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Spring and acceleration 
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#1
Dec3012, 07:24 PM

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1. The problem statement, all variables and given/known data
The olympic commissioners decided to buy and sell gold, hoping to set a new profit record. They conducted all their business in an elevator. They bought and sold at the same price per ounce. An archaic unit of force, which measured with a verticle spring scale scale suspended from the ceiling. They always bought their gold when the elevator had a downward acceleration of 2 m/s^2 and always sold when the acceleration was 2.5 m/s^2 upward. Evaluate their percentage profit based on their buying profit.  means a loss. 3. The attempt at a solution Alright so i know the following things: a of buying: 2 m/s^2 a of selling : 2.5 m/s^2 and there is something to do with springs... but how would i start off? 


#2
Dec3012, 11:36 PM

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Don't worry about the springs. All that matters is that the scale doesn't measure mass, it measures the force. So consider the free body diagram of the gold. For a given mass of gold, what weight will be measured when accelerating up? Down?



#3
Jan213, 03:14 PM

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#4
Jan213, 03:41 PM

P: 21

Spring and acceleration
Umm is it like this..
so the t represents the price per ounce so Fg = mg mgt = price per ounce so when accelerating up it would be like this ma[up]+mg[up] = mg[down]t so a[up]+g[up] = g[down]t right? and when accelerating down ma[down]+mg[down]t = mg[up] a[down] + g[down]t = g[up] but how would i find the profit? 


#5
Jan213, 09:05 PM

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When accelerating upwards, what price will they sell m ounces of gold for? 


#6
Jan213, 09:38 PM

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#7
Jan213, 09:55 PM

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#8
Jan213, 09:58 PM

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#9
Jan213, 10:05 PM

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#10
Jan213, 10:30 PM

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#11
Jan213, 11:12 PM

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first, to avoid some possible confusion, I'm going to modify what I said before. Suppose the nugget weighs mg when not accelerating. (I wrote 'm' before.) This means that gravity exerts a downward force mg on the nugget, and in order to achieve equilibrium the spring balance has to push up with force mg. That leads the balance to record a weight of mg ounces.
When the nugget, sitting on the spring balance, is accelerating upwards at rate a, what force must the spring balance be exerting on it? 


#12
Jan413, 06:30 PM

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#13
Jan413, 08:05 PM

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#14
Jan513, 07:14 PM

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ma + mg = mg 


#15
Jan513, 07:21 PM

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force on spring balance = ma+mg So, if the price is p per ounce, the fair price for this nugget is mgp. What price will they actually sell it for? 


#16
Jan513, 07:26 PM

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So if they are selling the price for mgp than they are actually just selling it for the acceleration they go up by mass



#17
Jan513, 07:32 PM

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Suppose the lift is not accelerating. We weigh the nugget and it weighs mg. The fair price for this nugget is therefore mgp. Now the lift accelerates upwards at rate a. You have calculated that the balance will now show the weight as mg+ma. If a buyer buys based on the weight the balance shows now, how much will the buyer pay? 


#18
Jan513, 07:38 PM

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