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A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betweet the pickle and the floor is 3.8N, what is the work done to move the object

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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
hope this is correct
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