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).


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.

1 Answer

  • 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:

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