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# 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 ?

Arun Kumar IIT Delhi
7 years ago
Loss in KE = |work done by friction|
$1/2mv^2*3/4=\mu mg*x\\ x=3v^2/8\mu g\\ v_{final}=v_{initial}-\mu gt_{0}\\ x=v_{initial}t_{0}-1/2\mu gt_{0}^2$
equate both x to get
$\mu$

Arun Kumar
IIT Delhi
Parul
11 Points
3 years ago
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•
Rishi Sharma
10 months ago
Dear Student,

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