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Homework Help: Solving for variables

  1. Aug 12, 2005 #1
    a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from v_initial to vfinal , undergoing displacement s given by s=x_final-x_initial.

    Express the acceleration in terms of v_initial, v_final, and s:
    that was easy.. i got


    since F=ma

    the question is Give an expression for the work in terms of m, v_initial, v_final ?

    and i cant seem to eliminate the 's' because i end up getting

    W=m*a(see above) *s and i cant get eliminate it? any ideas?
  2. jcsd
  3. Aug 12, 2005 #2
    the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial
    and we are supposed to express that in m_vinitial and v_final

    and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula

    fint{t*dt between a and b} = b^2-a^2/t
  4. Aug 12, 2005 #3


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    can't you use ,

    [tex] v_f^2 - v_i^2 = 2as[/tex] ?
  5. Aug 12, 2005 #4
    the s is still in the equation though
  6. Aug 12, 2005 #5


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    [tex]\int_{vi}^{vf} mv dv = m[v^2/2]_{vi}^{vf}[/tex]

    The m doesn't get squared.
  7. Aug 12, 2005 #6


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    substitute for "as" from one eqn into t'other.
  8. Aug 12, 2005 #7
    oo thats right stupid parentheses.. but i still cant get the first part?
  9. Aug 12, 2005 #8


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    W = Fs = Mas

    vf² - vi² = 2as ==> as = ½(vf² - vi²)


    W = Mas = M*½(vf² - vi²)
    W = (M/2)(vf² - vi²)
    Last edited: Aug 13, 2005
  10. Aug 12, 2005 #9
    o wow yea that makes alot more sense than what i had, i used the wrong eq. lol thank youuu!!
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