Final speed after dragging crate

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AI Thread Summary
A 96 kg crate is pulled with a constant force of 350 N across a frictionless surface for 15 m, followed by a surface with a coefficient of friction of 0.25 for another 15 m. The discussion focuses on calculating the final speed of the crate after both segments. The initial approach involved using work-energy principles, but there was confusion regarding the application of work and the need for acceleration calculations. A suggested alternative method involves calculating acceleration using F=ma for the first segment and then applying kinematic equations, factoring in friction for the second segment. The conversation highlights the importance of correctly applying physics principles to solve for the final velocity.
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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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