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## Homework Statement

An ideal spring is hung vertically from the ceiling. When a 2.0

*kg*mass hangs from it at rest, the spring is extended 0.06

*meters*from its relaxed state. An upward external force is then applied to the block to move it upward a distance of 0.16

*meters.*While the block is being raised by the force, the work done by the spring is:

## Homework Equations

Force from a spring= -kd

Work due to a spring= .5(k)(xi)^2-.5(k)(xf)^2

## The Attempt at a Solution

I believe to find the magnitude of the spring constant i rearrange F= -kd to become k=F/d (i read that we can neglect the negative in this case? correct me if im wrong)

so F=mg=(2.0kg)(9.8m/s^2)=19.6N

19.6N/.06m=326.667N/m=k

Finally, W=.5(326.667N/m)(.06m)^2 - .5(326.667N/m)(.10m)^2= -1.04J

Did i find the spring constant correctly? I feel like using mg/d was wrong.

Also, when i use the

*work done by spring*equation, is xi and xf the distance away from equilibrium?