If the maximum acceleration of a moving platform to keep a box of mass 5 kg on it without sliding is \(3ms^{-1}\) , then the static friction between the box and floor of the platform is \((g = 10ms^{-2})\)
KEAM - 2026,
Answer: C
The coefficient of static friction is 0.30.
The box accelerates with the platform only because of static friction, which can supply at most \(\mu mg\). So the maximum acceleration without sliding is \(a_{max} = \mu g\).
\[ \mu = \frac{a_{max}}{g} = \frac{3}{10} = 0.30 \]
The mass of 5 kg cancels and is not needed. (The acceleration is printed with the unit m s-1; it is an acceleration of 3 m s-2.) 0.15 and 0.4 do not equal \(a/g\).
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