Work and motion strategies to a problem

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The discussion centers on understanding the work-energy theorem in the context of a ball thrown into the air. The calculated work done by gravity at the ball's maximum height is -112.5 J, derived from the change in kinetic energy. However, confusion arises when using the formula vf^2 = vi^2 + 2ad to find distance, leading to an incorrect calculation of work as 13524 J. The error stems from mistakenly using a mass of 30 kg instead of the correct 0.25 kg. Clarifying these values is crucial for accurately applying the work-energy theorem.
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Homework Statement



So the problem is getting at a basic understanding of the work energy theorem, and just asks for the work done by gravity at a ball's max height if the ball (with mass 0.25 kg) is thrown into the air with initial velocity of 30 m/s. using W=change in KE, the answer is -112.5 J

But if I use the formula vf^2 = vi^2 + 2ad, to determine d, I get d as approx. 46, and so W=294 x 46 = 13524 J ??



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Where did 294 come from? And 46 what? (units!)
 
right, 46 meters. and the 294 (N) was the force calculated as mxa = 30 kg x 9.8 m/s2
 
You said the mass is .25 kg, not 30 kg
 
haha, right. not sure why my brain did that.
 

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