- #1
urtalkinstupid
- 261
- 0
Dictionary.com said:Work is defined as: The transfer of energy from one physical system to another, especially the transfer of energy to a body by the application of a force that moves the body in the direction of the force. It is calculated as the product of the force and the distance through which the body moves and is expressed in joules, ergs, and foot-pounds.
[tex]\textcolor{red}{W=\vec{F}dcos \theta}[/tex]
If the object is not moved by the force applied, I'm sure we can all agree that [itex]\textcolor{red}{W=0~Joules}[/tex], ergo no work was done. Does this also imply that there was no energy output?
Scenario 1:
A man applies [itex]\textcolor{red}{4N}[/itex] of force on an object. In the end, he moves the object a total of [itex]\textcolor{red}{2m}[/itex] horizontally. So from this we can say: [itex]\textcolor{red}{W=(4N)(2m)cos(0)}[/itex]. This gives us: [itex]\textcolor{red}{W=8~Joules}[/itex]. Agree? There was work done, and there was also energy output by the man in a total amount of [itex]\textcolor{red}{8~Joules}[/itex].
Scenario 2:
A man applies the same amount of force on another object. This object is much more massive than the first object. He eventually gets tired of trying to move the object, and can not go on. So, he accomplished to move the object [itex]\textcolor{red}{0m}[/itex] across a horizontal surface. So, from this we can say: [itex]\textcolor{red}{W=(4N)(0m)cos(0)}[/itex]. This gives us: [itex]\textcolor{red}{W=0~Joules}[/itex]. There was no work done. Does that mean there is no energy output since the amount of work done was [itex]\textcolor{red}{0~Joules}[/itex]? Obviously, this man used energy to TRY and push the object. He was tired, because he output all his energy trying to move the boulder, but the equation says otherwise.
Scenario3:
[tex]\textcolor{red}{\vec{F}=G\frac{M_1M_2}{r^2}}[/tex]
The Earth pulls on the moon with a certain amount of force. The moon pulls back with that same amount of force. This allows the moon to orbit the earth. Does the force that keeps the moon in orbit apply work?
[tex]\textcolor{red}{G=6.67300x10^{-11}N-m^2/kg^2}[/tex]
[tex]\textcolor{red}{M_1(Earth)=6x10^{24}kg}[/tex]
[tex]\textcolor{red}{M_2(Moon)=7x10^{22}kg}[/tex]
[tex]\textcolor{red}{r_{distance}=3.844x10^8m}[/tex]
After calculating all that out, [itex]\textcolor{red}{\vec{F} \approx 18.9x10^{19}N}[/itex] That's a lot of force in the earth-moon system. It would seem logical to say that this force outputs energy inorder to keep the moon in orbit, right? If that were the case, where is this energy coming from? To get around this work is introduced into the situation. The moons orbit is not perfectly circular, but its eccentricity is 0.0549. That is neglible. So, if we plug and chug in the work equation we get [itex]\textcolor{red}{W=0~Joules}[/itex]. We know that is not true though, because energy is used to keep the moon in robit. The reason why most scientists believe that no work is done is because if energy were output, there would be no source of this energy to drain, ergo violating the laws of conservation of energy.
There is obviously something wrong with this. No, this is not an attempt to try and go against physics. I just want answers.
I want just a sipmle explanation. Not scrutiny from everyone.