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Sagot :
As we know,
- [tex] \boxed{ \boxed{potential \: \: energy = \dfrac{1}{2} k {x}^{2} }}[/tex]
where,
- k = 136 N/m
- x = 12.5 cm = 0.125 m
now, let's solve to find potential energy :
- [tex]p = \dfrac{1}{2} \times 136 \times 0.125 \times 0.125[/tex]
- [tex]p = \dfrac{68 \times 125 \times 125} {1000 \times 1000} [/tex]
- [tex]1.0625 \: \: \: joules[/tex]
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