TI-M: Re: Re: Re: Re: RE: physics
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TI-M: Re: Re: Re: Re: RE: physics
Yes, I know. a) I was tired last night b)the original post would require
that there be only one significant figure in the answer (3 kg for mass of
the first block), so the answer should have been given as 0.2 m.
Oh well.
:)
--Mike
----- Original Message -----
From: "Daniel Bishop" <danb2k@hotmail.com>
To: <ti-math@lists.ticalc.org>
Sent: Friday, November 10, 2000 11:11 AM
Subject: 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 :)
> > >
> > >
> >
> >
>
>
References: