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Homework Help: Power, the rate of energy transfer

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A 613.0 kg mass is placed on a forklift that can generate 950 W of power. What is the constant speed of the forklift while lifting this load?

    2. Relevant equations
    P= E/T
    e=energy in joules
    T= time

    3. The attempt at a solution
    not enough information therefore i couldnt do it. pls help
  2. jcsd
  3. Apr 28, 2010 #2
    Hint: [tex] P = \frac{dW}{dt} [/tex]

    Where W is the work.
  4. Apr 28, 2010 #3
    time and work are not given, only have Watt and i have no idea wat to do with Kg and i need to find velocity...dunno
  5. Apr 28, 2010 #4
    What is the equation for the work done in this problem?
  6. Apr 28, 2010 #5
    W= FD but i dont have the distance
  7. Apr 28, 2010 #6
    Right, and what is F?
  8. Apr 28, 2010 #7
    Watt, power
  9. Apr 28, 2010 #8
    No, F is the force doing the work W = FD, what force is doing the work?
  10. Apr 28, 2010 #9
    i know what you're getting into but i have try that attempt before, even so I don't understand how to find the velocity with FG
  11. Apr 28, 2010 #10
    I'm trying to help you lol. If you knew what I was getting at you would already have the answer. You know what W is so what is dW/dt?
  12. Apr 28, 2010 #11
    velocity but i don't have the time?
  13. Apr 28, 2010 #12
    Would you please show your work. I cannot help you otherwise. Show what you have done. W is work not velocity.
  14. Apr 28, 2010 #13
    ok i think i get it you divide the FG by the power to find time and then divide the Fg by the time>? i think
  15. Apr 28, 2010 #14
    UGH. PLEASE WRITE EQUATIONS. Let's start over, this is not that difficult you can do it I promise, but I don't want to just give you the answer. You know derivatives right?

    [tex] P = \frac{dW}{dt} [/tex]

    Where W is the work done. So power is the rate of change of work with respect to time. So if you can write out the equation for work then all you have to do is take the derivative of it (dw/dt) and that will be equal to the power.
  16. Apr 28, 2010 #15
    so 950w= 6007N/dt and then i use it to find time and i know velocity?
  17. Apr 28, 2010 #16
    [tex] W = F_g x = mgx[/tex]

    What is dW/dt?
  18. Apr 28, 2010 #17
    velocity, i got 950m/s. i got the time by dividing dw by P. i think tis should be right
  19. Apr 28, 2010 #18
    yo man don't worry i about it. i screwed up, the answer suppose to be 0. 158m/s. thanks for your help though
  20. Apr 28, 2010 #19
    That is 2000 mph!! DO YOU KNOW DERIVATIVES?? This is what you are going to do:

    FIND dw/dt (W = mgx) where x is the change in height and POST WHAT YOU GET FOR THE DERIVATIVE.
  21. Apr 28, 2010 #20
    Remember mg is a constant, but x is not.
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