Question

a particle moves on a rough horizontal ground with some initial velocity is v0 if 3/4 of its kinetic energy is lost due to friction in time t0 if coefficient of friction between the particle and the ground is xv0/4gt0 find x ​

Answer 1

i hope you understand....

Answer 2

Answer:

⇒μ=2gt0v0

Explanation:

2gt0v0

Initial Kinetic Energy =21mv02

Let final velocity is v, ⇒ Final Kinetic Energy=21mv2=21mv02−43×21mv02

⇒v=2v0

Using v=u+at, Acceleration a=−2t0v0

Retardation, a=2t0v0

Let friction coefficient is μ ⇒a=μg

⇒μ=2gt0v0