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

Question

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

Tardigrade Admin · Accepted Answer

KE lost is (3/4) th, therefore, KE left is (1/4) th. Hence, velocity of particle reduces from v0 to (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 $v_{0} .$ If $\frac{3}{4}$ th of its kinetic energy is lost due to friction in time $t_{0}$ , the coefficient of friction between the particle and the ground is

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

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

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

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

$K E$ lost is $\frac{3}{4}$ th, therefore, $K E$ left is $\frac{1}{4}$ th.
Hence, velocity of particle reduces from $v_{0}$ to $\frac{v _{0}}{2} = v_{0} - μg t_{0}$
or $μ = \frac{v _{0}}{2 g t _{0}}$

Q. A particle moves on a rough horizontal ground with some initial velocity v0​. If 43​ th of its kinetic energy is lost due to friction in time t0​, the coefficient of friction between the particle and the ground is

2gt0​v0​​

4gt0​v0​​

4gt0​3v0​​

gt0​v0​​