Question

A spring has a spring stiffness constant, k, of 440 N/m. How much must the spring be stretched to store 25 J of potential energy?

Answer 1

E p = 25 J,  k = 440 N/m
E p = 1/2 k x² 
25 J = 1/2 · 440 N/m x²
25 J = 220 N/m · x²
x² = 25 : 220 = 0.113636
x = √0.113636
x = 0.337 m
The string must be stretched 0.337 m. 

Answer 2

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 $\dfrac{1}{2}\times k \times x^2.$  

Therefore, $\dfrac{1}{2}\times k \times x^2=25\\\dfrac{1}{2}\times 440\times x^2=25\\x^2=0.113\ m\\x=0.337\ m=33.7\ cm.$

The spring is stretched by 33.7 cm.

Hence, this is the required solution.