Question

A block slides down an inclined surface of inclination 30 degrees with horizontal .Starting from rest it covers 8 M in first two seconds.find the coefficient of kinetic friction between the two.
please tell the question with detailed steps and tell me the basics of this question also.Please tell fast

Answer 1

see diagram.

let the coefficient of kinetic friction = μ.      Frictional force Ff acts upwards along the inclined surface.      Ff = μ * Normal reaction N

Since the block is not moving perpendicular to the incline. So forces arebalanced in that direction.
                            mg Cos 30° = N
           So Ff = μ * N = μ * m * g Cos 30°

Resultant Force along the incline downwards :   mg Sin 30° - Ff
                so mass m * acceleration a  = m g 1/2 - μ * m * g √3 /2 
                           a =  (1 - μ √3) * g/2

let g = 10 m/sec²

formula for distance traveled along the incline:  s = u t + 1/2 a t²
 
             8 meters = 0 + 1/2 [ (1 - μ √3) * 10/2 ]  *  2²
                8 = 10 (1 -  μ √3)
              
                    1 - μ √3  = 8/10 = 0.8
                      μ √3  = 0.2

                    μ = 0.1155
           

Answer 2

$f_{k}$ =  μN

So we have N = mgcos30

$f_{k}$ =  μmgcos30

now the 
s = ut + 1/2at²
8 = 1/2a . 4
a = 4 m/s²

so now we get a equation for external forces

ma + μmgcos30 = mgsin30

cancelling m

a + μgcos30 = gsin30
μ = (gsin30 - a)/gcos30
μ = (5 - 4)/5√3    (g = 10 m/s²)
μ = 1/5√3
μ = 0.1154