Answer:
The spring is stretched by 33.7 cm.
Explanation:
It is given that,
Spring constant, k=440 N/m.
Energy stored in the stretched spring, E=25 J.
Now, Spring potential energy when it is stretched x meters is [tex]\dfrac{1}{2}\times k \times x^2.[/tex]
Therefore, [tex]\dfrac{1}{2}\times k \times x^2=25\\\dfrac{1}{2}\times 440\times x^2=25\\x^2=0.113\ m\\x=0.337\ m=33.7\ cm.[/tex]
The spring is stretched by 33.7 cm.
Hence, this is the required solution.
