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avatar+1832 

 

 

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

Best Answer 

 #7
avatar+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
avatar+32822 
+10

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
avatar+1832 
0

So his answer is incorrect too right ? 

 Nov 28, 2015
 #3
avatar+1832 
0

And please explain this question for me 

 

 Nov 28, 2015
 #4
avatar+32822 
+15

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
avatar+115747 
0

Thanks Alan :)

 Nov 29, 2015
 #6
avatar+1832 
0

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
avatar+32822 
+15
Best Answer

"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
avatar+1832 
+5

Now I get it . 

Thank you Alan 

 Dec 3, 2015

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