Question

A spring with a force constant of 5.5 N/m hasa relaxed length of 2.54 m. When a mass isattached to the end of the spring and allowedto come to rest, the vertical length of thespring is 3.52 m.Calculate the elastic potential energystored in the spring.Answer in units of J.

Answer 1

Answer:

E = 2.6411 J

Explanation:

we know that elastic potential energy could be calculated by:

$E = \frac{1}{2}kx^2$

where k is the constant of the spring and x is the deform of the spring.

Now x can be calculated as:

x = 3.52m-2.54m

x = 0.98 m

Finally, replacing the data, we get:

$E = \frac{1}{2}(5.5 N/m)(0.98m)^2$

E = 2.6411 J