Work - Kinetic energy theorem Need more fundamental understanding

[1MileCrash]

I'm trying to find a good way to think about work.

From what I can tell, it's what the object experiences, not really what the force does, correct?

If a train is heading west and superman grabs it, stops it, and starts pushing it in the opposite direction until it's at the same speed as it was before, he's done zero work right?

And this is because before it stops, we have W = FDcos(180) = -FD, while after it stops we had W = FDcos(0) = FD? For a net work of FD - FD = 0.

Which coincides with W = K(final) - K(initial)

This makes perfect sense to me (assuming I got that right) but it just feels much weirder when we're talking about changing it's direction in a different way (not just the reverse.)

If the train were going along +x and superman was drunk and decided to push it until it were going along +y at the same speed (and no longer x displacement at all), he's still done 0 work, but how do I show it as above? It's not as simple as two things canceling out.

My thinking is that this would be more complex to "write out" because the angle relating the force to displacement would constantly change while this diagonal force were applied, not just once as above. If that is the case... would there need to be some function for a relationship between that angle and the duration of the force's application to integrate, pretending we didn't know that the work would come out to zero?

I realize all of this is unnecessary to actually solve problems, but I'm not one of those "just take it as axiom" kind of people.

 

[Opus_723]

In the first case we slowed the train down to a stop. And then we applied force to accelerate it to its original speed, but in a new direction. It had an initial +x velocity, and then a final -x velocity. So there is a Δv with a magnitude of -2x. But to visualize it, we can break that Δv into two vectors of magnitude -1x. The first slows the train to a stop. The second accelerates it to its original speed in the opposite direction.

For a point particle that goes from a +x velocity to a +y velocity, you can break up the Δv joining them into one with components -x and +y. Think of it as slowing the particle down to a stop, and then applying force to speed it up to its original speed, but in the +y direction. This is hard to visualize (and not particularly realistic) for a train, since you would picture having to turn the whole thing around.

 

[Alpharup]

First u shud understand why the work done be superman in the first case is zero. When the superman tries to reverse the direction keeping the velocity(v) same, he does zero work. this is because of thermodynamical laws. The train can't change its velocity from v to -v directly. It has to travel in all the velocities between v and -v when superman applies force F. When superman tries to stop the train, work is done by the train on his force. This work increases the internal energy of the superman. Now avoid dissipative forces like friction. The train's velocity is zero and now superman uses his newly gained energy and does work on the train. This work must be equal to the initial kinetic energy of the train if it has to travel with velocity -v. thus the total work done is 0.

Now in the second case work done by superman is not zero. Please avoid the work-kinetic energy theorom by only considering initial and final velocities here since it is misleading. When he tries to change direction from +X to +Y, he uses centripetal force. he changes the angle by ∏/2. its not necessary that he shud change the change the velocity to 0 as in case 1 inorder to bring the change in direction. let us assume that he changes the velocity's direction by keeping the velocity constant. let this velocity(magnitude) be original velocity v. When the train turns it takes a curved path. let the mass of train be m. let the imaginary radius of curved track be r. so the centripetal force is mv^2/r. the torque is mv^2/*r=mv^2. so the work done by superman = torque *change in angle=mv^2*∏/2.initial kinetic energy of the train was 1/2*mv^2.so total work done by superman is =mv^2*∏/2-1/2*mv^2. now let us consider that superman changes the direction of train by bringing its velocity to zero. when he brings the velocity to 0, his internal energy increases as 1/2mv^2. he applies centripetal force and the energy he uses is mv^2*∏/2. he also has to increase the velocity to v. so he also loses the kinetic energy(1/2mv^2) gained from the train to increase its velocity to v. so net work done in second part of this case is mv^2*∏/2. So he should work to change its direction keeping its speed same.

Hope this helps.

 

[Opus_723]

I believe no work is done by centripetal forces, since they act perpendicular to the motion of the object. A simple question to test your idea is this: If work is done on the train, but it's speed and therefore kinetic energy does not change, then where did the energy you supplied go? Clearly no net work is done. Also, centripetal forces don't produce torque, since they act radially through the axis of rotation, and thus have no lever arm. There is no torque on an object in uniform circular motion because the object has no angular acceleration.

 

[PJC01]

I completely understand your desire for a deeper understanding of the concept of work. The work-energy theorem is an important principle in physics that relates the work done on an object to its change in kinetic energy. You are correct in stating that work is the energy an object experiences, not just what a force does.

In the example of Superman pushing a train, the work done on the train is indeed zero. This is because, as you pointed out, the force and displacement are perpendicular, resulting in a cosine of 90 degrees and a work of zero. The train's kinetic energy remains the same, so the change in kinetic energy is also zero. This is represented in the equation W = K(final) - K(initial).

When considering a change in direction, as in the example of the train being pushed from +x to +y, the work done may not be as straightforward. In this case, the angle between the force and the displacement is constantly changing, so we cannot simply use the cosine function to find the work. Instead, we would need to integrate the work done over the entire displacement. This would involve determining the force at each point along the displacement and the angle between the force and displacement, and then using calculus to find the total work done.

However, as you mentioned, this level of complexity is not always necessary to solve problems involving work. In many cases, we can use simpler methods, such as considering the work done by individual forces and then adding them together. But for a more fundamental understanding, it is important to recognize the relationship between work and kinetic energy, and how it applies in various situations.