Question

a block of mass m is released on the top of a smooth inclined plane of length x and inclination @ Horizontal surface is rough. If block comes to rest after moving a distance d on the horizontal surface, then coefficient of friction between block and surface is

Answer 1

Answer:

the coefficient of friction is depends upon the angle between inclined plane and their surface

Answer 2

Answer:

μ  =  x sin @ / d

Explanation:

Given that

Inclination =  @ =θ

Length of incline surface = x

Length of horizontal surface before coming to rest = d

A to B:

At position A speed of block is zero

The speed of block at point B is V

From diagram

h = x sin θ

From work power energy theorem

m g h = 1 /2 m V²

V² = 2 g h = 2 g x sin θ

$V=\sqrt{ 2 g x sin \theta}$        ---------1

B to C:

We know that

Friction Fr = μ m g

The final velocity of block is zero

From work power energy theorem

- μ m g  .d =  -  1 /2 m V²

μ  g  .d =    1 /2  V²

V² =  2  μ  g  .d

From equation 1

V² =  2 g x sin θ =  2  μ  g  .d

x sin θ =   μ    .d

 μ  =  x sin θ / d

μ  =  x sin @ / d

This is  coefficient of friction between block and surface.