Question

A spring requires 12 J12 J to stretch the spring from 8 cm8 cm to 10 cm10 cm, and an additional 48 J48 J to stretch the spring from 10 cm10 cm to 14 cm14 cm. What is the natural (unstressed) length of the spring?

Answer 1

6cm

Explanation:

Given

12J to stretch from 8cm to 10cm ---- Expression 1

Additional 48J from 10cm to 14cm ---- Expression 2

Let l represents the length of the spring

Let k represent string constant,

We can then write ( from expression 1)

12 = integral of kx dx

The lower bound being 8 - l

And the upper bound being 10 - l

Integrating kx dx

We have

½kx²

= ½k[x²]

= ½k[(10 - l)² - (8 - l)²].

So, 12 = ½k[(10 - l)² - (8 - l)²] ------ Equation 1

From expression 2, we can write

48 = integral of kx dx

The lower bound being 10 - l

And the upper bound being 14 - l

Integrating kx dx

We have

½kx²

= ½k[x²]

= ½k[(14 - l)² - (10 - l)²].

So, 48 = ½k[(14 - l)² - (10 - l)²] ------ Equation 2

Divide Equation 2 by Equation 1, so we get

48/12 = ½k[(14 - l)² - (10 - l)²] / ½k[(10 - l)² - (8 - l)²] ----- ½k cancel out ½k

So, w have

4 = [(14 - l)² - (10 - l)²] / [(10 - l)² - (8 - l)²]

Recall that a² - b² = (a + b)(a - b).

So,

4 = [(14 - l - 10 + l) (14 - l + 10 - l)] / [(10 - l - 8 + l) (10 - l + 8 - l)]

4 = [(4)(24 - 2l)]/[(2)(18-2l)]

4 = (96 - 8l)/36 - 4l) ---- Multiply both sides by 36 - 4l

4(36 - 4l) = 96 - 8l

144 - 16l = 96 - 8l ----- Collect Like Terms

144 - 96 = 16 - 8l

48 = 8l ----- Divide both sides by l

48/8 = l

6 = l --- Rearrange

l = 6

So, the natural length of the string is 6cm