| Topic: |
Science > Physics |
| User: |
"Jesper" |
| Date: |
11 Oct 2003 07:35:27 AM |
| Object: |
Can somebody solve this problem? |
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
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| User: "Tom Potter" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 10:32:05 AM |
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"Jesper" <jesper@i-dont-want-spam.oek.dk> wrote in message
news:bm8tcn$jln$1@news.net.uni-c.dk...
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
A spring scale measures force,
whereas a balance scale measures mass.
Weight is another name for force.
If a mass is dropped on a spring scale,
and the scale is critically damped,
the indicator will move from zero
(Assuming that the scale has been zeroed."
to a point which indicates the force between the mass
and the Earth (Assuming that the measurement was made on the Earth.).
It may "overshoot" once before the indicator settles on the reading.
Force is equal to mass times acceleration,
where the acceleration in this case if "g".
That's the small "g', which is the acceleration
of the mass at that point in the Earth's gravitational field.
If the scale is over-damped,
the scale will slowly wise from zero to the correct reading.
Note that the reading will depend upon the mass, and
where on the Earth's surface you make the measurement.
If the scale is over-damped,
the scale will overshoot, oscillate,
and eventually settle on the correct reading.
The amount of overshoot depends upon
the "impulse" and the amount of damping.
For details on damped systems,
and how to compute how much overshoot there might be,
see that following URL:
http://www.samconsult.biz/Science/Dynamics/0503%20-%20Frame.htm
Damping converts the oscillations in a system into heat.
A certain percentage of the oscillation is converted to heat each cycle.
The magnitude of the oscillation divided by the loss per cycle
is called the "Q" of the system. Underdamped systems
are high "Q" systems. Note that a 100 "Q" system
would lose one percent of its' magnitude per cycle.
--
Tom Potter http://tompotter.us
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| User: "Tom Potter" |
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| Title: Re: Can somebody solve this problem? |
12 Oct 2003 05:05:11 AM |
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"Tom Potter" <tdp@hotsheet.com> wrote in message news:<bm97rj$g8fh9$1@ID-188019.news.uni-berlin.de>...
"Jesper" <jesper@i-dont-want-spam.oek.dk> wrote in message
news:bm8tcn$jln$1@news.net.uni-c.dk...
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
A spring scale measures force,
whereas a balance scale measures mass.
Weight is another name for force.
If a mass is dropped on a spring scale,
and the scale is critically damped,
the indicator will move from zero
(Assuming that the scale has been zeroed."
to a point which indicates the force between the mass
and the Earth (Assuming that the measurement was made on the Earth.).
It may "overshoot" once before the indicator settles on the reading.
Force is equal to mass times acceleration,
where the acceleration in this case if "g".
That's the small "g', which is the acceleration
of the mass at that point in the Earth's gravitational field.
If the scale is over-damped,
the scale will slowly wise from zero to the correct reading.
Note that the reading will depend upon the mass, and
where on the Earth's surface you make the measurement.
If the scale is over-damped,
the scale will overshoot, oscillate,
and eventually settle on the correct reading.
The amount of overshoot depends upon
the "impulse" and the amount of damping.
For details on damped systems,
and how to compute how much overshoot there might be,
see that following URL:
http://www.samconsult.biz/Science/Dynamics/0503%20-%20Frame.htm
Damping converts the oscillations in a system into heat.
A certain percentage of the oscillation is converted to heat each cycle.
The magnitude of the oscillation divided by the loss per cycle
is called the "Q" of the system. Underdamped systems
are high "Q" systems. Note that a 100 "Q" system
would lose one percent of its' magnitude per cycle.
I see I made a typo in one of the paragraphs.
instead of
"If the scale is over-damped,
the scale will overshoot, oscillate"
it should read
"If the scale is UNDER-DAMPED
the scale will overshoot, and oscillate"
Note that if you dump a hundred pound sack of grain
on a spring scale, from one foot, two feet,
or any height that does not cause damage to the scale,
it will eventually read one hundred pounds.
The indicator may come up slowly if the
scale is overdamped, and it may oscillate
about the average weight if the scale is underdamped,
but the scale will end up reading one hundred pounds.
Note that if a one hundred pound meteor hits the scale,
you no longer have a scale, and the concept of
"what does the scale show?" is irrelavant.
--
Tom Potter
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| User: "Martin Hogbin" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 08:00:23 AM |
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"Jesper" <jesper@i-dont-want-spam.oek.dk> wrote in message news:bm8tcn$jln$1@news.net.uni-c.dk...
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
Are you asking what will be the maximum reading on the spring
scale?
Martin Hogbin
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| User: "Uncle Al" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 10:42:24 AM |
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Jesper wrote:
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
Hook's law vs. initial KE plus mgh less air resistance, followed by
GmM/r^2 at equilibrium assuming the scale didn't inelastically deform
on impact.
If it is a long fall, you must integrate over GM/r^2 and allow for
local static reference frame vs. continuing expansion of the universe,
http://arXiv.org/abs/gr-qc/9810085
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "The Ghost In The Machine" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 03:00:08 PM |
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In sci.physics, Uncle Al
<UncleAl0@hate.spam.net>
wrote
on Sat, 11 Oct 2003 08:42:24 -0700
<3F8824E0.64EEDBE9@hate.spam.net>:
Jesper wrote:
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
Hook's law vs. initial KE plus mgh less air resistance, followed by
GmM/r^2 at equilibrium assuming the scale didn't inelastically deform
on impact.
If it is a long fall, you must integrate over GM/r^2 and allow for
local static reference frame vs. continuing expansion of the universe,
Not to mention atmospheric resistance. :-) Of course, after it
melts/explodes and disintegrates, is it really a block?
Or a meteorite?
http://arXiv.org/abs/gr-qc/9810085
--
#191,
It's still legal to go .sigless.
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| User: "Sam Wormley" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 07:42:33 AM |
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Jesper wrote:
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
Show us what you have done so far.
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| User: "Sam Wormley" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 08:30:47 AM |
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Sam Wormley wrote:
Jesper wrote:
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
Show us what you have done so far.
Since the problem doesn't specifically ask for the reading of the scale
when the spring is maximally compressed, I would assume that the reading
will converge to the mass of the block.
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| User: "John C. Polasek" |
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| Title: Re: Can somebody solve this problem? |
11 Oct 2003 11:18:58 AM |
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On Sat, 11 Oct 2003 14:35:27 +0200, "Jesper"
<jesper@i-dont-want-spam.oek.dk> wrote:
A block is allowed to fall freely in the gravitational field of the earth
(from a distance h and with initial velocity 0) on a spring scale. What
will the spring scale show when the block has fallen on the spring scale?
They cant possibly want damping factor etc., but for the furthest
deflection H, dropped from h
v =sqrt(2gh) as it hits the hook
At that point it has kinetic and potential energies:
.5m*2gh + mgH = .5Kh^2
You should be able to do something with that.
The scale calibration is
w = Kx
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
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