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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? solve using work-energy theorem

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  1. anonymous
    • 4 years ago
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    ma=1/2*kx^2 10*9.8=1/2*700*x^2 x approximately equals 0.529m

  2. Dandandan
    • 4 years ago
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    wer did u get this -> ma=1/2*kx^2

  3. Dandandan
    • 4 years ago
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    i mean the ma..

  4. anonymous
    • 4 years ago
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    ma=potential energy

  5. anonymous
    • 4 years ago
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    F=ma

  6. anonymous
    • 4 years ago
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    a is g

  7. anonymous
    • 4 years ago
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    gravity

  8. Dandandan
    • 4 years ago
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    oww.. okok, yeah, i can remember it now,, gheez thanks !! ..

  9. Dandandan
    • 4 years ago
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    can i still ask one more??

  10. anonymous
    • 4 years ago
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    kk

  11. 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?

  12. anonymous
    • 4 years ago
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    (mu_k)mg(delta d)=1/2*kx^2

  13. anonymous
    • 4 years ago
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    (mu_k)(0.5)(9.8)(1)=1/2*(100)(0.20)^2

  14. anonymous
    • 4 years ago
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    (mu_k)=4/9.8

  15. anonymous
    • 4 years ago
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    basically all the kinetic energy of the spring is converted into heat/friction

  16. Dandandan
    • 4 years ago
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    okok xD

  17. Dandandan
    • 4 years ago
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    wait. whats this --> (mu_k)

  18. anonymous
    • 4 years ago
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    coefficient of friction

  19. Dandandan
    • 4 years ago
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    i think its just mu..

  20. Dandandan
    • 4 years ago
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    aww. okok, k stands for kinetic xDD

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