Maximum height a mass can be thrown off a machine

[animasian]

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In the EK 1001 questions in MCAT Physics, #441:

An ideal machine has a mechanical advantage of 2; you step on Platform 1 and you can lift a maximum weight of 2 times your own weight on Platform 2. The max height Platform 2 can be raised is half the height that Platform 1 can be lowered.

If a very small mass were placed on platform 2 while on the floor, and a very large mass were placed on platform 1 while raised, to what maximum height above the floor could the small mass be thrown when platform 2 stops abruptly?

The answer key says that "No matter how great the mass on platform 1, it is accelerated by gravity and could not change velocity faster than 10 m/s^2. Since platform 2 must move half the distance of platform 1 in the same time, it's max acceleration is half that or 5 m/s^2." Using this information we can calculate the initial velocity and maximum height of the mass when it is thrown off the platform.

What I don't understand is why max acceleration is 0.5*g? I thought an ideal machine performs constant work so F1d1 = F2d2 --> since d2 is half of d1, then F2 is twice F1 and since F = ma then acceleration of platform 2 should be greater. A little confused here...

 

[Blueprint MCAT Tutor]

Interesting question! When they say "very small mass" and "very large mass" I assume that's their way of telling us to ignore the torques involved? That is, when you drop the large mass, it doesn't fall at gravity - the see-saw would resist it. But I think they want us to assume some sort of perfect ideal situation.

So in that case, my intuition would agree with you, to do this as a conservation of energy problem. In general when you're stumped by a physics problem, I'm always a fan of going back to conservation of energy and dimensional analysis.

The one problem we run into, though, is that to solve this as a conversation of energy problem, we would want to say m1 x g x h1 = m2 x g x h2

That is, all of the gravitation potential energy in the first mass is transferred into the second mass but we can't really do that here since we don't know m1 or m2 and we don't know the proportions between them. We're only told that one is much bigger than the other.

The other problem with approaching this as a conservation of energy problem is that it doesn't work intuitively - if we picture this scenario, the big mass would fall towards the ground and slam into it. Only a part of its energy would get transferred over to the little mass. That is, while it's falling pushing on the lever, some of its energy is getting transferred over but it would still crash into the ground going pretty fast - it would still have KE when it hits the ground.

So we think it through and neither the algebra nor the intuitions align with doing this is a conservation of joules problem.

We're forced to fall back to the kinematic explanation that EK gives. With equal t and half the distance, then the average velocity for the smaller mass must be half as much and half the acceleration.

It feels weird and counter-intuitive but I can't (in my early morning-pre-coffee-rambles) think of another explanation.