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?

1 answer

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