A 0.250 kg block on a vertical spring with a spring constant of 5.00 x 103 N/m is pushed downward compressing the spring 0.100 m. When relea

Question

A 0.250 kg block on a vertical spring with a spring constant of 5.00 x 103 N/m is pushed downward compressing the spring 0.100 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above its point of release

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Thiên Hương 4 years 2021-08-02T07:20:25+00:00 1 Answers 220 views 0

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  1. linhdan
    0
    2021-08-02T07:21:43+00:00
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    Answer:

    The height at point of release is 10.20 m

    Explanation:

    Given:

    Spring constant k = 5 \times 10^{3} \frac{N}{m}

    Compression x = 0.10 m

    Mass of block m = 0.250 kg

    Here spring potential energy converted into potential energy,

       mgh = \frac{1}{2} kx^{2}

    For finding at what height it rise,

      0.250 \times 9.8 \times h = \frac{1}{2} \times 5 \times 10^{3} \times (0.10) ^{2}                ( ∵ g = 9.8 \frac{m}{s^{2} } )

      h = 10.20 m

    Therefore, the height at point of release is 10.20 m

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