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25 October, 14:28

A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. The system is then released from rest. What is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?

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  1. 25 October, 16:18
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    kinetic energy is 2.02 J

    Explanation:

    Given data

    k = 600 N/m

    angle = 30°

    spring = 0.10 m

    mass = 2 kg

    to find out

    kinetic energy of the block

    solution

    we know initial energy that is

    Energy 1 = 1/2 * kx²

    put here k = 600 and x = 0.1

    energy 1 = 1/2 * 600 (0.1) ²

    energy = 3

    potential energy PE = mgh

    PE = 2 * 9.8 * 0.1sin30

    PE = 0.98 J

    so we know final energy is

    E = KE of block + PE

    so

    kinetic energy = 3 - 0.98

    kinetic energy = 2.02 J
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