KCET · Physics · Laws of Motion
Maximum acceleration of the train in which a \( 50 \mathrm{~kg} \) 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
\( g=10 m s^{-2} \) )
- A \( 5.0 \mathrm{~ms}^{-2} \)
- B \( 3.0 \mathrm{~ms}^{-2} \)
- C \( 1.5 \mathrm{~ms}^{-2} \)
- D \( 15 m s^{-2} \)
Answer & Solution
Correct Answer
(B) \( 3.0 \mathrm{~ms}^{-2} \)
Step-by-step Solution
Detailed explanation
Given, mass of box \(=50 \mathrm{~kg}\); coefficient of static friction \(=0.3 ; g=10 \mathrm{~ms}^{-2}\)
Train moving along this direction
In order that the box remains stationary, we must have \(m a=\mu N\)
\(\Rightarrow m a=\mu m g\)
\(\Rightarrow a=\mu g=0.3 \times 10=3\)
Therefore, maximum acceleration of the train \(=3 \mathrm{~ms}^{-2}\)
Train moving along this direction
In order that the box remains stationary, we must have \(m a=\mu N\)
\(\Rightarrow m a=\mu m g\)
\(\Rightarrow a=\mu g=0.3 \times 10=3\)
Therefore, maximum acceleration of the train \(=3 \mathrm{~ms}^{-2}\)
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