A block of mass 10 kg is released on rough incline plane inclined at an angle 30 degree. Block start descending with acceleration 2 m/s² . Kinetic friction force acting on block is ( take g= 10m/s² )

m= 10 kg
g = 10 m/sec2
θ = 30
a = 2m/sec2
f = ??

formula => f = m g sin (theta) - ma
                     substituting the required ,
      f = 10 * 10 * sin (30) - 10 * 2
      = 100 * 1/2 - 20
     = 50 - 20
     = 30 kg m/sec2
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F= mgsintheta - frictional force
==> Frictional force= mgsintheta - ma
Substituting the given values,
f = 10*10*1/2 - (10*2)
= 30N
 
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This may be lot of help u

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30N...1
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30N...
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m= 10 kg
g = 10 m/sec2
??= 30
a = 2m/sec2
f = ??

formula => f = m g sin?(theta) - ma
? ? ? ? ? ? ? ? ? ? ?substituting the required ,
? ? ? f = 10 * 10 * sin (30) - 10 * 2
? ? ? = 100 * 1/2 - 20
? ? ?= 50 - 20
? ? ?= 30 kg m/sec2
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30N.............
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