Answer:
![\Delta W=24.1162\ J](https://img.qammunity.org/2021/formulas/physics/college/3a7wp2nz9q8bbmfqhvqmnonvxvnja1sxqn.png)
Step-by-step explanation:
Given:
- work done to stretch the spring,
![W=4.5\ J](https://img.qammunity.org/2021/formulas/physics/college/atjyhidn6be0n6v313j7724jr7f1af8upa.png)
- length through which the spring is stretched beyond equilibrium,
![\Delta x=2.3\ cm=0.023\ m](https://img.qammunity.org/2021/formulas/physics/college/gtdyx9e78gdo9q9vm5t5r7bwkvdtl74yxu.png)
- additional stretch in the spring length,
![\delta x=3.5\ cm=0.035\ m](https://img.qammunity.org/2021/formulas/physics/college/t2qiirdfzz11piozn1wi86dtlo08wmhdve.png)
We know the work done in stretching the spring is given as:
![W=(1)/(2) * k.\Delta x^2](https://img.qammunity.org/2021/formulas/physics/college/ee9n5yq4zgdmdf2n82i52ne3tw5akstm0l.png)
where:
k = stiffness constant
![4.5=0.5* k* 0.023^2](https://img.qammunity.org/2021/formulas/physics/college/g36cpj5vu4i4qs97fn9r5i5hg9rkeahl71.png)
![k=17013.2325\ N.m^(-1)](https://img.qammunity.org/2021/formulas/physics/college/mjzboge1wrz973ac94ywcl7vj0qvzsk6c4.png)
Now the work done in stretching the spring from equilibrium to (
):
![W'=0.5* k.(\Delta x+\delta x)^2](https://img.qammunity.org/2021/formulas/physics/college/ffrjgjqyt3wwzb0qd0cy7x7n30p7t2y6gt.png)
![W'=0.5* 17013.2325* 0.058^2](https://img.qammunity.org/2021/formulas/physics/college/bcbjkjvp2frzmbsnla7fglbegxndb3mc0x.png)
![W'=28.6162\ J](https://img.qammunity.org/2021/formulas/physics/college/mdi4l7vfckw0c6w33sx12fxkpufzht4ea7.png)
So, the amount of extra work done:
![\Delta W=W'-W](https://img.qammunity.org/2021/formulas/physics/college/r52tgxjxppi2sqh1m7zuybdumkwdv378u6.png)
![\Delta W=28.6162-4.5](https://img.qammunity.org/2021/formulas/physics/college/4rzsmgme1tfx333bmzk1ec72zibg6qnlf1.png)
![\Delta W=24.1162\ J](https://img.qammunity.org/2021/formulas/physics/college/3a7wp2nz9q8bbmfqhvqmnonvxvnja1sxqn.png)