Question

A particle moves on a rough horizontal ground with some initial velocity sayv0v0. If 3/4th3/4th of its kinetic energy is lost in friction in time t0t0. Then coefficient of friction between the particle and the ground is -

Answer 1

Answer:

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