Question

A body of mass m moves with the velocity v on a surface whose friction co-efficient is x . If the
body covers distance s before coming to rest, then v will be
(A) square root of 2xgs (B)square root of xgs
(C) square root of xgs/2
(D) square root of 3xgs

explain with answer

Answer 1

  from the laws of motion:       2 a s =  v² - u²

        =>       v² =  - 2 a s    as  initial velocity = v  and  final is 0.

      friction force = f =  m a = - x m g

              =>     a = - x g  substituting in the above equation for v, we get 

              v = √(2 x g s) 

==================================

   Kinetic energy of the body = 1/2 * m * v²

  Work done by friction when it stops the body = Force * distance

              = (μ m g) * (- s) 

   work done by friction = - initial kinetic energy + final kinetic energy

         - μ m g s =  0 - 1/2 m v²

    hence,        v² = 2 μ g s     

 so         v = √(2μgs)

Answer 2

Answer:

final velocity=0

initial velocity=v

work is done against friction