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

Loss in KE = |work done by friction|

3/4 energy is lost means 1/4 energy is left.So velocity becomes v•/2Now, v•/2=v• -ug×t•So u= v•/ 2gt•

Dear Student,
Please find below the solution to your problem.

Initial Kinetic Energy = 0.5​mv0^2​
Let final velocity is v,
⇒ Final Kinetic Energy = 0.5​mv^2
= 0.5​mv0^2​ − 0.75​×0.5​mv0^2​
⇒ v = v0/2​​
Using v = u+at, Acceleration a = ​v0/−2t0​​
Retardation, a = ​v0/2t0​​
Let friction coefficient is μ ⇒ a = μg
⇒ μ = ​v0/2gt0​​

Thanks and Regards