Finding Mass and Distance using apparent Weight


by iHate Physics
Tags: apparent, distance, mass, weight
iHate Physics
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#1
Oct6-07, 05:15 PM
P: 23
1. The problem statement, all variables and given/known data
Henry gets into an elevator on the 50th floor of a building and it begins moving at t=0s The figure shows his apparent weight over the next 12s.



What is Henry's mass?
How far has Henry traveled at t =12s

2. Relevant equations

F=ma
w=mg
weight apparent = w(1+ a/g)


3. The attempt at a solution

Well so far, all I know is that the elevator is moving down since the apparent weight has gone up. I have no idea how to approach the problem and I have tried using the third formula that I have listed but no luck. Do I even use the apparent weight equation? Because I need the acceleration, and no information has been given on that.
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Astronuc
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#2
Oct6-07, 06:30 PM
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Well the way I would read that, is that the elevator accelerates downard initially for 2 seconds, since Henry is lighter, then descends at constant speed (from which one can get Henry's weight), then decelerates (Henry is heavier), and then stops (from which one again get's Henry's weight).

One needs to determine when to apply W = m(g-a) and W = m(g+a).
iHate Physics
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#3
Oct6-07, 06:43 PM
P: 23
Oh, I see, so you would you would use 750 N as his weight when determining the mass. Thanks for explaining the graph to me. Even our prof never explained this kinda graph to us. To me this is new material. k, Lemme try to find the distance now. I would need to find the acceleration first right?

iHate Physics
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#4
Oct6-07, 07:11 PM
P: 23

Finding Mass and Distance using apparent Weight


ouch, ok nevermind... Anyone have any tips on how to approach this when finding distance?
Astronuc
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#5
Oct6-07, 07:21 PM
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For an approximation, one could assume constant acceleration.

Then there are three parts of the problem to find distance - the initial drop at constant acceleration, the descent at constant velocity, and the descent during deceleration.

See this reference - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
iHate Physics
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#6
Oct6-07, 07:35 PM
P: 23
but wouldn't i need the velocity as well? i have acceleration as 1.962 m/s.
Astronuc
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#7
Oct6-07, 07:38 PM
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Start with an initial velocity of zero at t = 0s.

Use the acceleration to find the distance and velocity at t = 2s. That velocity is the initial velocity and constant velocity for the period t =2 s to t = 10 s.

Determine the distance for that period.

Then use that intial velocity and deceleration for the final period 10 s to 12 s, when velocity goes to zero because the elevator stops.
iHate Physics
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#8
Oct6-07, 07:49 PM
P: 23
Do i add all the distances up?
Astronuc
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#9
Oct6-07, 07:55 PM
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Quote Quote by iHate Physics View Post
Do i add all the distances up?
Sure, why not?
iHate Physics
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#10
Oct6-07, 07:57 PM
P: 23
Argh, i dunno what im doing wrong but the answer i keep getting is wrong. Is my acceleration correct?
Astronuc
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#11
Oct6-07, 08:06 PM
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About 1.96 m/s2 is correct. Here I'm assuming a step change in acceleration. From the curve, it looks like it changes rapidly from 1.96 to 0 between 2 and 2.2 s, so that could be a source of error, but it should be relatively small.

Please write out the formulas one is using.
iHate Physics
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#12
Oct6-07, 08:06 PM
P: 23
ok well i got 62.784 m as the total distance traveled after 12 secs. Is that correct?
iHate Physics
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#13
Oct6-07, 08:07 PM
P: 23
oh well I used the formula i mentioned in the beginning

Apparent Weight = mg(1+a/g)
iHate Physics
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#14
Oct6-07, 08:09 PM
P: 23
Basically for t=0 to t=2 i got 3.924 m
t=2 to t=10 i got 31.392m
t=10 to t=12 i got 27.468m

and i got the total to be 62.784m

whoops i should start editing my posts instead of continuously posting. Sorry.
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#15
Oct6-07, 08:10 PM
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Quote Quote by iHate Physics View Post
oh well I used the formula i mentioned in the beginning

Apparent Weight = mg(1+a/g)
It seems you've done that part correctly.

Please write the formulas for distance and velocity at 2 s, and then at 10 s, and then at 12 s, so we can see how you determine the cumulative distance.
iHate Physics
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#16
Oct6-07, 08:18 PM
P: 23
k well i did for

t=0 to t=2

x = (0)(2) +1/2 (1.962)(2)^2

x = 3.924m

Vf^2 = (0)^2 +2(1.962)(3.924)
Vf = 3.924

is that correct so far?
iHate Physics
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#17
Oct6-07, 08:25 PM
P: 23
then for t = 2 to t = 10 Constant Velocity so a = 0

x = 3.924 (8) + 1/2 (0)(8)^2 Took t=8 since 10-2 = 8

x = 31.392m

then for t=10 to t=12

x = (3.924)(2) + 1/2 (-1.962)(2)^2

x = 3.924m

and the total i got was 39.24 m ( Realized i had an error in there so i changed it)
iHate Physics
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#18
Oct7-07, 08:56 AM
P: 23
oh cool it was right. Thanks Astronuc.


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