IACalculate Maximum Compression Distance of a Spring with Vertical Placement

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SUMMARY

The discussion centers on calculating the maximum compression distance of a spring when a 1.70 kg book is dropped from a height of 0.70 m onto it. The user applied the conservation of energy principle, using the equation mg(y+x) = 1/2kx^2, where k is the spring constant. After substituting the values, including a spring constant of 1400 N/m, the user calculated the maximum compression distance to be 0.142 meters. The reasoning was confirmed as correct by another participant, although the spring constant was initially omitted.

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I just need to make sure I understand how to do this and that I am doing everything correctly. Could someone check my work please?


You place the spring vertically with one end on the floor. You then drop a 1.70 kg book onto it from a height of 0.70 m above the top of the spring. Find the maximum distance the spring will be compressed.
So
k_1 + U_1 + W_{other} = k_2 + U_2
I have that

k_1=0, U_1= mgy_1, W_o=0, k_2=0, U_2=1/2kx^2

So the issue is that I have to the potential energy to the start of the spring plus the distance the spring compresses.

To solve this I need to adjust the equations to look like

mg(y+x)=1/2kx^2

This works out to a homogenous quadradic that looks like

1/2kx^2-mgx-mgy=0

Plugining in values and solving gets me

0.142 meters.

TIA

T
 
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You didn't include the spring constant, so I can't check your arithmetic, but your reasoning looks faultless to me. Good job. :)
 
Diane_ said:
You didn't include the spring constant, so I can't check your arithmetic, but your reasoning looks faultless to me. Good job. :)

oops...sorry about that.

It should be 1400N/m...and thanks :smile:
 

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