Post a New Question


posted by .

A spring with k = 54 N/cm is initially stretched 1 cm from its equilibrium length.

(a) How much more energy is needed to further stretch the spring to 2 cm beyond its equilibrium length?
(b) From this new position, how much energy is needed to compress the spring to 2 cm shorter than its equilibrium position?

  • Physics -

    PE1 =k•x1²/2
    PE2 = k•x2²/2
    W = ΔPE =PE2 –PE1 =k(x2² - x1²)/2 = ….
    where k =54 N/m, x1 =0.01 m,
    x2 =0.01+0.02 = 0.03 m.

    this compression will be done by the work of elastic force of the spring. When we release the spring which is stretched by 0.03 m, the released energy is k•(0.003)²/2 which is large than k•(0.02)²/2.

  • Physics -

    I'm still a little confused. When I use the equation:
    W = k(x2^2 - x1^2)/2, I get 0.0216 J. This doesn't seem correct.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question