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# Find coefficient of kinetic friction

Asked by ashutosharnold1998 24th February 2020, 10:14 AM
Figure shows the different forces acting on a block moving on a inclined plane of angle 30o

mg = weight of block. m is mass and g is acceleration due to gravity

N = Normal reaction foce by inclined surface at the contact region of block and inclined plane
This normal force equals the force component mg cos30 which is vertical to incline plane.

Hence friction force = μN = μ mg cos30

where μ is friction coefficient

Hence the net force acting on the block for its motion = mgsin30 - μ mg cos30 = mg [ (1/2) - μ (√3/2) ]

Hence acceleration a = force / mass = (g / 2 ) [ 1- √3 μ ]  ...........................(1)

if the block travels 12 m in ints initial 3 sec duration of time starting from rest,

then its acceleration a  is obtained from, S = (1/2)at2

Hence we get,  12 = (1/2) a × 3 × 3  , i.e., a = 8/3  ..........................(2)

By equating eqn.(1) and eqn.(2), we get  friction coefficient μ as

8/3 = (g / 2 ) [ 1- √3 μ ]

By substituting g = 9.8 m/s2 in above equation, we get, μ = 0.26
Answered by Expert 24th February 2020, 4:36 PM
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