The Work-Energy Theorwm and Kinetic Energy

[pstfleur]

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?

 

[Doc Al]

Both hammer and bullet start from rest.

 

[pstfleur]

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

 

[pstfleur]

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..