A particle moves on a rough horizontal ground with some initial velocity say v0. If (3/4)th of its kinetic energy is lost in friction in time t0, then coefficient of friction between the particle and the ground is:

Question

A particle moves on a rough horizontal ground with some initial velocity say v0. If (3/4)th of its kinetic energy is lost in friction in time t0, then coefficient of friction between the particle and the ground is: (A) (v0/2gt0) (B) (v0/4gt0) (C) (3v0/4gt0) (D) (v0/gt0)

Tardigrade Admin · Accepted Answer

3 / 4 text th energy is lost, i.e., 1 / 4 text th kinetic energy is left. Hence, its velocity becomes v0 / 2 under a retardation of mmg in time t0. (v0/2)=v0-μ g t0 or μ=(v0/2 g t0)

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

$\frac{v _{0}}{2 g t _{0}}$

$\frac{v _{0}}{4 g t _{0}}$

$\frac{3 v _{0}}{4 g t _{0}}$

$\frac{v _{0}}{g t _{0}}$

$3/ 4^{th}$ energy is lost, i.e., $1/ 4^{th}$ kinetic energy is left.
Hence, its velocity becomes $v_{0} /2$ under a retardation of $mm g$ in time $t_{0}$ .
$\frac{v _{0}}{2} = v_{0} - μg t_{0}$
or $μ = \frac{v _{0}}{2 g t _{0}}$

Q. A particle moves on a rough horizontal ground with some initial velocity say v0​. If (3/4)th of its kinetic energy is lost in friction in time t0​, then coefficient of friction between the particle and the ground is:

2gt0​v0​​

4gt0​v0​​

4gt0​3v0​​

gt0​v0​​