- #1
Ethan0718
- 39
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I'm not sure if I'm right ..
Here's my thought:
1. The First Law of Thermodynamics is the only single law which could derive energy formula. Furthermore, we can't, solely, use Newton's law to derive Kinetic Energy. It will make us fail to explain the existence of thermal energy.(*)
2. In order to derive energy formula, we must define or determine our system. Without system, there's no meaning of energy.
3. Now, I would like to derive three kinds of common energy, especially, the gravitational potential energy:
(a)Kinetic energy of a mass point
(b)Elastic energy of an ideal spring (mass→0)
(c)Finally, Gravitational potential energy of the gravitational field
4. As you will see, we can derive any energy term of any kinds of system by the first law of thermodynamics.
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Am I correct?
Thank you very much... This problem confuses me so much...
Here's my thought:
1. The First Law of Thermodynamics is the only single law which could derive energy formula. Furthermore, we can't, solely, use Newton's law to derive Kinetic Energy. It will make us fail to explain the existence of thermal energy.(*)
2. In order to derive energy formula, we must define or determine our system. Without system, there's no meaning of energy.
3. Now, I would like to derive three kinds of common energy, especially, the gravitational potential energy:
(a)Kinetic energy of a mass point
(b)Elastic energy of an ideal spring (mass→0)
(c)Finally, Gravitational potential energy of the gravitational field
4. As you will see, we can derive any energy term of any kinds of system by the first law of thermodynamics.
Kinetic Energy of a mass point
∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System : a mass point
Assumption:
(a)Q=0
(b)observer is in the inertial frame so that NSL is valid.
Process: it's acted by so many forces and .. moves!
∴ΔE = ƩFi,xΔxi + Q
∴ΔE = ƩFi,xΔxi = maxΔx
∴ΔE = Δ([itex]\frac{1}{2}mv^{2}[/itex])
∴E[itex]\;\equiv\;\frac{1}{2}mv^{2}+C[/itex]
For convenience, we let C equal zero so that E = [itex]\frac{1}{2}mv^{2}[/itex]
∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System : a mass point
Assumption:
(a)Q=0
(b)observer is in the inertial frame so that NSL is valid.
Process: it's acted by so many forces and .. moves!
∴ΔE = ƩFi,xΔxi + Q
∴ΔE = ƩFi,xΔxi = maxΔx
∴ΔE = Δ([itex]\frac{1}{2}mv^{2}[/itex])
∴E[itex]\;\equiv\;\frac{1}{2}mv^{2}+C[/itex]
For convenience, we let C equal zero so that E = [itex]\frac{1}{2}mv^{2}[/itex]
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Elastic Energy of an ideal spring
∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System: an ideal spring (m→0)
Assumption:
(a)Q=0
(b)the left end is connected with wall so that this end is fixed and cannot be moved.
(c)the origin of coordinate is at right end when length of the spring is original length.
Process: pull the spring, and it's stretched from (x1+L0) to (x2+ L0)
∴ΔE = Wleft end + Wright end + Q
∴ΔE = 0 + ∫(+kx)dx + 0
(integration interval: x1→x2. Drawing a FBD of the spring can help you understand why it is positive sign before kx)
∴ΔE = [itex]\frac{1}{2}kx^{2}_{2}-\frac{1}{2}kx^{2}_{1}[/itex]
∴E[itex]\;\equiv\;\frac{1}{2}kx^{2}+C[/itex]
For convenience, we let C equal zero so that E = [itex]\frac{1}{2}kx^{2}[/itex]
∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System: an ideal spring (m→0)
Assumption:
(a)Q=0
(b)the left end is connected with wall so that this end is fixed and cannot be moved.
(c)the origin of coordinate is at right end when length of the spring is original length.
Process: pull the spring, and it's stretched from (x1+L0) to (x2+ L0)
∴ΔE = Wleft end + Wright end + Q
∴ΔE = 0 + ∫(+kx)dx + 0
(integration interval: x1→x2. Drawing a FBD of the spring can help you understand why it is positive sign before kx)
∴ΔE = [itex]\frac{1}{2}kx^{2}_{2}-\frac{1}{2}kx^{2}_{1}[/itex]
∴E[itex]\;\equiv\;\frac{1}{2}kx^{2}+C[/itex]
For convenience, we let C equal zero so that E = [itex]\frac{1}{2}kx^{2}[/itex]
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Gravitational Potential Energy of the gravitational field
∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System: the gravitational field
Assumption:
(a)Q=0
(b)the positive y-axis direction is upward.
Process:
A ball is falling down from yinitial to yfinal
∴ΔE = Wball + Wearth
(the ball and Earth are called surroundings)
∴ΔE = (+mg)(Δyball) + (-mg)(Δyearth)
(+mg): the force between ball and the gravitational field.
It's acted by ball, and, according to Newton's Third Law, acted on gravitational field.
It's upward because the ball is pulled downward by gravitational field.
(-mg):the force between Earth and the gravitational field.
It's acted by earth, and, according to Newton's Third Law, acted on gravitational field.
It's downward because Earth is pulled upward by gravitational field
P.S. You can draw the Free Body Diagram of Gravitational Field to understand what I'm talking about.. I know it's a little bit weird but I think it's plausible.
∴ΔE = (+mg)(yf - yi) + (-mg)(0)
∴ΔE = mgyf - mgyi
∴E[itex]\;\equiv\;mgy+C[/itex]
For convenience, we let C equal zero so that E = mgy
(*) See "work and heat transfer in the presence of sliding friction" by Bruce Arne Sherwood.∵ ΔE = W + Q
(W=net work done by external forces or surroundings ; Q=heat from surroundings)
System: the gravitational field
Assumption:
(a)Q=0
(b)the positive y-axis direction is upward.
Process:
A ball is falling down from yinitial to yfinal
∴ΔE = Wball + Wearth
(the ball and Earth are called surroundings)
∴ΔE = (+mg)(Δyball) + (-mg)(Δyearth)
(+mg): the force between ball and the gravitational field.
It's acted by ball, and, according to Newton's Third Law, acted on gravitational field.
It's upward because the ball is pulled downward by gravitational field.
(-mg):the force between Earth and the gravitational field.
It's acted by earth, and, according to Newton's Third Law, acted on gravitational field.
It's downward because Earth is pulled upward by gravitational field
P.S. You can draw the Free Body Diagram of Gravitational Field to understand what I'm talking about.. I know it's a little bit weird but I think it's plausible.
∴ΔE = (+mg)(yf - yi) + (-mg)(0)
∴ΔE = mgyf - mgyi
∴E[itex]\;\equiv\;mgy+C[/itex]
For convenience, we let C equal zero so that E = mgy
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Am I correct?
Thank you very much... This problem confuses me so much...
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