Post a New Question


posted by .

an ideal spring of negligible mass is 12.00 cm long when noting is attached to it. When you hang a 3.15kg weight from it, you measures its length to be 13.40 cm.
if you wanted to store 10.0 J of potential energy in this spring, what would be its total length?
Assume that it continues to obey hooke's Low.

  • physics -

    First find K of the spring:
    Mg = Kx
    here, x = elongation of the spring
    = 13.40 - 12.00
    = 1.40 cm
    = 0.014 m

    K = Mg/x
    = 3.15*9.8/0.014 = ??

    Having got K, find X from:
    KX^2/2 = 10.0J
    where X is the elongation when stored PE is 10.0 Joules.
    The total length of spring then would be:
    X + 0.12 m

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question