Ask Question
19 April, 09:40

It takes 2.58 J of work to stretch a Hooke's-law spring 15.6 cm from its unstressed length. How much the extra work is required to stretch it an additional 9.39 cm? Answer in units of J.

+4
Answers (1)
  1. 19 April, 12:26
    0
    Extra work required = 4.04 J

    Explanation:

    Hooks Law: Hooks law states that provided the elastic limit is not exceeded, the force applied to an elastic material is directly proportional to the extension.

    E₁ = 1/2ke² ... Equation 1

    Where E = Energy required to stretch or compress the spring, k = force constant of the spring, e = extension.

    making k the subject of the equation in equation 1

    k = 2E/e² ... Equation 2

    Given: E = 2.58 J, e = 15.6 cm = 15.6/100 = 0.156 m.

    Substituting these values into equation 2

    k = 2*2.58 / (0.156) ²

    k = 212.03 N/m

    k = 5.75 N/m.

    With an additional extension of 9.39 m, therefore e₂ = 15.6 + 9.39 = 24.99 cm

    = 0.2499 m

    using,

    E₂ = 1/2k (e₂) ²

    E₂ = 1/2 (212.03) (0.2499) ²

    E = 6.62 J

    E ≈ 6.62 J

    Extra work required = E₂ - E₁

    Extra work = 6.62 - 2.58

    Extra work = 4.04 J
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “It takes 2.58 J of work to stretch a Hooke's-law spring 15.6 cm from its unstressed length. How much the extra work is required to stretch ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers