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

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SUMMARY

The discussion centers on the application of the work-energy principle in physics, specifically regarding the kinetic energy of a wagon. Bill's use of 500J of work, rather than the net work of 300J, is correct because the total work done includes overcoming friction, which is essential for calculating the kinetic energy. The net work, which accounts for friction, is indeed the change in kinetic energy, but the initial work must also be considered to understand the complete scenario. The mass of the wagon is a crucial variable that influences the final speed calculation.

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  • Understanding of the work-energy principle
  • Basic knowledge of kinetic energy equations
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  • Ability to perform calculations involving mass and speed
NEXT STEPS
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  • Learn how to calculate kinetic energy using the formula KE = 0.5 * m * v^2
  • Explore the effects of friction on motion and energy calculations
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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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