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

Answer:

1575 J

Explanation:

The elastic potential energy of a string is given by:

$E=\frac{1}{2}k \Delta x^2$

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

$E=\frac{1}{2}(1400 N/m)(1.5 m)^2=1575 J$

Answer 2

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