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

The spring is stretched by 33.7 cm.

Step-by-step 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
`(1)/(2)* k * x^2.`

Therefore,
`(1)/(2)* k * x^2=25\\(1)/(2)* 440* 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.

Answer 2

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.