EK Physics #440

[ilovemcat]

---

Members do not see this ad.

I have a few questions for the ideal machine described in questions #437 to #441.

I know that when the mass 2 (object) is less than the mass 2 (engineer), the engineer would accerate at some fraction of gravity (due to the upward force of the mass).
I also know that when the mass 2 (object) is exactly twice the mass of the engineer, the acceration is zero and the engineer falls at constant velocity. (Please correct if I'm wrong)

Now, what would happen when mass 2 is greater than twice the mass 1 (engineer)? This would mean the engineer would experience a net upward acceration, but given the fact that his platform is attached to the machine, his acceration would then be zero (because he can't move). Does that sound right?

The reason I bring these scenarios up is because I'm trying to understand the concept of mechanical advantage. The problem explains that this machine has a mechanical advantage of 2. What would happen when someone exceeds the maximum mechanical advantage of a system. For instance, what would happen when a weight greater than twice the weight of the engineer were placed on platform 2. Would this prevent the machine from working?



A entirely different scenario I wanted to consider is the work and mechanical energy of this machine. Questions #438, #440, and #441 all provide examples when the mass placed on object 2 is less than the mass of the engineer. For one of the objects, we are asked to consider the amount of work done on it by the input force engineer. This is where I'm confused. When is the amount of work actually equal? I'm so use to seeing situations where the amount of work is conserved for an ideal system, but based on the solutions to some of these questions, it seems that isn't the case.

Take for instance question #440,

Why couldn't the work on object 2 (smaller mass) simply be the change in potential energy of the engineer? The way the book solved this was by using conservation of mechanical energy to come up with this equation:

KE(b)f + PE(b)f = PE(e)i - KE(e)f

I thought as an object is accelerating downward from some initial position, it is losing Gravitational PE. That energy is converting into Kinetic Energy, which results in the mass of the engineer hitting the floor at some final velocity. How could you explain in the equation above. I thought for the engineer, PE(e)i minus KE(e)f would equal zero?

This is really really confusing me. I've spent so much time trying to understand this and still can't seem to get it right. Please help.

 

[ilovemcat]

By the way, the question(s) I'm asking aren't really focused to one question in particular but the whole range of questions provided for that figure in the book. It's sort of impossible to explain it without having the EK book though 🙁

 

[shffl]

Take for instance question #440,

Why couldn't the work on object 2 (smaller mass) simply be the change in potential energy of the engineer? The way the book solved this was by using conservation of mechanical energy to come up with this equation:

KE(b)f + PE(b)f = PE(e)i - KE(e)f

I thought as an object is accelerating downward from some initial position, it is losing Gravitational PE. That energy is converting into Kinetic Energy, which results in the mass of the engineer hitting the floor at some final velocity. How could you explain in the equation above. I thought for the engineer, PE(e)i minus KE(e)f would equal zero?

Correct me if I'm wrong but this is what I arrived at:

As the engineer descends from top of the lift to bottom, his PE is converted to KE but not ALL of the PE is converted into KE. Some of the PE is used to do work to raise the block of mass. This is why the equation is

KE(b)f + PE(b)f = PE(e)i - KE(e)f

If you want to solve for PE of engineer:

KE(b)f + PE(b)f + KE(e)f = PE(e)i

which makes sense because all of the PE of the engineer has been converted into work to raise block and KE of engineer, so the sum of all that will equal PE of the engineer.

If the engineer was to descend on the lift without any block of mass on the other side, then the final KE in that situation would be a lot more different than the ones with a block of mass. The reason being the engineer would probably descend slower if there is a block of mass since v is lower then KE is lower. Where did the differences in KE go? The amount is probably identical to work used in raising the block of mass.

Hope this helps 👍