Asked by Anonymous
A 150 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 420 N. For the first 21 m the floor is frictionless, and for the next 21 m the coefficient of friction is 0.30. What is the final speed of the crate?
Answers
Answered by
bobpursley
Work put into the crate: force*distance=420*42 Joules
Work used by friction=150g*.3*21
KEfinal= workputintocrate-workusedonFriction.
velocityfinal= sqrt(2*KEfinal/mass)
Work used by friction=150g*.3*21
KEfinal= workputintocrate-workusedonFriction.
velocityfinal= sqrt(2*KEfinal/mass)
Answered by
Anonymous
11.51
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