How Much Work to Compress a Spring Further by 7cm?

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Homework Help Overview

The problem involves calculating the additional work required to compress a spring further by 7cm after an initial compression of the same distance, with a focus on the relationship between force, displacement, and work done on springs.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the work done in compressing the spring using the equations for work and spring force. There is an attempt to derive the spring constant and compare the work for different compression distances.

Discussion Status

Some participants express confusion regarding the comparison of work equations and the implications of the results. There is an ongoing exploration of how to correctly interpret the work done for different compression distances, with some guidance provided on the relationships between the variables involved.

Contextual Notes

Participants note discrepancies between their calculations and the expected answer from the textbook, prompting questions about the assumptions made in the problem setup.

aleferesco
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Homework Statement



In order to compress a spring by 7cm from its natural length, 20J of work has to be done. How much additional work should be performed in order to compress the spring by an additional 7cm?


Homework Equations



F= -Kx (spring's equation)

W=1/2(Kx)x = [k(x)^2]/2 (Work equation and spring's equation)


The Attempt at a Solution



20J = [K (7)^2]/2 (Solving for k, but I get the same answer of 20J when substituting k in the work equation)

Although the answer seems to be 20J as you are stretching it an additional 7cm, that answer is wrong. The answer in the back of the book says it is 60J.
 
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Compare W1 = 1/2k(x1)^2 with W2 = 1/2k(x2)^2 = 1/2k(2x1)^2
 
I'm sorry I don't understand the comparison.

1/2k(x2)^2 = 1/2k(2x1)^2

1/2k(7)^2 = 1/2k [2(7)^2]

24.5k = 98k

After this I have no idea what to do... am I supposed to have it like this

W2 = 98k/24.5k (the k gets canceled out)

W2 = 4

Can you explain it a bit, thanks
 
aleferesco said:
I'm sorry I don't understand the comparison.

1/2k(x2)^2 = 1/2k(2x1)^2

This is the work needed to compress the spring by x2 = 2x1 (x1 = 7cm, in the example).

1/2k(7)^2 = 1/2k [2(7)^2]
This is incorrect. They (obviously) are not equal. Instead you want to compare them.

24.5k = 98k
The left side is the work needed (in terms of k) to compress the spring by 7cm. We know it's equal to 20J.

How much bigger is 98k? (Which is the work needed to compress the spring by 2*7=14cm.) By what factor is it bigger? Thus, how much energy does it correspond to? Thus, how much additional energy is needed?
 
Thanks a lot!, I got the answer
 

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