Ch5solutions

# Ch5solutions - i Â X:r~Â(o â€œâ€˜3â€˜ V64" â€œE 9 Â...

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Unformatted text preview: i Â» ,, X -.:r~Â§(o â€œâ€˜3â€˜ V64" _ â€œE 9 , Â» â€œXe->1; = V34 â€™2'â€œ? * WW; . L Problem 5-2 solution a) False The net force is not zero because there is one force still operating: the sliding friction force slowing the motion down b) False Since the friction force is constant, there is constant deceleration from the instant the block leaves the instructorâ€™s hands c) True The weight only depends on gravity and mass: W=mg d) True This is the wonderful fact that sliding friction depends only on the normal force and the coefficient of friction as given by F11: = ,ukN e) False They are equal, but by the second law. The normal force on the block is equal and opposite to the normal force on the table. The weight of the block is and now folks, the drum roll please, equal and opposite to the pull of the block on the earth. Omigosh! f) True By Newtonâ€™s third law, this is another action-reaction pair. It acts on the table in equal and opposite direction to the friction force on the block. This tends to want to move the table forward. g) True This is exactly the constant force model weuse for that sliding friction h) True If thereâ€™s no sliding thereâ€™s no sliding friction. The sliding friction goes from a constant value to zero in an instant. i) False The net vertical force has been zero all along. ...
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Ch5solutions - i Â X:r~Â(o â€œâ€˜3â€˜ V64" â€œE 9 Â...

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