# 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

2
by rajusetu
is tghe block sliding with acceleration?
not tghe it is the

2014-10-12T09:36:48+05:30

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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

sir exams are approching
wait. let me see
3 * square root of 3/5 ??? it cannot be.. it is more than 1. coefficient of friction is usually less than 1.
mu = (1/3) * root(3) / 5 perhaps
2014-10-12T10:32:37+05:30

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=  μN

So we have N = mgcos30

=  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

plz mark as best
but answer is 3*square root of3/5
it was given in text book