Net work and kinetic energy (pushing a wagon to accelerate it)

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The discussion centers on the confusion regarding the use of work versus net work in calculating kinetic energy for a wagon. The original poster questions why their teacher used 500J of work instead of the 300J of net work they calculated. It is clarified that the net work done on the wagon corresponds to the change in kinetic energy. The impact of friction is acknowledged, as it does affect the wagon's acceleration, but the kinetic energy equation focuses on the net work applied. The importance of specifying the mass of the wagon for accurate calculations is also highlighted.
aqryus
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Homework Statement
Bill does 500J of work on a wagon, friction does work of -200J. What is the final speed of the wagon if it starts at rest?
Relevant Equations
w=fdcostheta
Ek=1/2mv^2
I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why isn't that accounted for in the kinetic energy equation to find speed? Thank you
 
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You are correct. The net work done on the wagon equals the change in the kinetic energy of the wagon. What did you get for the answer? Was the mass of the wagon specified?
 
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