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Work done by a spring

  • Thread starter aleferesco
  • Start date
  • #1
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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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
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Compare W1 = 1/2k(x1)^2 with W2 = 1/2k(x2)^2 = 1/2k(2x1)^2
 
  • #3
28
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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 cancelled out)

W2 = 4

Can you explain it a bit, thanks
 
  • #4
Doc Al
Mentor
44,882
1,129
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?
 
  • #5
28
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Thanks a lot!, I got the answer
 

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