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)

Question

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) (A) 5.0 ms-2 (B) 3.0 ms-2 (C) 1.5 ms-2 (D) 15 ms-2

Tardigrade Admin · Accepted Answer

Key Idea Motion is possible when m a ≥ μ m g. The maximum static frictional force =μs m g should be equal to force F=m a <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d46e625a629240edd94aa7550fdfa29d-.png" /> ∴ F=m a=μs m g ⇒ a=μs g Here, μs=0.3 . g =10 ms -2 a=(0.3) × 10 ⇒ a=3 ms -2

Q. Maximum acceleration of the train in which a $50 k g$ 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 $g = 10 m s^{- 2}$ )

$5.0 m s^{- 2}$

$3.0 m s^{- 2}$

$1.5 m s^{- 2}$

$15 m s^{- 2}$

Key Idea Motion is possible when $ma \geq μ m g$ .
The maximum static frictional force $= μ_{s} m g$ should be equal to force $F = ma$

$∴ F = ma = μ_{s} m g$
$\Rightarrow a = μ_{s} g$
Here, $μ_{s} = 0.3. g = 10 m s^{- 2}$
$a = (0.3) \times 10$
$\Rightarrow a = 3 m s^{- 2}$

Q. Maximum acceleration of the train in which a 50kg 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=10ms−2)

5.0ms−2

3.0ms−2

1.5ms−2

15ms−2