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DominantVampire

  • one year ago

kindly help please. A Block Of Mass 12 kg

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  1. DominantVampire
    • one year ago
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    a block of mass 12 kg is released from rest om a friction-less incline of the angle theta=30 below the block is a spring having spring constant 10^4 N/m. the block stops momentarily when it compresses the spring by 4 cm. how far does the block move down the incline from its rest position to this stopping point

  2. DominantVampire
    • one year ago
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    |dw:1434185406478:dw|

  3. DominantVampire
    • one year ago
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    no body knows perhaps, sighs.

  4. anonymous
    • one year ago
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    do you know how to use energy conservation??

  5. DominantVampire
    • one year ago
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    yes.

  6. DominantVampire
    • one year ago
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    Ki+Ui=Kf+Uf ?

  7. anonymous
    • one year ago
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    yep... initial energy will be the potential energy of the block and the final energy will be the spring extension.. can you relate?

  8. anonymous
    • one year ago
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    compression**

  9. DominantVampire
    • one year ago
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    the Ki is 0 and Uf would be zero as well i guess

  10. anonymous
    • one year ago
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    Uf would not be zero as the spring is compressed and is having potential energy its like the GPE is getting converted to spring energy

  11. anonymous
    • one year ago
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    GPE is gravitational potential energy

  12. DominantVampire
    • one year ago
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    yes.

  13. DominantVampire
    • one year ago
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    the spring constant is 10,000N/m ?

  14. anonymous
    • one year ago
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    yes

  15. DominantVampire
    • one year ago
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    will it be mgh=1/2 kx^2

  16. anonymous
    • one year ago
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    yes

  17. DominantVampire
    • one year ago
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    and plug in the values? h is not given

  18. anonymous
    • one year ago
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    h is what you have to find!!

  19. DominantVampire
    • one year ago
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    v have to find out h?

  20. DominantVampire
    • one year ago
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    yes, sorry

  21. DominantVampire
    • one year ago
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    so it becomes like h = [(1/2)kx^2][mg]

  22. DominantVampire
    • one year ago
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    correct?

  23. anonymous
    • one year ago
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    is mg in the denom?

  24. DominantVampire
    • one year ago
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    h = [(1/2)kx^2] / [mg]

  25. anonymous
    • one year ago
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    then correct

  26. DominantVampire
    • one year ago
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    the h is 6.80 m

  27. DominantVampire
    • one year ago
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    now what

  28. DominantVampire
    • one year ago
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    sin(30) = h/x to determine the hypotenuse

  29. DominantVampire
    • one year ago
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    @ganeshie8

  30. DominantVampire
    • one year ago
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    now h/sin30=x ?????

  31. DominantVampire
    • one year ago
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    the ans is 13.6 ????

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