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:

Correct option is

A 2gt0v0

Initial Kinetic Energy = 21mv02

Let final velocity is v, ⇒ Final Kinetic Energy=

21mv2 = 21mv02 43 x 21mv02

⇒v=2v0

Using v = u + at, Acceleration a = -2t0v0

Retardation, a = 2t0v0

Retardation, a = 2t0v0

Let friction coefficient is u⇒ a = µg

⇒ μ = 2gt0v0