The minimum value of the coefficient of friction between the object and the floor of the truck which makes rolling of the object possible is $μ \geq \frac{a}{k g}$ . what will be k?

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Solution

The maximum value of frictional force is $2 μ m g$
$∴ \frac{2}{3} m a \leq 2 μ m g$
$\Rightarrow μ \geq \frac{a}{3 g} .$
Hence, the answer is $μ \geq \frac{a}{3 g} .$

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The minimum value of the coefficient of friction between the object and the floor of the truck which makes rolling of the object possible is μ≥akg. what will be k?

The maximum value of frictional force is 2μmg∴23ma≤2μmg⇒μ≥a3g.Hence, the answer is μ≥a3g.

The minimum value of the coefficient of friction between the object and the floor of the truck which makes rolling of the object possible is $μ \geq \frac{a}{k g}$ . what will be k?

The maximum value of frictional force is $2 μ m g$
$∴ \frac{2}{3} m a \leq 2 μ m g$
$\Rightarrow μ \geq \frac{a}{3 g} .$
Hence, the answer is $μ \geq \frac{a}{3 g} .$