Conservation of energy hard problem.

In summary, an object of mass m slides a distance d down an incline of angle (theta). While sliding, it contacts an unstressed spring of negligible mass. Find the initial separation d between object and spring.
  • #1
Neon32
68
1

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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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".
 
  • #6
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##).
 
  • #7
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.
 
  • #8
Nope it doesn't matter its just a symbol as you say for the displacement of the spring.
 
  • Like
Likes Neon32
  • #9
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.
 
  • Like
Likes Delta2

FAQ: Conservation of energy hard problem.

What is the conservation of energy hard problem?

The conservation of energy hard problem refers to the philosophical and scientific debate surrounding the concept of energy conservation in the universe. It questions whether energy is truly conserved in all physical processes, or if there are exceptions or loopholes that allow for energy to be created or destroyed.

Why is the conservation of energy hard problem important?

The conservation of energy hard problem is important because it challenges our fundamental understanding of the laws of physics. It also has implications for our understanding of the universe and the possibility of perpetual motion machines and other technologies that violate the principle of energy conservation.

What are some proposed solutions to the conservation of energy hard problem?

Some proposed solutions to the conservation of energy hard problem include the idea of hidden variables in quantum mechanics, the concept of non-locality, and the possibility of energy being transferred to other dimensions or parallel universes.

What evidence supports the conservation of energy?

The principle of energy conservation is supported by numerous experiments and observations in physics, such as the law of conservation of momentum and the first law of thermodynamics. These laws have been consistently observed and tested, providing strong evidence for the conservation of energy.

How does the conservation of energy relate to the concept of entropy?

The conservation of energy is closely related to the concept of entropy, which is a measure of the disorder or randomness in a system. The second law of thermodynamics states that in any closed system, the total entropy will always increase over time. This is consistent with the principle of energy conservation, as energy is constantly being dispersed and becoming less organized, leading to an increase in entropy.

Back
Top