A basket of negligible weight hangs from a vertical spring scale of force constant 1500N/m .?

If you suddenly put a 3.00kg adobe brick in the basket, find the maximum distance that the spring will stretch.

Δlmax = ?cm

I tried 19.6 cm but its wrong 🙁

B) If, instead, you release the brick from 1.00m above the basket, by how much will the spring stretch at its maximum elongation?

Δl = ?cm

2 Answers

  • let x metres be the vertical extension:

    conservation of energy:

    loss in potential energy = gain in spring p.e

    => mgx = ½kx²

    => x = 2mg/k = 2x3x(9.81)/1500 = 0.03924 m = 3.924 cm

    b) conservation of energy:

    mg (1+x) = ½kx²

    => 1500x² – 6gx – 6g = 0

    or x² – .03924x – 0.004 = 0

    solving x = [0.03924 ± 0.13243]/2 = 0.086 m = 8.6 cm

    hope this helps

  • Try 1.96 cm

    (1/2)kx^2=(1/2)mv^2

    (1/2)mv^2=mgh

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