Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 years ago

Work to Stretch a Spring?

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 37.5 N, the spring is stretched by 15.6 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 13.9 cm from that position (in J).

Update:

If you could please explain why you chose to use specific formulas, that would be very helpful! I'm trying to learn how to better solve these problems on my own.

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  • Jim
    Lv 7
    4 years ago
    Favorite Answer

    F = kx {Hooke's Law for a spring in its elastic range}

    3.75 = k(0.156)

    k ≈ 24.0 N/m

    SPE = 1/2kx² {definition of Spring Potential Energy}

    at position 1: SPE = (0.5)(24.0)(0.156)² = 0.292032 J

    at position 2: SPE = (0.5)(24.0)(0.156 + 0.139)² = 12.0(0.295)² = 1.0443 J

    work/energy required to stretch spring from position 1 to position 2: = 1.0443 - 0.292032 ≈ 0.752 J ANS

    • Dark4 years agoReport

      Jeez ur so active just read this from last year:

      https://answers.yahoo.com/question/index?qid=20150616151750AAhJosQ

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