The Work-Energy Theorwm and Kinetic Energy

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Homework Help Overview

The discussion revolves around the work-energy theorem and kinetic energy in the context of a hammer throw and a bullet's motion. Participants are exploring the calculations related to the work done in launching both objects, specifically focusing on their kinetic energy at the moment of release.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the kinetic energy of the hammer and bullet using the formula KE=(1/2)mv^2, but questions how to determine the initial kinetic energy of the hammer since it starts from rest while being whirled. Other participants confirm that both objects start from rest and engage in recalculating values.

Discussion Status

The discussion is active, with participants providing calculations for the kinetic energy of the hammer and confirming initial conditions. There is an acknowledgment of errors in calculations, and some participants are working through the implications of these findings without reaching a final consensus.

Contextual Notes

Participants are navigating the constraints of the problem, including the initial conditions of both the hammer and bullet, and the need to clarify the initial kinetic energy of the hammer as it is whirled before release.

pstfleur
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1. The hammer throw is a track-and0field event in which a 7.3-kg ball (the hammer), starting from rest, is whirled around in a circle several times and released. It than moves upward on the familiar curvinf path of projectile motion. In one throw, the hammer is given a speed of 29 m/s. For comparison, a .22 caliber bullet has a mass of 2.6g and, starting from rest exits the barrel of a gun with a speed of 410 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet.



2. KE=(1/2)mv^2
W= KE(final)-KE(initial)



3. I think I know my KE final, which would be just plugging in the numbers-
(1/2)*(7.3kg)*(29^2)=KE(final) of the Hammer... But How would I get my KE(initial) since I don't know what the initial velocity of the hammer is as its being whirled around in a circle.
Can someone help?
 
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Both hammer and bullet start from rest.
 
Doc Al said:
Both hammer and bullet start from rest.
OK WOW, clearly I input some wrong numbers in my calculator..

Thanks for the help Doc..Sorry for wasting ur time
 
I guess Ill go ahead and answer the rest of the Question...
KE=(1/2)mv^2
W=Ke final- Ke initial

W= (1/2)(7.3kg)(29^2)
=3069.65 or 3.1*10^3 for the Hammer..

You can use the same to Find KE final for the Gun..
 

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