a 25 kg object is accelerating from rest to a distance of 6 m in 4 seconds across a level floor. The friction is 3.8 N - what is the total work........ i started off by trying to find fa but it seems to give me an absurd answer.. maybe if u could help me by telling me the steps i need to do or how to do this.. please, please, please:(
The fricion component of work is 3.8*6 joules. Now you can figure final velocity (assuming uniform acceleration) as twice average veloctiy (avgvel=6m/4sec), so Vf=12/4 m/s or 3m/s
Final Ke=1/2 m vf^2, so add that to friction work.