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Find coefficient of kinetic friction
Asked by ashutosharnold1998 | 24 Feb, 2020, 10:14: AM
Expert Answer
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 Thiyagarajan K | 24 Feb, 2020, 04:36: PM
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