# Ideal Spring

## 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?

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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?
the k value is correct.
regarding work done by the spring -pl. check whether the choice of xi and xf is correct....
in the sector of extended portion the external force was supporting the wt. so what work was done by the spring -it came back to the relaxed state...i am just trying to think ....alongwith you

• Le_Anthony
haruspex
Homework Helper
Gold Member
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)
Yes, that's ok. In fact, the force is always opposite to the displacement, so technically it should be k=-F/d.

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.
That all looks correct. Are you surprised by the minus sign?

• Le_Anthony