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DanDanDan

  • 4 years ago

A vertical spring with a force constant of 700 N/m is initially at equilibrium. A 10kg mass is attached to the end of the spring, causes it to elongate. What will be the maximum elongation/deformation of the spring? use work-energy theorem

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  1. IsTim
    • 4 years ago
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    W=E?

  2. anonymous
    • 4 years ago
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    Assuming this is a vertical arrangement. \[W_s = {1 \over 2} k x^2\]\[E_m = mgx\]

  3. Dandandan
    • 4 years ago
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    so \[(1/2) kx^2=mgx ?! \]

  4. IsTim
    • 4 years ago
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    Yup.

  5. Dandandan
    • 4 years ago
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    i asked this earlier but that guys said ma=1/2*kx^2.. is it the same ?!

  6. IsTim
    • 4 years ago
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    I dunno. It doesn't look like it, but I recognize that configuration.

  7. Dandandan
    • 4 years ago
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    A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m. What is the coefficient of friction between the block and the tabletop?

  8. AravindG
    • 4 years ago
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    hey salini1!

  9. anonymous
    • 4 years ago
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    |dw:1328026340799:dw| by work-energy theorem change in k.e=0 as when we consider that last point when spring has zero velocity,that is when it is pulled backward so that is equal to Net WORK DONE=work done by spring force-mge where e is the extension of spring that is haow mg*e=1/2ke^2

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