1. Mar 11, 2008 #1

   msattar07

   

   1. The problem statement, all variables and given/known data
   How does the work required to stretch a spring 2cm compare with the work required to stretch a spring 1 cm?
   
   
   2. Relevant equations
   ?
   
   
   3. The attempt at a solution
   
   i think its twice the work

    

   msattar07, Mar 11, 2008
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3. Mar 11, 2008 #2

   Hootenanny

   

   Staff Emeritus

   Science Advisor

   Gold Member

   Assuming that the two springs are the same and Hookean, you are wrong.

    

   Hootenanny, Mar 11, 2008
4. Mar 11, 2008 #3

   mysqlpress

   

   P.E. of a spring is 1/2kx^2 , which is directly proportional to the square of extension from the natural length

    

   mysqlpress, Mar 11, 2008
5. Mar 11, 2008 #4

   Oerg

   

   you can derive the expression for the potential enegy of the spring by integrating the hooke law over the distance moved.

    

   Oerg, Mar 11, 2008

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