What is the Relationship Between Spring Stretching Distance and Applied Work?

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The discussion focuses on understanding the relationship between spring stretching distance and the work applied in a physics problem. Key equations include Fs = kx for force and Es = 1/2 kx^2 for elastic potential energy. The problem involves analyzing how a weight affects the stretch of a spring, with the hint suggesting dividing the spring into segments to determine individual stretches. By applying the concept of proportionality, it is concluded that the weight of a fraction of the rod will stretch the spring accordingly. The final calculations confirm the total stretch based on the distribution of weight across the spring segments.
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Homework Statement


See image attachment

Homework Equations


Fs = kx
Es = 1/2 kx^2
ET1 = ET2 which is Eg1 + Ek1 + Es1 + work applied = Eg2 + Ek2 + Es2 + friction

The Attempt at a Solution


grade 12. Please help i never did anything like this.
 

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This is a force/equilibrium problem, not an energy problem.

A hint to start you off: If you look at diagram I and imagine the spring cut into three pieces, how much does each piece stretch?
 
each piece will stretch 1 cm
 
firezap said:
each piece will stretch 1 cm
Good. So now you know that the weight of the complete rod will stretch one of the spring pieces by 1 cm. How much will the weight of 1/3 of the rod stretch one of those spring pieces?

Use this reasoning to figure out how much each spring stretches in diagram II.
 
1/3 x 1cm = 1/3cm
is it 1/3cm + 2/3cm + 3/3cm = 2cm
 
firezap said:
1/3 x 1cm = 1/3cm
is it 1/3cm + 2/3cm + 3/3cm = 2cm
Good!
 
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