Calculating Final Speed with Given Work

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SUMMARY

The discussion focuses on calculating the final speed of a 4.00 kg block initially traveling at 15.0 m/s after work is done on it. For part (a), when 200 J of work is added, the final kinetic energy is calculated using the equation KE = 1/2mv^2, resulting in a higher final speed. In part (b), when -200 J of work is done, the final speed decreases. The correct application of the kinetic energy formula is crucial for accurate results.

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markmil2002
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1. A 4.00 kg block is traveling 15.0 m/s. (a) How fast is the block traveling if 2.00 x 10^2 J of work is done on the block? (b) If -2.00 x 10^2 of work is done on the block instead, what would be the final speed?



Homework Equations


KE = 1/2mv^2



The Attempt at a Solution


a KE = 4.5 x 10^2 + 2.0 x 10^2
b 4.5 x 10^2 J - 2.0 X 10^2
 
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Also, the Advanced Physics forum is more for problems from upper level undergraduate (300-400 level) physics course. Please post problems like these in the Intro Physics Forum.
 
Well I got the question wrong with the work that I did, so I waned to know what I did wrong.
 

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