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Homework Help: Temp rise due to hammering nails

  1. Jun 12, 2006 #1
    Here's my question.....

    The 1.20 kg head of a hammer has a speed of 6.5m/s just before it strikes a nail and is brought to rest. Estimate the temperature rise of a 14 gram iron nail generated by ten such hammer blows done in quick succession. Assume the nail absorbs all the energy.

    The picture shows a hammer hitting a nail with the swings being horizontal(as if you were nailing into a wall.......not the floor).

    So, I can't seem to get my formulas set to get past a certain point. This is what I've done so far(which is not much).

    W= 4.186 J = 1 cal of heat
    W = Fd
    F=ma------where I get stuck......I've got velocity, but not acceleration.

    Thanks for any help.....I'd be glad to work it with some help....or maybe I can work it a little further until I get stuck again(if I do).

  2. jcsd
  3. Jun 12, 2006 #2

    Doc Al

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    Staff: Mentor

    How much kinetic energy does the hammer possess before it hits the nail? What happens to that energy?
  4. Jun 12, 2006 #3
    Ok.....well I kinda see where you are going, but still need a bit more direction.


    KE(hammer)=25.35 J -----So I think this energy is now transfered into the nail, with some lost to heat?? Or, do I have to recaculate the KE for the nail(using its mass)?

    I'm also not real sure how this energy(J) is going end up giving me a temp measurment.....will I eventually relate it to the specific heat of iron?

    Thanks again and I'll keep working....

  5. Jun 12, 2006 #4

    Doc Al

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    What's the speed of the nail after you hit it with a hammer? (Hint: Nothing to calculate.)

  6. Jun 12, 2006 #5
    The speed of the hammer would be directly transfered to the nail.....so it would be 6.5m/s. With that in mind(and the nails mass), I found KE(nail)=.296 J

    Do I multiply 10 swings into both the KE of the nail and the KE of the hammer?? I'm thinking I do, it just seems obvious. Then I'm thinking I have to Joule measurements....so then do I multiply them both by the specific heat of iron.....and add them together to obtain a final amount of heat created?

    Thanks again!!!
  7. Jun 12, 2006 #6

    Doc Al

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    No! You bang a nail into the wall, does it goes shooting through the wall? (Let's hope not.) No, when the hammer hits, the nail moves a bit, then hammer and nail both stop. The KE of the nail is zero. (That's why I said there is nothing to calculate. :smile: )

    All of the KE of the hammer gets absorbed by the nail as thermal energy.
  8. Jun 12, 2006 #7
    Ok.....that does make sense, as I think of it more and more.

    So, now I am thinking that I multiply my KE(hammer) by 10 swings to find total KE. Then do I use an equation similar to U(internal energy) = N(1/2mv^2).....where N is number of molecules?? This doesn't seem like the right equation though. Also, I can't yet understand why they gave me the mass of the nail(and maybe that is for later in the problem).
  9. Jun 12, 2006 #8
    [tex] \frac{1}{2} m v^2 = mc \theta[/tex]
    Last edited: Jun 12, 2006
  10. Jun 12, 2006 #9

    Doc Al

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    It's not. Think specific heat of iron, which you mentioned earlier. (You'll need the mass of the nail.)
  11. Jun 12, 2006 #10
    Ok, well I am thinking this now.

    KE(total) = 253.5 J

    Q = mc*(change in T)

    Change in T = Q/mc

    253.5 J * .014 Kg * 450 J/Kg*deg C = 40.24 deg C

    Even close?
  12. Jun 12, 2006 #11
    yeah I knew they did heat up......as I used to frame houses/apartment buildings for a few yrs.......
  13. Jun 12, 2006 #12
    nevermind......ha......just noticed the problem was odd, so I checked it in the back, and I have the correct answer!!!

    Thanks a ton Doc Al!!

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