Energy stored while loading a spring

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Homework Help Overview

The discussion revolves around a problem related to plotting a Force vs. extension graph and estimating the energy stored in a spring during the loading process. Participants are exploring the interpretation of the area under the graph in relation to energy calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the markscheme's phrasing regarding the area under the graph, seeking clarification on its meaning. Participants discuss how to calculate the area under the graph at specific points, particularly when the extension is at half the maximum value.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the calculation of the area under the graph. There is a recognition of a mistake in one participant's calculation, leading to further clarification and discussion.

Contextual Notes

There is an emphasis on understanding the specific area under the graph at given points rather than the total area, which may indicate a focus on incremental energy storage during the loading process.

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I am doing a question which asks to plot a graph of Force vs. extension; it then asks how can the graph be used to estimate the energy stored during the loading process. I thought that this would simple be the area under the graph, however the markscheme says "the area under the graph at any given point." Can anyone explain the reason for this please?

Thanks
 
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This means the area under the graph up to the given point. For instance if you have a graph that looks like a triangle, one corner at the origin, sloping up to some maximum x and maximum y; if you want to know the energy stored when the extension (or compression) is at x/2 (half the maximum) you only want to know the area up till that point, i.e. 1/8*x*y
 
thanks, where did you get 1/8 xy from? wouldn't it be 1/4 xy?

Thanks
 
np,
1/2*base*height right? base = x/2 ; height = y/2 --> 1/8xy
 
Yes, my mistake. Thank you.
 

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