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October 24, 2025, 05:22:12 am

Author Topic: Spring and hand question  (Read 921 times)  Share 

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taiga

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Spring and hand question
« on: June 10, 2010, 05:42:45 pm »
Well I thought about it some more. For the spring question given the gravitational potential energy is dissipated through the hand due to it being gently lowered down, what would be the scenario if it was not brought down by the hand, and just let go?

Where does the energy go? If it were to be dissipated into the environment, what is the relevance of the hand in the question whatsover?

Thinking about it, if the hand changes the way the energy is transferred, could this be indicating that the original spring constant really was 100N/m, but due to the hand being there, the characteristics of the spring were changed, hence making it 50N/m in the end?

I'm sure the concept would be beyond the scope of the course, but I can't think of any other way to look at it.

The energy change in the system however would remain at 4J though.


I have attached a full copy of the exam, it's in PDF so hopefully better than the one currently up :P
« Last Edit: June 10, 2010, 06:21:56 pm by taigastyle »
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mark_alec

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Re: Spring and hand question
« Reply #1 on: June 10, 2010, 05:45:49 pm »
Can you post the question so that those who didn't sit the exam could chime in on your question?

taiga

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Re: Spring and hand question
« Reply #2 on: June 10, 2010, 06:11:38 pm »
Can you post the question so that those who didn't sit the exam could chime in on your question?

Attached a PDF :)
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a.zirek

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Re: Spring and hand question
« Reply #3 on: June 10, 2010, 07:21:00 pm »
Look, if the hand were to be SUDDENLY pulled away, the spring would extend MORE than 0.4m and then slowly oscillate until it reaches its equilibrium point.
It's like bungee jumping: the first time you jump, you go down real low, the next time not as low, until you reach an equilibrium position.
In this question, the spring is already lowered into equilibrium position.
If people were to use mgh=o.2kx^2, then they had to know the first time the spring came to rest momentarily.

Now, given that the spring constant is 50Nm, we can do the following calculations.
mgh=0.5kx^2
2 x 10 x h = 0.5 x 50 x h^2
20h = 25h^2
25h^2 - 20h = 0
5h^2 - 4h = 0
h(5h-4) = 0
h = 0.8

Therefore, if the spring was SUDDENLY released, it would fall down to a height of 0.8m and oscillate until it reaches equilibrium at 0.4m
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mark_alec

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Re: Spring and hand question
« Reply #4 on: June 11, 2010, 12:12:39 am »
If the mass were dropped, it would indeed oscillate. If the spring were ideal and the system lossless, it would continue to oscillate forever, but if there was any energy loss, it would eventually settle down to the equilibrium position.