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
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?
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