A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to further stretch

Question

A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to further stretch the spring to 5.79 cm beyond its equilibrium length?

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Ben Gia 4 years 2021-08-17T19:28:07+00:00 1 Answers 28 views 0

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  1. Jezebel
    0
    2021-08-17T19:29:08+00:00
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    Answer:

    2.31J

    Explanation:

    the energy for a spring system is given by:

    E=\frac{1}{2} kx^2

    where k is the spring constant: k=15.3N/cm=1530N/m and x is the distance stretched from the equilibrium position.

    In the first case x=1.81cm=0.0181m

    so the energy to stretch the spring 1.81cm is:

    K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J

    and for the second case,  the energy to stretch the spring 5.79cm:

    x=5.79cm=0.0579m

    K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J

    so to answer a) we must find the difference between these energies:

    2.56J-0.25J=2.31J

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