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Help with momentum problem -- a delivery man pushes a 5kg box along a smooth surface

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A UPS delivery man pushes a 5kg box (A) along a smooth surface where the coefficient of kinetic friction, [µ][/k], is equal to 0.20. The box slides for 1.5m before colliding with a 3kg box (B). The boxes then move together as one unit on a rough surface with [µ][/k]= 0.9 for 0.8m. Determine the initial velocity of the larger box. This problem was on the test that I took today and I'm not sure if I got it correct, or not.

    2. Relevant equations
    ∑F=ma
    [F][/fric]=[µ][/k][F][/N]
    v[f]^2=v[0]^2+2aΔx
    p=p'
    g=9.8m/s[2]
    3. The attempt at a solution
    ∑F[y]=m[A]a[y]
    F[N]-F[G]=0
    F[N]=F[G]=m[A]g=(5)(9.8)=49N
    F[fric]=(.2)(49)=9.8N
    ∑F[x]=m[A]a[x]
    F[fric]=(5)a[x]
    a[x]=1.96m/s[2]
    v[f]^2=v[0]^2+2aΔx
    0=v[0]^2+2(1.96)(-1.5)
    v[0][smooth skid]=2.42m/s =v
    ∑F[y]=m[A+B]a[y]
    F[N]-F[G]=0
    F[N]=F[G]=m[A+B]g=(5+3)(9.8)=78.4N
    F[fric]=(.9)(78.4)=70.56N
    ∑F[x]=m[A+B]a[x]
    F[fric]=(5+3)a[x]
    a[x]=8.82m/s[2]
    v[f]^2=v[0]^2+2aΔx
    0=v[0]^2+2(8.82)(-.8)
    v[0][A+B][skid]=3.75m/s=v'
    p=p'
    m[A]v[A]+mv=(m[A]+m)v'
    (5)v[A]+(3)(2.42)=(5+3)(3.75)
    v[A][beg skid on rough]=4.56m/s
    v[f]^2=v[0]^2+2aΔx
    (4.56)^2-2(1.96)(-1.5)=v[0][A]^2
    v[0][A]=3.86m/s (or something close to that.)

    I'm pretty sure I got this one wrong. I wonder how much partial credit I'm going to get...
     
  2. jcsd
  3. Dec 10, 2014 #2

    haruspex

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    I'm afraid I would give no credit because your working is unintelligible. You need to make it much easier for others to follow:
    - use latex, or at least proper super/subscripting
    - define your unknowns
    - use unique labels for unknowns
    - keep all the working symbolic, only plugging in numbers at the final step
    - be prepared to write some words to describe what you are doing along the way
     
  4. Dec 10, 2014 #3
    I'm still working out how to subscript/ superscript.
     
  5. Dec 10, 2014 #4

    haruspex

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    The easy way is to select the text to be shifted and click the X2, X2 symbols above the text area.
    Using latex, type a pair of hash symbols (#), then the latex text, and another pair of hash symbols to return to normal text. Within the latex, use _ for subscript and ^ for superscript. E.g. x_1 y^2 becomes ##x_1 y^2 ##. If multiple characters are to go in the shifted text, bracket using { and }. So x_{final} becomes ##x_{final}##.
    You can also develop a more complex latex string using the Latex Preview button. This opens up a region for you to type the latex in (including the hash symbols), then you can click Show Preview to see how it will look. If it's not right, edit the text and click show again. When happy with it, copy and paste into the post body.
    If someone else posts latex and you want to see how they did it, right click on it, and Show Math As-> Tex Cmds. You can copy and paste out of the window that comes up.
     
  6. Dec 10, 2014 #5
    I've attached a word document with my solution. I have no idea what the correct answer is.
     

    Attached Files:

  7. Dec 10, 2014 #6

    haruspex

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    It may be because I use OpenOffice, not Word, but the algebra is completely missing when I open it.
     
  8. Dec 10, 2014 #7
    So what do you see when you open it?
     
  9. Dec 10, 2014 #8

    haruspex

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    As far as I can tell, all the text except any algebra.
     
  10. Dec 10, 2014 #9

    haruspex

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    How about having another go at posting in the usual way (not as an attachment), using the subscripts and superscripts.
    Since you are trying to find an initial condition, run the problem backwards. What must the speed be just after the collision?
     
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