This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full 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
- Spring '08