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Help! Work/Energy Problem

  1. Oct 2, 2004 #1
    Hi guys! I need some help :)

    A projectile of mass 0.709 kg is shot straight up with an initial speed of 18.7 m/s. (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only 7.58 m, determine the magnitude of the average force due to air resistance.

    So I figured out how to do part a) by setting KE=PE and solving for h. I just don't know what to do to solve for part b. I have been trying to find the work and the plug the work into the W=FdCos (theta) equation, but I can't seem to get the right answer.

    THANKS!
     
  2. jcsd
  3. Oct 2, 2004 #2

    Pyrrhus

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    let me see your second part, show the work.
     
  4. Oct 2, 2004 #3
    I have a question first, for part a I set KE=PE....so I did (.5*.709*(18.7^2)) and I got 123.965. I took that value and divided that by (.709*9.8) and I got 17.84m...which is close to what the real answer should be (17.66111 m)...do you know why it's off or is it because I am using an equation that I shouldn't???

    Okay, for part b:
    I know what W=FdCos (theta) and also that W=mgh (PE)...So I was thinking of just plugging in (.709*9.8*7.58) which equals 52.667J. Should I plug that into the W=Fdcos (theta) or not??
     
  5. Oct 2, 2004 #4

    Pyrrhus

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    The first part is right

    Applying Law of Conservation of Mechanical Energy

    [tex] K_{o} = \Omega_{maxHeight} [/tex]

    For the second part

    Conservative Systems Work:

    [tex] W_{conservative} = -\Delta \Omega [/tex]

    Because the work done by air resistance forms an angle of 0 degrees with the displacement vector, it will be positive.

    [tex] Fd = -\Delta \Omega [/tex]

    [tex] F = -\frac{\Delta \Omega}{d} [/tex]

    Also because the initial Yo is 0, it will be mgY.
     
    Last edited: Oct 2, 2004
  6. Oct 2, 2004 #5
    Is your equation F=delta omega/d...is the delta omega the addition of the change in potential and kinetic energy? And are you trying to say that because the object starts from the ground, that the initial PE is O...so you only use the final?
     
  7. Oct 2, 2004 #6

    Pyrrhus

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    [tex] \Delta \Omega [/tex] is the way i was taugh to represent Potential Energy, and it's the change of Potential Energy, mgy - 0, in this case. Btw y = d, i forgot to use the same variable :smile:
     
  8. Oct 2, 2004 #7
    Okay so I can say that the change in potential energy is (.709*9.8*7.58) which gives me 52.66 J.....I just don't know what distance to divide by though...grr...i guess i better search my book again :/ thanks for all of your help though :) I appreciate it.
     
  9. Oct 3, 2004 #8
    I got it! I didn't realize that I had to factor in air resistance and I found a similar problem which dealt with air resistance...so yay! thanks again! :)
     
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