 # If 6 J of work is needed to stretch a vegetars8t 2021-12-16 Answered
If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?
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Let the natural length be l
Then we can write
$6={\int }_{10-l}^{12-l}kxdx$
$6={\left[\frac{1}{2}k{x}^{2}\right]}_{10-l}^{12-l}$
$6=\frac{1}{2}k\left[{\left(12-l\right)}^{2}-{\left(10-l\right)}^{2}\right]\to$ Equation 1
And
$10={\int }_{12-l}^{14-l}kxdx$
$10={\left[\frac{1}{2}k{x}^{2}\right]}_{12-l}^{14-l}$
$10=\frac{1}{2}k\left[{\left(14-l\right)}^{2}-{\left(12-l\right)}^{2}\right]\to$ Equation 2
Divide Equation 2 by Equation 1
$\frac{10}{6}=\frac{{\left(14-l\right)}^{2}-{\left(12-l\right)}^{2}}{{\left(12-l\right)}^{2}-{\left(10-l\right)}^{2}}$
Recall that: ${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$
$\frac{5}{3}=\frac{\left(14-l-12+l\right)\left(14-l+12-l\right)}{\left(12-l-10+l\right)\left(12-l+10-l\right)}$
$\frac{5}{3}=\frac{\left(2\right)\left(26-2l\right)}{\left(2\right)\left(22-2l\right)}$
$\frac{5}{3}=\frac{13-l}{11-l}$
Cross Multiply
$55-5l=39-3l$
Add 5l on both sides
$55=39+2l$
Subtract 39 drom both sides
$16=2l$
Divide both sides by 2
$8=l$
Result: Natural length of the spring is 8 cm

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