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# The Work Done By The Spring Force .

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Did he calculates the work done by the spring force correctly !

Because he wrote (Sf - Si )^2 , but is must be (Sf^2 - Si^2 ) .

physics
Nov 28, 2015

#7
+32822
+15

"Here how did you get 3^2  ? it must be ( 5^2 - 3^2 ) according to ( Sf^2 - Si^2 ) ! ​ "

Sf and Si are the distances from the unstretched position.  So Sf = 2-5 (i.e. 3)  and Si = 5-5 (i.e. 0)

Dec 1, 2015

#1
+32822
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Yes.

$$U_s=\int_{s_1}^{s_2}(-kx)dx \rightarrow U_s=-\frac{k}{2}(s_2^2-s_1^2)$$

Nov 28, 2015
#2
+1832
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So his answer is incorrect too right ?

Nov 28, 2015
#3
+1832
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And please explain this question for me

Nov 28, 2015
#4
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The work done on the spring to get it from 5" to 2" is k*3^2/2

The work done on the spring to get it from 5" to 4" is k*1^2/2

Hence the work done on the spring to get from 2" to 4" is k*(1^2 - 3^2)/2 or -k*(3^2 - 1^2)/2  (it's the work done by the spring that's positive here of course, so the work done on the spring is negative).

Nov 29, 2015
#5
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Thanks Alan :)

Nov 29, 2015
#6
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The work done on the spring to get it from 5" to 2" is k*3^2/2

Here how did you get 3^2  ? it must be ( 5^2 - 3^2 ) according to ( Sf^2 - Si^2 ) !

Dec 1, 2015
#7
+32822
+15

"Here how did you get 3^2  ? it must be ( 5^2 - 3^2 ) according to ( Sf^2 - Si^2 ) ! ​ "

Sf and Si are the distances from the unstretched position.  So Sf = 2-5 (i.e. 3)  and Si = 5-5 (i.e. 0)

Alan Dec 1, 2015
#8
+1832
+5

Now I get it .

Thank you Alan

Dec 3, 2015