Gaining Insight into Energy Conservation: A New Perspective

[daniel_i_l]

I was thinking about that question (what is energy) and I realized that I could easily understand what energy is and why it is conserved if I thought of it as the total force needed to get a mass to the position that it is in. For example:

Kinetic energy - inorder to give a mass speed you need to accelerate it, the force would be F = ma, and the integral of mad(x) is m*a*x = m * a * 1/2at^2 = m * 1/2a^2t^2 = 1/2mv^2
Potential energy - F = mg (the same as KE) and the integral of mgd(x) is m*g*x (mgh) This is true for other kinds of energy too. So really the energy of a mass is simply the sum of all the forces needed to get it to the position and speed that it is in (this is why work is the change in energy). And that is why energy is always conserved if you don't add any external forces to it - cause if you don't add any force then the total force (energy)will always stay the same. (this isn't a new theory or anything, just a way of looking at things)
So, is this obvious and I'm stupid for not noticing it before , interesting but not connected to reality, enlightning...

 

[daniel_i_l]

Is this true?

 

[KingNothing]

Yes, your logic is fine. No, it's not stupid or obvious. Everybody has a point in time where they figure out a way to think of it that makes complete sense to them. Congratulations:).

 

[michealsmith]

here i would like to suggest a site tell me what u think http://home.pacifier.com/~ppenn/whatis2.html

 

[El Hombre Invisible]

I was thinking about that question (what is energy) and I realized that I could easily understand what energy is and why it is conserved if I thought of it as the total force needed to get a mass to the position that it is in. For example:
Kinetic energy - inorder to give a mass speed you need to accelerate it, the force would be F = ma, and the integral of mad(x) is m*a*x = m * a * 1/2at^2 = m * 1/2a^2t^2 = 1/2mv^2

This will give you the change in kinetic energy, but not the total kinetic energy, since the initial velocity of the body is not considered here. m*a*x = F*x which is work - i.e. the work done by an unbalanced force accelerating a body over a distance of x is equal to the change in the body's kinetic energy. Likewise in your potential example, the work done in moving a body from one equipotential to another is equal to the change in potential energy. So just be aware that you're dealing with changes in energy, not total mechanical energy.

 

[daniel_i_l]

Thanks guys!
Yes, I was aware that I was only dealing with the changes - the total energy depends in were you start from, or were your referance spot is.