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.

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

raphealnwobi raphealnwobi · Answer 2 · 2022-03-03T12:31:20+01:00

The extra work is required to stretch it an additional 9.39 cm is 4.04 J

Work

The work done in stretching a spring is given by:

W = (1/2)kx²

where k is the spring constant and x is the extension.

Given W = 2.58 for x = 15.6 cm = 0.156

2.58 = (1/2)*k * 0.156²

k = 212 N/m

For an additional 9.39 cm, x = 15.6 + 9.39 = 0.2499, hence:

W = (1/2) * 212 * 0.2499² = 6.62 J

Extra work required = 6.62 - 2.58 = 4.04 J

The extra work is required to stretch it an additional 9.39 cm is 4.04 J

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