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Finding Mass and Distance using apparent Weight 
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#1
Oct607, 05:15 PM

#2
Oct607, 06:30 PM

Admin
P: 21,911

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(ga) and W = m(g+a). 


#3
Oct607, 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?



#4
Oct607, 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?



#5
Oct607, 07:21 PM

Admin
P: 21,911

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.phyastr.gsu.edu/hbase/mot.html 


#6
Oct607, 07:35 PM

P: 23

but wouldn't i need the velocity as well? i have acceleration as 1.962 m/s.



#7
Oct607, 07:38 PM

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P: 21,911

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. 


#8
Oct607, 07:49 PM

P: 23

Do i add all the distances up?



#10
Oct607, 07:57 PM

P: 23

Argh, i dunno what im doing wrong but the answer i keep getting is wrong. Is my acceleration correct?



#11
Oct607, 08:06 PM

Admin
P: 21,911

About 1.96 m/s^{2} 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. 


#12
Oct607, 08:06 PM

P: 23

ok well i got 62.784 m as the total distance traveled after 12 secs. Is that correct?



#13
Oct607, 08:07 PM

P: 23

oh well I used the formula i mentioned in the beginning
Apparent Weight = mg(1+a/g) 


#14
Oct607, 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. 


#15
Oct607, 08:10 PM

Admin
P: 21,911

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. 


#16
Oct607, 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? 


#17
Oct607, 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 102 = 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) 


#18
Oct707, 08:56 AM

P: 23

oh cool it was right. Thanks Astronuc.



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