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Homework Help: Speed of block after 2 forces acted on it (Mechanics)

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A 1.5kg block is at rest on a table. The coefficient of friction between the block and the table is 0.15. A force of 6N is applied to the block for 2 seconds followed by a force of 1N (in the same direction) for 3s.
    (a)What is the speed of the block at the end (t=5)?
    (b)What is the total distance the block moves up to t=5?

    m=1.5kg
    v1=0m/s
    CofF=0.15
    F1=6N
    t1=2s
    F2=1N
    t2=3s


    2. Relevant equations

    I really don't even have an idea which formulas to use :confused:
    F=ma
    Ff=(coefficient of friction)(FN)
    Ek=(mv2)/2
    W=FdcosA
    vf=vi+at
    d=vit+1/2at2
    2ad=vf2-vi2
    d=vt



    3. The attempt at a solution

    Ff=CofF(FN)=0.15*1.5kg*9.8m/s2=2.205N

    Fnet=ma --> a=F/m --> a=(F1+F2-Ff)/m

    a=3.1967m/s2

    ??????????????

    I only (really) need help with part (a). I've tried about a dozen solutions and none of them have worked. I've been working on this question for hours, PLEASE help.
     
  2. jcsd
  3. Feb 10, 2010 #2
    Be careful with the net force. You have to separate two different events: first you have force [tex]F_{1}=6 \textsc{N}[/tex] acting for 2 seconds and then [tex]F_{2}=1 \textsc{N}[/tex] for 3 seconds.

    Now, it's not clearly stated if those forces are applied together after time interval of 2 seconds. If we assume that they are applied one after another then you have

    [tex]F_{net_{1}}=F_{1}-F_{fr}=ma_{1}[/tex]
    [tex]F_{net_{2}}=F_{2}-F_{fr}=ma_{2}[/tex]

    From there we get accelerations. Distances and speeds are then easily calculated using given time intervals.

    Otherwise, we have

    [tex]F_{net_{1}}=F_{1}-F_{fr}=ma_{1}[/tex]
    [tex]F_{net_{2}}=F_{1}+F_{2}-F_{fr}=ma_{2}[/tex]
     
  4. Feb 10, 2010 #3
    Thank you, you are a life saver:smile:
     
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