A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block

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Arun · Answer

Given, m = 2 kg θ = 30 0 μ = 0.6 R = m g cos ​θ Therefore, force of friction, F = ​μ R = ​μ m g cos ​θ = 0.6 X 2 X 9.8 X cos 30 0 = 10.16 N Regards Arun (askIITians forum expert)

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A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block

A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block

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A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block

A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block

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Other Related Questions on Mechanics

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