Energy conservation question

  • #1
Pseudo Statistic
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6
In the attached picture below I am compressing a spring...
In picture 2, I've compressed the spring a distance x and thus the elastic potential is 1/2 kx^2 and the total energy in the system AT THAT POINT is 1/2 kx^2 + mgh, isn't it? (m is the mass of the ball and h the height from my reference level, the floor) Or am I wrong in thinking this?
Assuming energy conservation, let's say I let go of the spring and it's at that point where it has JUST gone over the edge of the table with its constant horizontal velocity Vx; would the total energy at that point be equal to 1/2 mVx^2 (where m is the mass of the ball) or would I be wrong in assuming this?
Thanks for any help.
 

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Answers and Replies

  • #2
Doc Al
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Pseudo Statistic said:
In the attached picture below I am compressing a spring...
In picture 2, I've compressed the spring a distance x and thus the elastic potential is 1/2 kx^2 and the total energy in the system AT THAT POINT is 1/2 kx^2 + mgh, isn't it? (m is the mass of the ball and h the height from my reference level, the floor) Or am I wrong in thinking this?
Your thinking seems correct. That would be the total mechanical energy of the ball/spring/earth system.

Assuming energy conservation, let's say I let go of the spring and it's at that point where it has JUST gone over the edge of the table with its constant horizontal velocity Vx; would the total energy at that point be equal to 1/2 mVx^2 (where m is the mass of the ball) or would I be wrong in assuming this?
You forgot the gravitational PE (mgh). The total mechanical energy at that point would be its KE + mgh.
 
  • #3
Pseudo Statistic
391
6
Alright so, the mghs in the equality would cancel out leaving us with 1/2mv^2 = 1/2 kx^2, wouldn't it?
Thanks for the reply.
 
  • #4
Doc Al
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That's right. The spring PE is transformed into KE. (Since the height never changes, the gravitational PE remains constant.)
 
  • #5
Pseudo Statistic
391
6
Alright, thanks, but one more thing...
The kinetic energy would be equal to half of the mass times by the constant horizontal velocity the ball would experience when under projectile motion squared, correct?
 

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