A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m, how much elastic potential energy is stored in th…

Question

asked Jan 22, 2018 107k views

2 Answers

Stephane Mathis · Answer 1 · 2018-01-27T14:18:00+0000

Answer:

1575 J

Step-by-step explanation:

The elastic potential energy of a string is given by:

$E=(1)/(2)k \Delta x^2$

where

k is the spring constant

$\Delta x$ is the stretching/compression of the string with respect to its natural length

For the spring in the problem, we have:

- Spring constant:
k=1400 N/m

- Stretching:
$\Delta x=2.5 m -1.0 m=1.5 m$

Therefore, the elastic potential energy stored in the spring is

E=(1)/(2)(1400 N/m)(1.5 m)^2=1575 J