 2}$ )

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

$4 m$ $s^{- 2}$

$6 m$ $s^{- 2}$

$8 m$ $s^{- 2}$

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

Q. 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 (Takeg=10ms−2)

2m s−2

4m s−2

6m s−2

8m s−2