The Work-Energy Theorwm and Kinetic Energy

AI Thread Summary
The discussion revolves around calculating the work done to launch a hammer and a bullet using the work-energy theorem. The hammer, weighing 7.3 kg and thrown at a speed of 29 m/s, has its kinetic energy calculated as KE = (1/2)mv^2, resulting in approximately 3069.65 J. The initial kinetic energy for both the hammer and the bullet is zero since they start from rest. The bullet, weighing 2.6 g and exiting a gun at 410 m/s, can be analyzed similarly to find its kinetic energy. The conversation highlights the application of the work-energy theorem in determining work done in both scenarios.
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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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