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DominantVampire

  • one year ago

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

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  1. IrishBoy123
    • one year ago
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    Gravitational potential converts to spring potential as block slides and spring compresses. So balance the 2 taking account of slope.

  2. shamim
    • one year ago
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    Potential energy of compressed spiring=.5kx^2=.5*10^4*.04^2=?

  3. shamim
    • one year ago
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    Potential energy at the top of inclined plane=kinetic energy at the bottom of inclined plane=mgh=12*9.8*h

  4. shamim
    • one year ago
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    Potential energy of spring=kinetic energy of block

  5. shamim
    • one year ago
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    h=?

  6. shamim
    • one year ago
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    At last u hv to find out the length of inclined plane

  7. Pawanyadav
    • one year ago
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    MgHsin30 ⏰=0.5×10^4×4^2

  8. DominantVampire
    • one year ago
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    @sikinder is this formula above written right which as been suggested

  9. DominantVampire
    • one year ago
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    but the ans is totally diff. the h i get is 0.136m

  10. anonymous
    • one year ago
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    i was thinking that ur calculation method is right @DominantVampire

  11. DominantVampire
    • one year ago
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    is it?

  12. anonymous
    • one year ago
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    yes i think @DominantVampire

  13. DominantVampire
    • one year ago
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    alright. thankyou.

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