- #1
Le_Anthony
- 17
- 1
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 I am 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?