How Does a Spring's Work Relate to Its Stretch and Constants?

[uestions]

Homework Statement

How much work is done by the spring on the object as it stretches 0.02m? The spring has a k value of 270N/m and the object is 0.55 kg.

Homework Equations

U_i = U_f - W_s
W_s = 0.5kx_f² - 0.5kx_i²

The Attempt at a Solution

Assuming U_i is when the spring isn't stretched.
U_f = W_s
(.55kg)(9.8m/s/s)(-0.2m) = W_s
(.55kg)(9.8m/s/s)(-0.2m) = .5(270N/m)(0.2m)²
Except they aren't equal.
When W_s = 0.5kx_f² - 0.5kx_i² is used, the answer is -0.054J. What is wrong with my first method?

 

[paisiello2]

First, there seems to me to be some information missing in your solution attempt. You might want to include a diagram or at the very least state all the values given or calculated.

Second, for some reason you are equating the potential energy of gravity with the work done on the spring. While this may or may not be true, it is irrelevant to the question being asked.

 

[gneill]

The question doesn't mention anything about gravity, only that the spring stretches. Could be some other force doing that. It also doesn't mention that the mass is just allowed to fall... if that were the case it would continue to bounce up and down around the equilibrium point, not settle there.

Anyways, consider the principle of conservation of energy. What's the change in potential energy that the spring undergoes when it's stretched by the given amount?

 

[NascentOxygen]

This was discussed in a similar recent thread. When you allow a mass attached to a spring to fall, it first falls to around double its final steady position. At that point, you can compare the two potential energies.

The only way it can settle to a final steady equilibrium is by continually losing energy with each oscillation until motion ceases. Your energy balance hasn't accounted for that energy lost (mostly as heat, ultimately).