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a 1 kg mass is attached to a spring hanging vertically and hangs at rest in the equilibrium position. the spring constant of the spring is 1 n/cm. the mass is pulled downward 2 cm and released. what is the speed of the mass when it is 1 cm above the point from which it was released?

  • physics -

    Max spring potential energy
    = (1/2) k X^2 = (100/2)*(0.02)^2
    = 0.2 Joules

    When 0.01 m below the release point, the spring has 1/4 of its max potential energy, so 0.15 J is available for kinetic energy

    (1/2)MV^2 = 0.15 J can be solved for the speed V. I get 0.548 m/s

    It is permissible to neglect gravity in this problem when you regard kx as the force applied by the spring, including Mg, with x measured from the equilibrium position. g does not affect the answer or the period of vibration.

  • physics -

    the answer is 1.73 but i don't know how can every one help me ??

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