Not sure which formula to use! Deals with work and position

  1. 1. The problem statement, all variables and given/known data
    A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by:

    x = 0.484t3 - 33.6t

    Find W, the work done by this force during the first 1.49 s.


    2. Relevant equations
    F=mg
    F=ma
    W=F*distance


    3. The attempt at a solution
    To find the distance, I plugged in 1.49 into the formula they gave, and I got (-48.5). I know that to find work, the formula is W=F*s, so in order to find the force, I'm not sure which formula to use.

    I think I should be using F=ma, and to find the acceleration, I can take the second derivative of the given formula. However, I forget whether that is allowed in one-dimension. If not, then do I use F=mg? If I use F=mg, then the answer for work will come out positive. If I use F=ma, then the work comes out negative.

    Here are my results:

    W=Fs
    W=ma*s
    W=6.26*18.18*(-48.5)
    W=(-5519.6)J


    W=Fs
    W=mg*s
    W=6.26*(-9.81)*(-48.5)
    W=2978.4 J
     
  2. jcsd
  3. Delphi51

    Delphi51 3,410
    Homework Helper

    Yes! Do that and you will find that the acceleration varies with time so you can't use W=Fs; instead you have to integrate F*ds.

    Gravity is not involved in this. It would cause a constant acceleration and a distance formula of s = Vi*t + .5*a*t² rather than the cubic formula of this problem.
     
  4. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    hi iJamJL! :smile:
    if you need to find the force, then yes you use F = ma

    and yes the acceleration is the second derviative, even in three dimensions! :smile:

    (what is worrying you about that? :confused:)
    yes, that does give you the total distance

    but W = F*s only works if F is constant

    here, F isn't constant, so you would have to use W = ∫ F.ds

    however it might be simpler to use the work-energy theorem, and calculate the difference in kinetic energy :wink:
     
  5. Thanks for the replies!

    I didn't realize that until you pointed it out! I tried it out, but I think I made a mistake somewhere. Here's what I did:

    To find the final velocity, which for this case is at 1.49s, we take the first derivative at 1.49:
    x = 0.484t3 - 33.6t
    dx/dt= 2*(.484)t^2 - 33.6
    v= (-30.38)m/s

    W=ΔKE
    W= 1/2*m(v-final^2) - 1/2*m(v-initial^2)
    **v-initial is 0 because t=0, x=0**
    W=1/2*6.26*(-30.38)^2
    W=2888.8 J

    I tried entering that into my online homework system and it came out wrong. The system doesn't really care for the form that the answer is, as long as there are 3 significant digits. That means that I solved for this incorrectly! :cry:
     
  6. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    erm :redface:3*t2 :wink:
     
  7. lol I wrote that on my scrap paper, but typed and calculated it incorrectly!

    My end result came to:

    v=(.484)*3*(1.49^2) - 33.6
    =(-30.4)m/s

    W=1/2*m*v^2
    W=1/2*6.26*(-30.4^2)
    W=2892.6 J

    That's still coming out as incorrect.
     
  8. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    no, W = 1/2*m*v22 - 1/2*m*v12 :rolleyes:
     
  9. gneill

    Staff: Mentor

    Just because t=0 and x=0, that does not imply that v(0) = 0. Plug t=0 into your velocity formula.
     
  10. Wow, I'm bad at this. I initially thought that v-initial would be 0, but then I looked back and realized it's not. *facepalm*

    v(0)=(-33.6)m/s
    v(1.49)=(-30.4)m/s

    W=1/2*m*(v2^2) - 1/2*m*(v1^2)
    W=1/2*6.26(-30.4^2) - 1/2*6.26*(-33.6^2)
    W=2892.6 - 3533.6
    W=(-641) J

    Did I finally get the correct answer? :approve:
     
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