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adnanchowdhury

  • 3 years ago

Kinetic Energy and Potential Energy question: http://d.pr/i/vcK4 The gymnast descends from position B and regains contact with the trampoline when it is in its unstretched position. At this position, her centre of mass is 3.2 m below its position at B. Calculate her kinetic energy at the instant she touches the unstretched trampoline

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  1. timo86m
    • 3 years ago
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    mass*4.2*9.8 I think

  2. timo86m
    • 3 years ago
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    mass is 55

  3. adnanchowdhury
    • 3 years ago
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    nope. That's PE of when the person is at B.

  4. timo86m
    • 3 years ago
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    and i will be ke once the person reaches back

  5. timo86m
    • 3 years ago
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    |dw:1337888965008:dw|

  6. timo86m
    • 3 years ago
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    |dw:1337889467795:dw|

  7. timo86m
    • 3 years ago
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    |dw:1337889552978:dw|

  8. lalaly
    • 3 years ago
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    kinetic energy =3.2/4.2 *2264

  9. stormfire1
    • 3 years ago
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    @lalaly: Wow...that's much more shorter than the stuff I was about to post but it's dead-on.

  10. stormfire1
    • 3 years ago
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    Well, almost :) (3.2/4.2)*2264

  11. stormfire1
    • 3 years ago
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    At any rate, what lalaly did was calculate the total PE from point A to B given 4.2m, 55kg and g (mgh = 2264J). Her KE at point B will be equal to this. Once you know those two things, you should be able to see that (3.2/4.2)*2264J will be the amount of KE at the point where the girl just touches the unstretched trampoline.

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