posted by steve .
A 25.0 kg pickle is accelerated from rest through a distance of 6.0 m in 4.0 s across a
level floor. If the friction force between the pickle and the floor is 3.8 N, what is the
work done to move the object?
Add the KE increase and the work done against friction.
The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J
The friction work done is 6*3.8 = 22.8 J