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Hammer smashing nail

  1. Oct 17, 2005 #1
    A hammer of certain mass, say 0.5 kg, smashes a nail at a velocity of 6 ms^-1 which sticks in a table 0.01 m deep. Calculate the force due to friction caused by the table.

    This is the solution my teacher showed;

    The kinetic energy at the momement the hammer hits the nail is

    e = mv^2 / 2
    = 0.5 (6)^2/2
    = 9

    W = E = Fs

    9 = F(0.01)

    F = 900 N


    The fact is that i don't quite follow... My idea is to calculate the distance the nail would have covered before it reaches a constant velocity if there was no opposing force, let say s. If s>0.01 (which is obvious), then I would have to calculate the avegage force the nail has and then calculate the deceleration that is caused by the table. Am I right or not, or simply talking about the same thing?
     
  2. jcsd
  3. Oct 17, 2005 #2
    Hmm this is a strange question. What your teacher showed you looks correct although the notation could have been better. :tongue:

    ∆Ek = (1/2)m(v2^2-v1^2) = (1/2)(0.5)(6^2-0^2) = 9J

    Here you are assuming that it is an elastic collision where all Ek is conserved.

    ∆Ek = WEK

    WEK = FA∆d
    9 = 0.01FA
    900N = FA

    Here you have to assume that the nail is being driven into the table at a constant velocity which doesn't make sense realistically since the nail has to decelerate to stop. They should really state this in the question though because this is a big assumption.

    Fnet = FA+Ff
    ma = FA+Ff
    0m = 900N+Ff
    -900N = Ff

    Hope that helps.
     
    Last edited: Oct 17, 2005
  4. Oct 17, 2005 #3
    If what you were talking about was finding the acceleration of the hammer then calculating the force that would result from that decceleration then yes that would work as well, although i like the way your teacher did it much better.
    [tex]t=\frac{x}{v}[/tex] since your intial velocity is 6 m/s and your final velocity is 0 then the average velocity is 3 m/s input that v into the eq. above to get the time which it took for the hammer to stop, x is distance.
    t=.00333333
    acceleration is [tex]a= \frac{v_f -v_i}{t}[/tex]
    so you get a =-1800 m/s^2
    input that into good old [tex]f=ma[/tex]
    and you get f= -900
     
    Last edited: Oct 17, 2005
  5. Oct 17, 2005 #4
    what you are assuming the hammer accelerated on a distance of 0.01 which is a false assumption... and Erwin, I don't quite see. Do you mean the table needs the same energy as the nail in order to stop it?
     
  6. Oct 17, 2005 #5
    it isn't a false assumption the nail only moves as long as the hammer moves, if the nail moved .01 m then assuming the integrity of the nail remained constant the hammer moved .01m as well.
     
  7. Oct 17, 2005 #6
    I see what you mean but haven't you neglegted the mass of the nail?
    Edit: Alright I understand now. In fact I wasen't conscious that when an object hits another, the second object only has only kinetic energy, so if undisturbed remains at a constant speed.
    Then let say for any case the momemtum of object 1 is [tex] gi[/tex] and the momentum of object 2 is [tex]vm[/tex]
    [tex] gi = vm [/tex]
    [tex] v = \frac{gi}{m} [/tex]
    Also by conservation of energy, we know;
    [tex] \frac{v^{2} m}{2} = e [/tex]
    [tex] \frac{(\frac{gi}{m})^{2} m} {2} = e [/tex]
    So
    [tex] \frac{gi^{2}}{2m} = e [/tex]
    [tex] 2m = \frac{gi^{2}}{e} [/tex]
    and
    [tex] s = \frac{v^{2} - u^{2}}{2a} [/tex]
    [tex] a = \frac{-gi^{2}}{2sm^{2}} [/tex]
    [tex] F = ma = \frac{-g^{2}}{2sm} = \frac{e}{s} [/tex]
    I know this is absurd since it involves the nail dosen't accelerate (Erwin you were right). But since the mass of the nail is negligtable, then it can be taken that way without a great loss of precision.
     
    Last edited: Oct 18, 2005
  8. Oct 17, 2005 #7
    I like to think of it this way, the board the nail and the hammer are a system, into that system you have introduced a finite amount of energy, in this case it is the initial momentum of the hammer, and that energy transfers from the hammer to the nail to the board, note you assume all of the energy is conserved, none of it is lost. So if the hammer exerts a force on the nail all of that force must have been transferred, likewise if the nail exerts a force on the board all of the force must have been transferred. So in essence the force that the hammer puts on the nail is the same as the force that the nail puts on the board. Therefore, you can neglect the nail. However, it would work the same, the nail would have a much larger acceleration than the hammer and you would end up with the nail exerting the same amount of force that the hammer does.
     
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