A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in first two seconds. Find the coefficient of kinetic friction between the two.Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
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Solution :
From the Free body diagram :
R-mgCosθ=0
R=mg Cosθ-----------------(1)
For the block :]
u= 0 m/s
s=8m
t=2 sec
From second equation of motion : S=ut+1/2 at²
8=0+1/2 ax4
a=16/4=4 m/s2
Again, μR +ma-mgsinθ=0
From equation (1),
μmgcosθ+ma-mgsinθ=0
m(μgcosθ+a-g sinθ)=0
μx10 x cos30°=g sin30°-a
μx10 √3/2=10 x1/2 - a
5√3 μ=1
μ=(1/5√3)=0.11
Hence the coefficient of Kinetic friction between the two is 0.11.
From the Free body diagram :
R-mgCosθ=0
R=mg Cosθ-----------------(1)
For the block :]
u= 0 m/s
s=8m
t=2 sec
From second equation of motion : S=ut+1/2 at²
8=0+1/2 ax4
a=16/4=4 m/s2
Again, μR +ma-mgsinθ=0
From equation (1),
μmgcosθ+ma-mgsinθ=0
m(μgcosθ+a-g sinθ)=0
μx10 x cos30°=g sin30°-a
μx10 √3/2=10 x1/2 - a
5√3 μ=1
μ=(1/5√3)=0.11
Hence the coefficient of Kinetic friction between the two is 0.11.
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