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Average Retarding Force Problem

  1. Jan 14, 2008 #1
    [SOLVED] Average Retarding Force Problem

    1. The problem statement, all variables and given/known data

    An arrow of mass 45 g strikes a wood block target horizontally with a velocity of 70
    ms-1 and penetrates a distance of 280 mm. Ignoring air resistance, calculate (a) the time it takes for the arrow to stop moving, and (b) the average retarding force exerted on the arrow by the wood block target.

    2. Relevant equations

    v=u + at
    F= ma
    V2=u + 2as

    3. The attempt at a solution

    I worked out the time using the 1st formula listed (t=7.14 3sf) but when i go to work out the average retarding force using F=ma and get 0.44145 which seems wrong to me .Could someone please tell where i'm going wrong?
    I also worked out the total distance travelled by the arrow which is 250m including the 280mm penetration of the wood.
     
  2. jcsd
  3. Jan 14, 2008 #2

    Kurdt

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    For the first part I'd use the following since 7.14 seems a lot to me.

    [tex] s = \left(\frac{u+v}{2} \right) t [/tex]

    Once you have time you can work out the acceleration and then the force.
     
  4. Jan 14, 2008 #3
    I've worked out the total distance travelled by the arrow which is 250m . I found this out using the equation [tex] v^2=u^2 +2as [/tex] . I then substitute it into the equation you gave me and i still get 7.14 (3sf).

    u=70 , a=-9.81 s= 250 (including distance penetrated)
     
  5. Jan 14, 2008 #4

    Kurdt

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    It doesn't ask for the the time it takes for the arrow to travel the full distance. It asks how long it takes to stop moving. That is how long it takes to come to rest in 280mm.
     
  6. Jan 14, 2008 #5
    oh ok. me thinks i interpretated the question. Thanks.
    that means t=0.008s.
    I thought a= -9.81m/s since the effects of air/wind resistance are ignored?

    I think this is the right answer for retarding force: -393.75N
    I got it by working out a instead of using -9.81 , and i got -8750 which seems quite alot.
     
    Last edited: Jan 14, 2008
  7. Jan 14, 2008 #6

    Kurdt

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    -9.81 will be the vertical component of acceleration but we're only considering the horizontal motion. Now you have your time you can work out the acceleration the arrow underwent to cause it to stop. Then you can work out the force.

    Actually I've been a bit of a fool because you don't really need to work out the time. Its just since you said you'd worked it out in the first post I kind of got blinkered by it. Of course you could have worked out the acceleration directly from [itex]v^2 = u^2 +2as[/itex]. Never mind. You get the same answer in the end. :smile:
     
  8. Jan 14, 2008 #7
    Thanks for the help. I've been at uni for a year now and this stuff still puzzles me.
     
  9. Jan 14, 2008 #8

    Kurdt

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    You'll get there. I was pretty much puzzled initially because before that I'd never really worked at my education. Once I realised I had to start putting in some work it took a while to learn how to learn but I made it. As will yourself.
     
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