Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

Amy hangs a weight on a newtonmeter. The energy stored in the spring is 0.045J She then increases the weight by 2N?

While spring constant = 400N/m

Calculate the total extension?

Any help would be really appreciated, thanks!

3 Answers

Relevance
  • Anonymous
    5 months ago
    Favorite Answer

    Key formulae: F = kx and E = ½kx²

    k = 400N/m

    The energy initially stored in the spring is E₁ = 0.045J. So if the initial extension is x₁:

    E₁ = ½kx₁²

    0.045 = ½ * 400 * x₁²

    x₁ = √(2* 0.045/400) = 0.015m

    She then increases the weight by 2N. ΔF = 2N.

    Since x is proportional to F, increase in extension, Δx, is given by:

    ΔF = kΔx

    2 = 400Δx

    Δx =2/400 = 0.005 m

    Total extension = x₁ + Δx = 0.015 + 0.005 = 0.02m (=2cm)

    • Log in to reply to the answers
  • oubaas
    Lv 7
    5 months ago

    400/2*x^2 = 0.045

    x = √0.090/400 = 0.0150 m (1.5 cm)

    F = k*x = 0.015*400 = 6.0 N

    x' = (F+2)/k = 8/400 = 0.02 m (2.0 cm)

    • Log in to reply to the answers
  • 5 months ago

    If the spring constant is 400 N/m then the spring will be extended BY 1/200 m ( 0.5 cm)

    Now if the energy is 0.045 J

    and E = 1/2 k x^2 x = sqrt( 2E/k) = sqrt( 2*0.045/400) = 0.015 m

    Add the two to get a total extension of 0.020 m

    • Log in to reply to the answers
Still have questions? Get answers by asking now.