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Work,Power and Energy = big headache

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    1. A 1250 Kg piledriver falls from an initial height of 1.30m above the ground onto a 15kg spike the top of which is initially 0.30m above the ground. The top of the spike finishes at ground level. Determine the average force exerted on the piledriver by the spike.
    *Pd=Piledriver
    S=Spike

    mass of piledriver= 1250kg
    h1= 1.3m
    h2=0m
    g=9.8m/s2
    mass of spike=15kg
    h1=0.30m


    2. Relevant equations
    Fnet=F1+F2...

    3. The attempt at a solution

    Fnet=Fn(S) + Fg(Pd)
    = (15*9.8) + (1250*9.8)
    =12397 N

    Average= (FnS –FgPd) = FgPd

    2
    = FnS /2
    = (15*9.8) /2
    =73.5 N
    Fnet=Fn(S) + Fg(Pd)
    = (15*9.8) + (1250*9.8)
    =12397 N

    therefore, the average force applied on the piledriver by the spike is 73.5N


    * i really,reallly, really could use some assistance on this one. its grade 11 physics, so i know it can't be that hard. Any Suggestions?
     
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2

    Doc Al

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    Staff: Mentor

    The force of the spike on the piledriver--which you call Fn(S)--does not equal the weight of the spike. Furthermore, Fn is what you are trying to find, so you just can't assume what it is--you must solve for it.

    Do this. Find the acceleration of the piledriver during its interaction with the spike. (Use kinematics.) That will tell you the net force. Then you can solve for the average Fn(S).
     
  4. Apr 8, 2007 #3
    how do i find the acceleration if i'm not given the time or v2?
     
  5. Apr 8, 2007 #4

    Doc Al

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    Staff: Mentor

    You don't need time--you have the distance. You should know the final speed! :wink: And you can figure out the speed of the piledriver as it hits the spike.

    (You can also use energy methods to solve this.)
     
  6. Apr 8, 2007 #5
    so... if i use :
    Vf(squared)=Vi(squared)+ 2a(d)
    9.8(squared)= 0squared +2a (1)
    2a=96.04/2
    a=48.02 m/s2 ?

    *i always forget the value of v2
     
  7. Apr 8, 2007 #6

    Doc Al

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    Staff: Mentor

    The distance over which the spike interacts with the piledriver--d--is not 1 m. (1 m is the distance the piledriver falls before hitting the spike.)
     
  8. Apr 8, 2007 #7
    thanks. its 0.30m then because thats the point of contact? that would result in the a=160.067 m/s2 ?[did i mention i'm not that good at physics :-) ]
     
  9. Apr 8, 2007 #8
    ok, so then i take the acceleration, and do :
    Fnet=ma
    Fnet = (1250)(160.07)
    Fnet= 200083.3 N

    so then:
    Fave.=(Fg + Fnet) /2
    = (12397 +200083.3) /2
    =106240.15 N of force

    * i'm just getting confused as to what force i'm calculating for: for the piledriver or for the spike?
     
  10. Apr 8, 2007 #9

    Doc Al

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    Staff: Mentor

    Good. Note that this is the average net force on the piledriver, since we assumed that the acceleration and thus the force were constant.

    Now to find the average force (Fn) exerted by the spike on the piledriver, add up the forces:

    Fnet = Fn - mg

    and solve for Fn(s). Note that mg is negative, since it acts down; Fnet and Fn both act up, so they are positive. (All forces are already the "average" forces, since we had to pretend they were constant.)

    This is incorrect. (Perhaps you are confusing this with calculating the average speed for uniformly accelerated motion: Vave = (Vi + Vf)/2.)

    The only forces we are concerned with are the forces acting on the piledriver: the weight of the piledriver and the normal force exerted by the spike on the piledriver.
     
  11. Apr 8, 2007 #10
    has anyone told you that you are really helpful? [its true :) ]

    Fnet=Fn-mg
    200083.3=Fn-(1250*9.8)
    Fn=212333.3 N ?
     
  12. Apr 8, 2007 #11

    Doc Al

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    Staff: Mentor

    Looks good. (Round off your answer to a reasonable number of significant figures.)
     
  13. Apr 8, 2007 #12
    thanks so,so much !
     
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