A particle moves on a rough horizontal ground with some initial velocity say $v_{0}$ . If ( $\frac{3}{4}$ )th of its kinetic energy is lost in friction in time $t_{0}$ , then coefficient of friction between the particle and the ground is:

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$\frac{v_{0}}{{gt}_{0}}$

$\frac{v_{0}}{2 {gt}_{0}}$

$\frac{v_{0}}{4 {gt}_{0}}$

$\frac{3 v_{0}}{4 {gt}_{0}}$

answer is A.

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Detailed Solution

3/4h energy is lost, i.e., $\frac{1}{4} th$ kinetic energy is left.
Hence, its velocity becomes $\frac{v_{0}}{2}$ under a retardation of $μ$ mg in time $t_{0}$ .
$\frac{v_{0}}{2} = v_{0} - {μgt}_{0} or μ = \frac{v_{0}}{2 {gt}_{0}}$

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