Conservation of energy hard problem.

AI Thread Summary
The discussion revolves around solving a physics problem involving an object sliding down an incline and compressing a spring. The initial and final energy equations are set up correctly, leading to the relationship between the height difference and spring compression. Participants clarify that the choice of potential energy reference point does not affect the final result for the initial separation between the object and the spring. A minor notation difference regarding the spring displacement symbol does not impact the solution. The thread concludes with a consensus on the correctness of the approach and the final answer.
Neon32
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Homework Statement


An object of mass m starts from rest and slides a distance d down a frictionless incline of angle (theata). While sliding, it contacts an unstressed spring of negligible mass as shown in the Figure below. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring. (Use theta for (theta), g for acceleration due to gravity, and m, k and x as necessary.)

http://www.webassign.net/pse/p8-10.gif
p8-10.gif


Homework Equations


Initial energy=finnl energy
K.Ei+P.Ei=K.Ef+P.Ef

The Attempt at a Solution


Here is how I tried to solve it:

Initial energy=0+mgh1
Final energy=0+mgh2+1/2kx²

intial energy=Final energy
mgh1=mgh2+1/2kx²
mgh1-mgh2=1/2kx²
mg(h1-h2)=1/2kx² (1)
since h1-h2=(d+x)sin(theta)
By substituation in equation (1):

mg(d+x)sin(theta)=1/2kx²
then we can solve for d

I found a bit different answer in the answers sheet.
 
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Neon32 said:

Homework Statement


An object of mass m starts from rest and slides a distance d down a frictionless incline of angle (theata). While sliding, it contacts an unstressed spring of negligible mass as shown in the Figure below. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring. (Use theta for (theta), g for acceleration due to gravity, and m, k and x as necessary.)

http://www.webassign.net/pse/p8-10.gif

Homework Equations


Initial energy=finnl energy[/B]
K.Ei+P.Ei=K.Ef+P.Ef

The Attempt at a Solution


Here is how I tried to solve it:

Initial energy=0+mgh1
Final energy=0+mgh2+1/2kx²

intial energy=Final energy
mgh1=mgh2+1/2kx²
mgh1-mgh2=1/2kx²
mg(h1-h2)=1/2kx² (1)
since h1-h2=(d+x)sin(theta)
By substituation in equation (1):

mg(d+x)sin(theta)=1/2kx²
then we can solve for d
[/B]
I found a bit different answer in the answers sheet.


Your method looks good. The answer should be correct.
 
I agree. Perhaps post what the answer sheet says. The usual mistake is to forget the PE due to "x" but you got that right.
 
CWatters said:
I agree. Perhaps post what the answer sheet says. The usual mistake is to forget the PE due to "x" but you got that right.
Here is the answer in answer sheets. He made it in less steps than mine and didn't mention h1 and h2.
http://imgur.com/a/SdrlK
 
Neon32 said:
Here is the answer in answer sheets. He made it in less steps than mine and didn't mention h1 and h2.
http://imgur.com/a/SdrlK
That is the same result you derived; he just went ahead and actually solved for "d".
 
Just a note, in the answer sheet it shows clearly that your teacher choses another level of zero potential energy (you consider 0 potential energy at the base of the inclined, while your teacher puts the zero potential energy at height h2). But the answer should be independent of where we choose the zero potential energy to be and indeed both yours and your teacher method lead to the same result for d. (another note, your teacher uses ##\Delta x## instead of ##x##).
 
Delta² said:
Just a note, in the answer sheet it shows clearly that your teacher choses another level of zero potential energy (you consider 0 potential energy at the base of the inclined, while your teacher puts the zero potential energy at height h2). But the answer should be independent of where we choose the zero potential energy to be and indeed both yours and your teacher method lead to the same result for d. (another note, your teacher uses ##\Delta x## instead of ##x##).
I understood the first part about choosing zero potential energy but I don't get the second part. Does it matter if he say ##\Delta X or just X? In this problem it's just a symbol. As far as I can see it didn't affect the problem.
 
Nope it doesn't matter its just a symbol as you say for the displacement of the spring.
 
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Delta² said:
Nope it doesn't matter its just a symbol as you say for the displacement of the spring.

Thanks. Appreciated :).

This can be locked.
 
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