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?
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)
Answered by Anonymous
11.51
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