Calculating Velocity After a Force is Applied

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SUMMARY

The discussion focuses on calculating the velocity of a 100.0 kg wagon after a 2.0 N force is applied over a distance of 5.0 m, with initial motion at 0.50 m/s. When the force is applied in the direction of motion, the final velocity is calculated to be 0.67 m/s, with a final kinetic energy of 22.55 J. Conversely, when the force opposes the motion, the final velocity is determined to be 0.23 m/s, with a final kinetic energy of 2.55 J. The term 'internal energy' is discouraged in favor of using work-energy methods for clarity in physics problems.

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Homework Statement


Jennifer applies a force of 2.0 N upon a 100.0 kg wagon, which was initially moving at 0.50 m/s. This force acts over a distance of 5.0 m. Assume friction can be neglected.
a) What is the wagon's velocity after Jennifer's push if she pushes the wagon in the direction of motion?
b) What is the wagon's velocity after Jennifer's push if her push opposes the wagon's motion?


Homework Equations


KE=.5v2
lw-l=F[tex]\Delta[/tex]Xcos[tex]\theta[/tex]
w+=F[tex]\Delta[/tex]Xcos[tex]\theta[/tex]

The Attempt at a Solution


a) I got an initial KE of 12.5 J and an initial IE of 10 J. My ending KE was 22.55 J. My answer was a velocity of 0.67 m/s. Is this correct?

b) I got and initial KE of 12.5 J then my ending IE was -10 J and ending KE 2.55 J. My answer for the velocity was 0.23 m/s. Is this correct?

Thanks!
 
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As in your other post, you are getting correct answers, but refrain from using the term 'internal energy' when work done by non conservative forces are involved. Instead , use work energy methods...in part a, for example, the work done by the applied force is 10J, don't call it internal energy. What equation are you using when you refer to 'internal energy'?
 

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