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Potential Energy and Kinetic Energy Problem

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

    Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
    Vertical velocity up the mountain=0.0985m/s
    Horizontal velocity down the mountain=0.141 m/s

    A)Calculate the change in gravitational potential energy for Brandon and for Tom as they make their climb in the Rockies.
    Delta GPE for Brandon=
    Delta GPE for Tom=

    B)Calculate the average kinetic energy for Brandon and for Tom as they climb up the mountain and as they descend.

    KE for Brandon going up=
    KE Tom going up
    KE for Brandon coming down =
    KE for Tom coming down=

    2. Relevant equations
    A)GPE=mgh
    B)KE=1/2mv^2


    3. The attempt at a solution
    It seems to simple for question A is too simple to be true so I'm concerned about wether it is correct.
    M computations:
    GPE=mgh
    =(91.1kg)(9.8m/s^2)h and (11.2kg)(9.8m/s^2)h

    Question:What height should i use for h? Is this the right equation? What is meant by Change in gravitational energy,do I have to subtract them?


    For B)
    KE=1/2mv^2
    KE for Brandon going up
    KE=1/2(91.1kg)(0.0985 m/s^2)^2
    =.441

    KE Brandon going down
    KE=1/2(91.1kg)(0.141 m/s^2)^2
    KE=.905

    Whats the units for KE?
    Is this correct and if so should I do the same for Tom?
     
  2. jcsd
  3. Oct 6, 2007 #2

    learningphysics

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    change in gpe = [tex]mgh_2 - mgh_1 = mg(h_2-h_1)[/tex]

    so in your equation h refers to the change in height...

    For the kinetic energy part, you need the net velocity... you used only the horizontal and vertical component of velocity in your formula.

    "Vertical velocity up the mountain=0.0985m/s"

    so what is the magnitude of the net velocity? use trig...
     
  4. Oct 6, 2007 #3
    Thank you Learn,

    I dont understand because my avg velocity going up the mountain was 0.0985m/s. My avg velocity going down the mountain was 0.141m/s. But if I used my net velocity would that make my numbers the same for KE goin up and down for Brandon,becuase the only variable that changes is my mass. I'm really confused. Help clarify things for me please.
     
  5. Oct 6, 2007 #4

    learningphysics

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    This is what the question says:

    Vertical velocity up the mountain=0.0985m/s
    Horizontal velocity down the mountain=0.141 m/s

    The average velocity going up the mountain is not 0.0985m/s... the average vertical velocity going up the mountain is 0.0985m/s... Brandon has both vertical and horizontal velocity... what is the horizontal velocity while Brandon is going up the mountain? What is the net velocity going up the mountain (vector sum of horizontal and vertical velocities) ?
     
  6. Oct 6, 2007 #5
    We are getting there Learn,


    Can I find the horizontal velocity by doing my avg velocity of .705m/s times my time which is 22680s? Is my horizontal velocity equation v x t?

    Then in order to find my net velocity I add my horizontal and vertical velocities,Correct?
    But what does my net velocity solve for? I really need help with my forumulas because i am getting confused.
    My avg velocity is .705m/s if that helps.
     
  7. Oct 6, 2007 #6

    learningphysics

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    How did you get 0.705?
     
  8. Oct 6, 2007 #7
    I got 0.705 when i divided the length of their path(16000m)/ the time it took them to get there 22680s,which equals 0.705m/s. That was their average speed in m/s.
     
  9. Oct 6, 2007 #8

    learningphysics

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    jayced, I'm confused by the question itself... is this just the question or does this include stuff you calculated:

    "Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

    Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
    Vertical velocity up the mountain=0.0985m/s
    Horizontal velocity down the mountain=0.141 m/s"

    so all of this is part of the question? they gave 0.0985m/s and 0.141m/s in the question?
     
  10. Oct 6, 2007 #9
    It was one problem but my teacher refernced it for more questions. For the first part I calculated:

    A)What is the average angle of ascent the path makes with the horizontal?Theta=8 degre
    sin theta =2235m/16000m the 2235 came from 2746m-511m=2235
    sin-1(0.139)=8.03 degrees

    B)What is their average speed along the path in meters/second? average speed=.705 m/s
    Avg speed=d/t
    So I took 16000m/22680s=.705m/s

    C)What is their average vertical velocity? Vup=0.0985 m/s
    DeltaV/t=Average velocity
    2235m/22680s=0.0985 m/s

    D)After some food they hike back down to their starting point in T2=4.24 hours
    What is their average vertical velocity? Vdown=0.141 m/s
    DeltaV/t=avg vertical velocity
    2235m/15840s=0.141m/s

    This is what I calculated right above. Did I clarify it enough? The only thing thats given was :

    Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

    Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
    Vertical velocity up the mountain=0.0985m/s
    Horizontal velocity down the mountain=0.141 m/s
     
  11. Oct 6, 2007 #10

    learningphysics

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    ok. so the average velocity up the mountain is 16000/22680 = 0.705m/s. use this for the kinetic energy going up the mountain

    How did you get 15840s for the time down the mountain? 4.24*3600 = 15264s

    anyway for down the mountain, use 16000/time down the mountain... then use that velocity to get kinetic energy down the mountain.
     
  12. Oct 6, 2007 #11
    60mins=1hr 24mins= 24min*60s=1440s
    60mins* 4hr=240mins
    60s=1min
    240min*60s=14400 +1440s= total seconds =15840
     
  13. Oct 6, 2007 #12

    learningphysics

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    Oh, I see... it's not 4.24 hrs, but 4hrs, 24 minutes... okay, so 16000/15840 = 1.01m/s. use this for the kinetic energy down the mountain...
     
  14. Oct 6, 2007 #13
    Wow ,thanx so much learningphysics you helped me plenty and this is the best kept secret. Good luck and know that its greatly appreciated.
     
  15. Oct 6, 2007 #14

    learningphysics

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    :smile: you're welcome.
     
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