A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while mo

Question

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest. 1)If the initial speed of the box were doubled, how far x2 would the spring compress?

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Vodka 4 years 2021-09-05T05:58:03+00:00 1 Answers 20 views 0

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  1. danthu
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    2021-09-05T05:59:52+00:00
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    Answer:twice of initial value

    Explanation:

    Given

    spring compresses x_1 distance for some initial speed

    Suppose v is the initial speed and k be the spring constant

    Applying conservation of energy

    kinetic energy converted into spring Elastic potential energy

    \dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

    When speed doubles

    \dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

    divide 1 and 2

    \dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

    x_2=2x_1

    Therefore spring compresses twice the initial value

       

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