A body of mass 10kg lies on aroughhorizontalsurface when a horizontalforce of fn actson it gets an acceleration of5m/s2 and when the horizontal force is doubled it gets an acceleration of 18m/s2 if g=10m/s2 the coefficient of frictionis
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The answer I got is 0.8 . Let Ffri be the frictional force . So from the f.b.d , We get Fext– Ffri = ma
So we get F – Ffri = 10 kg x 5 m/s2 = 50 ……..(1)
Also 2F – Ffri = 180 ………(2)
(1) X 2 gives 2F – 2 Ffri = 100 …….(3)
Solving (2) and (3) , we get Ffri = 180 – 100 = 80
But Ffri = µN = µmg = 100µ
So 100µ = 80 or µ = 0.8
So we get F – Ffri = 10 kg x 5 m/s2 = 50 ……..(1)
Also 2F – Ffri = 180 ………(2)
(1) X 2 gives 2F – 2 Ffri = 100 …….(3)
Solving (2) and (3) , we get Ffri = 180 – 100 = 80
But Ffri = µN = µmg = 100µ
So 100µ = 80 or µ = 0.8
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