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What is the average force exerted on the putty by the surface?

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data

    A 0.43 kg piece of putty is dropped from a height of 2.8 m above a flat surface. When it hits the surface, the putty comes to rest in 0.22 s. What is the average force exerted on the putty by the surface?


    2. Relevant equations

    F[tex]\Delta[/tex]t=[tex]\Delta[/tex]P
    F[tex]\Delta[/tex]t=mvf-mvi
    [tex]\Delta[/tex]P=mvf-mvi

    (I dont think it is a collision, but if it is...any of those equations also)

    3. The attempt at a solution
    2.8m=(.22s)vo+.5(-9.8)(.22s)2
    vo=13.8m/s

    (.22s)F=(.43kg)(0m/s)-(.43kg)(13.8m/s)
    F=-27N

    But I know this isnt right...I just don't know what I did wrong...
     
  2. jcsd
  3. Nov 10, 2008 #2
    Energy conservation to the surface, then kinematics.

    [tex]U_0+K_0=U_f+K_f\rightarrow U_0=K_f[/tex], and

    [tex]F=ma=m\Delta v/\Delta t[/tex]...
     
  4. Nov 10, 2008 #3
    the .22s is how long the putty takes to come to a stop AFTER it hits the ground, not the time it takes from when it was released.

    How can you find the velocity of the putty JUST BEFORE it hits the ground?

    HINT: Use one of the kinematic equations that does not involve time.

    Casey
     
  5. Nov 10, 2008 #4
    Am I going to use the equation:
    v2=v02+2ax??
    Then it would be 0=v02+2(-9.8)(2.8).
    Vo= 7.4 m/s
    Is that what you are looking for?
     
  6. Nov 10, 2008 #5
    You're right for a start; here's my logic:

    [tex]U_0=K_f[/tex]

    Because it starts at rest, 2.8m above the ground, we know:

    [tex]U_0 = mgh = 11.8 J = mv^2/2 \rightarrow v = 7.41 m/s[/tex]

    Then,

    [tex]v_f=v_i+at\rightarrow a=[/tex]. Because force and acceleration are related, you can continue from here.
     
  7. Nov 10, 2008 #6

    I am not sure why you are solving for [itex]V_0[/itex]

    The putty is dropped and thus [itex]V_0=0[/itex] you need to solve for [itex]V_f[/itex]

    which is the Velocity of the putty immediately before it hits the ground

    You get the same number, but only by sheer luck.
     
  8. Nov 10, 2008 #7
    Ok thank you...I think im just going to wait and ask my teacher. =]
     
  9. Nov 10, 2008 #8

    LowlyPion

    User Avatar
    Homework Helper

    You've got part of it right.

    The Average Force will be given by the change in momentum divided by the time interval. Since the final v is 0 then all you need to do is figure the momentum

    V2 = 2*a*x = 2*(9.8)*(2.8 m)
     
  10. Nov 10, 2008 #9
    wow that actually helped a lot! thank you!
     
  11. Nov 10, 2008 #10
    The key to this problem was to break it up into two sub-problems.

    By identifying correctly that you needed to use Impulse and Momentum, you knew that you needed to find the Velocity of the putty just before it hit the ground.

    That Velocity could be found by using kinematics (the first sub-problem).

    On a side note, whenever you see forces (or moments/torques) and time involved, chances are pretty good that you can use Impulse and Momentum equations.

    Casey
     
  12. Nov 10, 2008 #11
    I got it...and the computer said my answer is right, so Im good to go! Thanks to everyone who helped! =]
     
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