Ch5solutions - i » ,, X -.:r~§(o “‘3‘ V64"...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: i » ,, X -.:r~§(o “‘3‘ V64" _ “E 9 , » “Xe->1; = V34 ’2'“? * 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 instructor’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 Newton’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 there’s no sliding there’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. ...
View Full Document

Page1 / 2

Ch5solutions - i » ,, X -.:r~§(o “‘3‘ V64"...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online