Question

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 the spring? 700 J 1050 J 1575 J 4375 J

Answer 1

The answer is C. 1575 J. I just took this review.

Answer 2

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`