A particle moves on a rough horizontal ground with some initial velocity say v0. If (34)th of its kinetic energy is lost in friction in time t0, then coefficient of friction between the particle and the ground is:

3/4h energy is lost, i.e., 14th kinetic energy is left.Hence, its velocity becomes v02 under a retardation of μmg in time t0.v02 = v0-μgt0 or μ = v02gt0

A particle moves on a rough horizontal ground with some initial velocity say v0. If (34)th of its kinetic energy is lost in friction in time t0, then coefficient of friction between the particle and the ground is:

3/4h energy is lost, i.e., 14th kinetic energy is left.Hence, its velocity becomes v02 under a retardation of μmg in time t0.v02 = v0-μgt0 or μ = v02gt0

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

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

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

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

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 $t_{0}$ .
$\frac{v_{0}}{2} = v_{0} - {μgt}_{0} or μ = \frac{v_{0}}{2 {gt}_{0}}$