"'Does the object continue to move after it comes to rest?'"
FYI, the answer is "yes". The slope is frictionless and the spring must be presumed to be ideal, since there is no indication in the problem that energy is lost in compression.
The object (always barring elephantine intervention, of course) will rise to the same height it started from. The first "at rest" refers to the moment when all the energy in the system is in the compressed spring, at which point the object is momentarily at rest.
The problem takes too much for granted, since it neglects to explicitly note that the spring is ideal, and also that there is no air friction. A good answer would note these assumptions, perhaps a bit sniffily.
8-)
(Sorry if this was obvious, Steve; if it was, you can assume that I include this for the other commenters for whom it wasn't. 8-)
I understand. It's just that in all of my years in engineering, I've never seen "come to rest" used that way. Are they saying that a pendulum (like the weight in this problem) is always coming to rest?
The 3,4,5 triangle should be automatic even for the fuzzy math people because they love that sort of thing. And just as funny are the patronizing comments that go along with the pictures.
"idiot"
It's just incredible that anyone might expect "artsy" types to learn algebra.
"Are they saying that a pendulum (like the weight in this problem) is always coming to rest?"
Well, since this is pretty much the moral equivalent of a pendulum....
8-)
I agree that "rest" is used a bit idiosyncratically here, since v -> 0 only instantaneously. But unlike engineering, physics is all about platonic solutions. Friction, boundary conditions, state transitions -- too messy.
With that mind-set, the use of "rest" is more understandable.
(I assume you would be less worried if the question were worded more like, "At the point at which the spring attains its maximum potential energy..."?)
Argggg... I received the exact same email this morning at work. I was going to post it this evening.
ReplyDeleteI like the one that asks:
ReplyDelete"Does the object continue to move after it comes to rest?"
SteveH: "I like the one that asks:
ReplyDelete"'Does the object continue to move after it comes to rest?'"
FYI, the answer is "yes". The slope is frictionless and the spring must be presumed to be ideal, since there is no indication in the problem that energy is lost in compression.
The object (always barring elephantine intervention, of course) will rise to the same height it started from. The first "at rest" refers to the moment when all the energy in the system is in the compressed spring, at which point the object is momentarily at rest.
The problem takes too much for granted, since it neglects to explicitly note that the spring is ideal, and also that there is no air friction. A good answer would note these assumptions, perhaps a bit sniffily.
8-)
(Sorry if this was obvious, Steve; if it was, you can assume that I include this for the other commenters for whom it wasn't. 8-)
I understand. It's just that in all of my years in engineering, I've never seen "come to rest" used that way. Are they saying that a pendulum (like the weight in this problem) is always coming to rest?
ReplyDeleteThe 3,4,5 triangle should be automatic even for the fuzzy math people because they love that sort of thing. And just as funny are the patronizing comments that go along with the pictures.
"idiot"
It's just incredible that anyone might expect "artsy" types to learn algebra.
"Are they saying that a pendulum (like the weight in this problem) is always coming to rest?"
ReplyDeleteWell, since this is pretty much the moral equivalent of a pendulum....
8-)
I agree that "rest" is used a bit idiosyncratically here, since v -> 0 only instantaneously. But unlike engineering, physics is all about platonic solutions. Friction, boundary conditions, state transitions -- too messy.
With that mind-set, the use of "rest" is more understandable.
(I assume you would be less worried if the question were worded more like, "At the point at which the spring attains its maximum potential energy..."?)