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Calculating work with kinematics and dynamics

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    the question is:
    A 62 kg person in an elevator is moving up at a constant
    speed of 4.0 m/s for 5.0 s. T / I C

    (b) Calculate the work done by the normal force on the person.
    (c) Calculate the work done by the force of gravity on the
    person.
    (d) How would your answers change if the elevator were
    moving down at 4.0 m/s for 5.0 s?


    2. Relevant equations
    W= F x Δd

    and i think

    Δd= (vf + vi/2)Δt
    fg=m.g
    3. The attempt at a solution

    for b) i did Δd= (vf + vi/2)Δt and got displacement which is 10 m and then i did fn=fg since fnet is 0 which means that they have the same force. and for that i got 607.6 N

    p.s the answer in the back is 12 kJ which i dont get?
     
  2. jcsd
  3. Nov 22, 2011 #2
    what if i told you that your equation for Δd is incorrect.

    Δd = (vf +vi)/2 * Δt

    see if that doesnt give you 12.1644 kJ
     
  4. Nov 22, 2011 #3
    it wouldnt matter though cause vi= 0 (i think)
     
  5. Nov 22, 2011 #4
    The initial velocity is not vi=0, however, vf=vi since the velocity is constant.
     
  6. Nov 22, 2011 #5
    you sure? ive never heard of that before but thanks.
     
  7. Nov 22, 2011 #6
    Technically, if vi=0, then during that time frame there was indeed an acceleration and the velocity is then not constant. However, by equating vi=vf we are merely indication that the velocity remains constant and a=0.
     
  8. Nov 23, 2011 #7
    thank you sandy.bridge

    vi does not equal 0

    the statement says constant velocity of 4 m/s

    thus vi = vf = 4 m/s
     
  9. Jul 16, 2013 #8
    the elevator is accelerating so fnet can not equal to 0.

    fn +fg=ma
     
  10. Jul 16, 2013 #9

    PhanthomJay

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    No, it is given during that particular 5 second time interval that the elevator is not accelertaing, that is, it is given that it is moving at constant speed.
     
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