4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the spring is 250 N/m. What is the potential energy of the

Question

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the
spring is 250 N/m. What is the potential energy of the spring?

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Tryphena 5 years 2021-07-14T20:02:25+00:00 2 Answers 13 views 0

Answers ( )

  1. thuthao
    0
    2021-07-14T20:03:30+00:00
    Reply

    31.25 hope this helps:)

  2. Doris
    0
    2021-07-14T20:03:58+00:00
    Reply

    Answer:

    31.25

    Explanation:

    The formula for the potential energy of a spring is \frac{1}{2}kx^2, where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get \frac{1}{2}\cdot 250\cdot 0.25=31.25. Hope this helps!

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