A particle moves on a rough horizontal ground with some initial velocity, say ν0. If 34th of its kinetic energy is lost due to friction in time t0, then coefficient of friction between the particle and the ground is

A particle moves on a rough horizontal ground with some initial velocity, say $ν_{0}$ . If $(\frac{3}{4}) th$ of its kinetic energy is lost due to friction in time t₀, then coefficient of friction between the particle and the ground is

A
$\frac{v_{0}}{2 g t_{0}}$
B
$\frac{v_{0}}{4 g t_{0}}$
C
$\frac{3 v_{0}}{4 g t_{0}}$
D
$\frac{v_{0}}{g t_{0}}$

Solution:

$\frac{3}{4}$ th of kinetic energy is lost. Hence, left kinetic energy is $\frac{1}{4}$ th.

$∴ v^{2} = \frac{v_{0}^{2}}{4} \Rightarrow v = \frac{v_{0}}{2} = v_{0} - a t_{0} = v_{0} - μ g t_{0}$

or $μ = \frac{v_{0}}{2 g t_{0}}$

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