Final speed after dragging crate

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SUMMARY

The discussion focuses on calculating the final speed of a 96 kg crate subjected to a constant horizontal force of 350 N over two segments: the first 15 meters on a frictionless surface and the next 15 meters with a coefficient of friction of 0.25. The correct approach involves using Newton's second law (F=ma) to find acceleration during both segments and applying kinematic equations to determine the final velocity. The initial velocity for the second segment is derived from the final velocity of the first segment, ensuring accurate calculations that account for friction in the second segment.

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Homework Statement



A 96 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15m the floor is frictionless and for the next 15 m the coefficient of friction is .25. What is the final speed of the crate?

Homework Equations



Change in K = Work
E + Ffr * d = 0
1/2mv^2-1/2mv^2 + F(fr) d

The Attempt at a Solution


I broke this up into two steps. The first 15 m I took the Work done, set it equal to 1/2mv^2 and got the velocity for the beginning of the second reference frame . Somethings not working though because I get the incorrect answer. Am I setting this up incorrectly?
 
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Umm..I don't think you need to calculate anything for the Work. I don't see the need for the work. Find the acceleration using F=ma and put it into the correct kinematic equations to find the Vf of the first 15 m. Then take into account friction for the second have and use the same kinematic equation with the Vi being the Vf of the first 15 m.
 

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