A pot of mass 1kg is raised vertically at a constant velocity with both the hands. The coefficient of friction between hand and pot is 0.2 .What is the force applied by each hand?

Question

Vikas TU · Answer

Let the normal force exerted by one hand be x.
Then, frictional force = 2x × 0.2 =0.4x
As, velocity is constant ; therefore net force is zero.
So, weight of pot is equal to frictional force.
: 0.4x = 9.8
x= *24.5 N

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A pot of mass 1kg is raised vertically at a constant velocity with both the hands. The coefficient of friction between hand and pot is 0.2 .What is the force applied by each hand?

A pot of mass 1kg is raised vertically at a constant velocity with both the hands. The coefficient of friction between hand and pot is 0.2 .What is the force applied by each hand?

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A pot of mass 1kg is raised vertically at a constant velocity with both the hands. The coefficient of friction between hand and pot is 0.2 .What is the force applied by each hand?

A pot of mass 1kg is raised vertically at a constant velocity with both the hands. The coefficient of friction between hand and pot is 0.2 .What is the force applied by each hand?

1 Answers

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