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 Mar 12th 2011, 07:33 PM #1 Junior Member   Join Date: Mar 2011 Posts: 25 force constant A spring is 17.0 long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to 25.0 , causing the spring to stretch to a length of 19.2 . a) I was able to find the force constant of the spring by using the k = F / x formula. 25/ .022 = 1136 N/m b) How much work was required to stretch the spring from 17.0 to 19.2 ? for this i used the kx^2 / 2, which gave me (1136)(.022)^2 / 2 = .275 J c) this is where i am having issues. How long will the spring be if the 25.0 force is replaced by a 50.0 force? I plugged the 50 into the F / k = x formula and got 50 / 1136 = .044, yet it is the wrong answer. did i misinterpret something? Thanks!
Mar 12th 2011, 09:55 PM   #2

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 Originally Posted by pre pt marc A spring is 17.0 long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to 25.0 , causing the spring to stretch to a length of 19.2 . a) I was able to find the force constant of the spring by using the k = F / x formula. 25/ .022 = 1136 N/m b) How much work was required to stretch the spring from 17.0 to 19.2 ? for this i used the kx^2 / 2, which gave me (1136)(.022)^2 / 2 = .275 J c) this is where i am having issues. How long will the spring be if the 25.0 force is replaced by a 50.0 force? I plugged the 50 into the F / k = x formula and got 50 / 1136 = .044, yet it is the wrong answer. did i misinterpret something? Thanks!
Just as a good habit to form, use the units when you report a number.

That 0.044 m is the distance the spring has been stretched from equilibrium. So you need to add 17 cm to your answer.

-Dan
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