TI-M: Re: Re: Re: RE: physics
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TI-M: Re: Re: Re: RE: physics
But it's still wrong.
(1) The uncertainty in d is likely to be much more than 1 µm, so writing six
significant digits is absurd.
(2) For values less than one, there should ALWAYS be a zero to the left of
the decimal point. NIST and ISO conventions require it: it makes the
difference between 23 and 0.23 easy to see.
So the answer should be reported as 0.23 m.
----- Original Message -----
From: "Mike Grass" <mike_grass@hotmail.com>
To: <ti-math@lists.ticalc.org>
Sent: Friday, November 10, 2000 01:00
Subject: TI-M: Re: Re: RE: physics
>
> Ooops, that should be .231481 meters, I just missed the decimal in my
> typing.
> ----- Original Message -----
> From: "Mike Grass" <mike_grass@hotmail.com>
> To: <ti-math@lists.ticalc.org>
> Sent: Thursday, November 09, 2000 11:45 PM
> Subject: TI-M: Re: RE: physics
>
>
> >
> > I'll bite....
> >
> > Ki is initial kinetic energy for the block at the top of the ramp, so 0.
> > Kf is final kinetic energy for the block right after it leaves the ramp.
> > Kfh is the final velocity on the horizontal surface, so 0.
> > Kih is the initial kinetic energy on the horizontal surface.
> > vf is the final velocity of the block after it leaves the ramp.
> > vi is the initial velocity on the horizontal surface.
> > m is the mass of the first block.
> > M is the mass of the second block.
> > N is the normal force.
> > h is the height.
> > g is gravity.
> > d is distance traveled on the horizontal surface (what we want to solve
> > for).
> > W is work
> > Wg is work done by gravity
> > Wn is work done by normal force (0, since normal force is perpendicular
to
> > the change in the position)
> > Wf is work done by friction (0 for the ramp, -(mu)*N*d for the
horizontal
> > surface)
> > Up is positive y
> > Right is positive x
> >
> > First, for the block sliding down the ramp:
> > (Work-kinetic energy theorem)
> > Kf-Ki = Sigma(W) with Ki = 0
> > = Wg + Wn + Wf
> > = Wg
> > Kf = Wg
> > Wg = m*g*h
> > Kf = 1/2*m*vf^2
> > 1/2*m*vf^2 = m*g*h
> > vf = sqrt(2*g*h)
> >
> > Now, because this is assumed to be a completely inelastic collision and
> > momentum is conserved:
> > m*vf = (m+M)*vi
> > m*(sqrt(2*g*h)) = (m+M)*vi
> > vi = (m*sqrt(2*g*h))/(m+M)
> >
> > So, we have the initial velocity of the system and we use the
Work-kinetic
> > energy theorem again:
> > Kfh-Kih = sigma(W) with Kfh = 0
> > -Kih = Wg + Wn + Wf, Wg = 0, Wn = 0
> > -Kih = -(mu)*N*d, with N = m*g
> > 1/2*(m+M)*vi^2 = (mu)*(m+M)*g*d
> > d = vi^2/(2*g*(mu))
> > = (m*sqrt(2*g*h)/(m+M))^2 * (2*g*(mu))
> > = (m^2*h)/((mu)(m+M)^2)
> >
> > And plugging in the values for m, h, (mu), and M, we get.....
> > 231481 meters
> >
> > I think that's right :)
> >
> >
>
>
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