Conservation of Mechanical Energy Requirement

AI Thread Summary
Conservation of mechanical energy occurs when the net work done by non-conservative forces is zero, while momentum conservation requires an external net force close to zero. Non-conservative forces can still be present, but their work must sum to zero for mechanical energy to be conserved. The discussion highlights that conservative forces are defined by a scalar potential function, and their curl must be zero. It is argued that time-dependent forces cannot be conservative due to their path-dependent nature, although they can be considered conservative at a fixed time if their spatial curl is zero. Understanding these principles is crucial for analyzing mechanical systems.
fog37
Messages
1,566
Reaction score
108
Hello Forum,
Conservation of mechanical energy ME= KE+PE happens when the net work done by the non conservative forces is zero. Conservation of total momentum, instead, happens when the external net force is zero (or close to zero).

In the case of mechanical energy, the non-conservative forces are also external forces to the system and can be nonzero. If the other external forces are conservative, the ME is conserved but not momentum.
But what we care about is the work they produce. How can nonconservative forces produce zero work? I guess their work may be finite but negligibly small. Maybe the total sum of the work done by the nonconservative force is zero...

Is that correct?

fog37
 
Physics news on Phys.org
First you should clarify the definitions. A conservative force is given by a time-independnet scalar potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Then you take Newton's 2nd Law, multiply it by ##\dot{\vec{x}}## and integrate over time to get the energy-conservation law
$$\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x})=\text{const}.$$
Then there are forces that do no work. An example is the motion in an external magnetic field. The Lorentz force is always perpendicular to ##\dot{\vec{x}}## and thus it doesn't contribute to the work.
$$\vec{F} \cdot \dot{\vec{x}}=q (\dot{\vec{x}} \times \vec{B}) \cdot \dot{\vec{x}}=0.$$
 
Sure, so there are two types of forces: conservative and non-conservative. Both types of force can perform work. What matters for the mechanical energy to be conserved is that the net work done by the nonconservative forces is zero. How can that happen? If the nonconservative forces act perpendicularly to the object's path or if the algebraic signs of the works produce a net zero work...

As you mention, conservative forces admit a potential function V(x), which is a function of position. Can a time-dependent force be conservative? I don't think so. But why? What is the rigorous argument to explain that?

Thanks!
fog37
 
Conservative force means that we can define a scalar potential function for that force. To be able to define a scalar function for a vector field the curl (wrt space coordinates) of the vector field should be 0 because curl of a gradient should be always 0. Since force can be written as gradient when we take curl of it, we should obtain 0.
 
Sure, wherever the curl F = 0 we can express F= Delta V, i.e. we can use a scalar potential V.

but what if the force F was F(t), i.e. a function of time?
 
I think you cannot talk about curl of a time dependent function (without fixing time) because now your force is four vector but curl is defined on 3D. Also notice that if you define a potential when you follow different paths in space between two points, you may not reach same potential due to time dependence which makes your force path dependent . So I don't think a time dependent force is conservative. However, in a fixed time your force can be considered as a conservative force if the curl of it wrt space coordinates 0.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top