1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Work-Energy Theorwm and Kinetic Energy

  1. Sep 27, 2008 #1
    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?
  2. jcsd
  3. Sep 27, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Both hammer and bullet start from rest.
  4. Sep 27, 2008 #3

    OK WOW, clearly I input some wrong numbers in my calculator..

    Thanks for the help Doc..Sorry for wasting ur time
  5. Sep 27, 2008 #4
    I guess Ill go ahead and answer the rest of the Question...
    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..
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: The Work-Energy Theorwm and Kinetic Energy