Spring and acceleration

In summary, the Olympic commissioners bought and sold gold in an elevator using a vertical spring scale suspended from the ceiling. They always bought and sold at the same price per ounce, using an archaic unit of force. They bought their gold when the elevator had a downward acceleration of 2 m/s^2 and sold when the acceleration was 2.5 m/s^2 upward. The percentage profit based on their buying profit can be evaluated by considering the weight of the gold when accelerating up or down. The spring balance must exert a normal force on the gold, which can be expressed as ma+mg=mg. The fair price for the gold is mgp, but the actual selling price will depend on the acceleration of the elevator.
  • #1
Please Wait
21
0

Homework Statement


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.


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?
 
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  • #2
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
haruspex said:
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?

They do not tell us the mass,all they tell us is that accelerating numbers in the elevator. However they are like they sold and bought at the same price per ounce.
 
  • #4
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
Please Wait said:
mgt = price per ounce
No, that would be the total price for m ounces.
when accelerating up it would be like this

ma[up]+mg[up] = mg[down]t
so a[up]+g[up] = g[down]t
That's a very confused way of writing it. On the left of the equals you have forces, on the right amounts of money. What do you mean by g[up] and g[down]?
When accelerating upwards, what price will they sell m ounces of gold for?
 
  • #6
haruspex said:
No, that would be the total price for m ounces.

That's a very confused way of writing it. On the left of the equals you have forces, on the right amounts of money. What do you mean by g[up] and g[down]?
When accelerating upwards, what price will they sell m ounces of gold for?

well they said that the selling and buying price per ounce stays the same in the question. what i meant by g[down] is gravity and g[up] i meant normal force.
 
  • #7
Please Wait said:
well they said that the selling and buying price per ounce stays the same in the question.
No, I mean if there really are m ounces of gold, no acceleration, what will they actually sell it for?
 
  • #8
haruspex said:
No, I mean if there really are m ounces of gold, no acceleration, what will they actually sell it for?

they will sell for only the weight of it right?
 
  • #9
Please Wait said:
they will sell for only the weight of it right?

i don't understand what you mean. Suppose they have a nugget of gold which, weighed in the absence of acceleration, tipped the scales at 'm ounces' (this being a force, m ounces of weight). What would that same nugget 'weigh' when nugget plus scales are undergoing an upward acceleration of a? What would they therefore sell the nugget for?
 
  • #10
haruspex said:
i don't understand what you mean. Suppose they have a nugget of gold which, weighed in the absence of acceleration, tipped the scales at 'm ounces' (this being a force, m ounces of weight). What would that same nugget 'weigh' when nugget plus scales are undergoing an upward acceleration of a? What would they therefore sell the nugget for?

they would weigh very little since we are going up against gravity.. i don't know specifically.
 
  • #11
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
haruspex said:
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?

The spring balance must be exerting the normal force
 
  • #13
Please Wait said:
The spring balance must be exerting the normal force
Yes, that's what it's called, but I mean how big is it? Write an expression for it in terms of m, a and g.
 
  • #14
haruspex said:
Yes, that's what it's called, but I mean how big is it? Write an expression for it in terms of m, a and g.

Fnet + Fn = Fg
ma + mg = mg
 
  • #15
Please Wait said:
Fnet + Fn = Fg
ma + mg = mg
I know what you mean, but you really must avoid writing equations so carelessly. If ma+mg=mg than ma=0, right? You should write this as
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
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
Please Wait said:
So if they are selling the price for mgp than they are actually just selling it for the acceleration they go up by mass
No.
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
haruspex said:
No.
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?

p(mg+ ma) he will pay whatever the weight the balance will show. since the balance shows the weight as mg+ ma as accelerating, and he buys it, to the weight then it would be p(mg+ ma) right?
 
  • #19
Please Wait said:
p(mg+ ma) he will pay whatever the weight the balance will show. since the balance shows the weight as mg+ ma as accelerating, and he buys it, to the weight then it would be p(mg+ ma) right?
Yes!
Now, the commissioners buy gold when the lift is accelerating downwards at rate a. What will they pay for the same nugget?
 
  • #20
Wouldn't it be the same thing except for the sign so like this
p(mg-ma)
 
  • #21
Please Wait said:
Wouldn't it be the same thing except for the sign so like this
p(mg-ma)

Right. So how much profit do the commissioners make?
 
  • #22
well interms of actual numbers.. they make a profit of 4.5 but like in percentage i am not sure how to find out.
 
  • #23
Please Wait said:
well interms of actual numbers.. they make a profit of 4.5 but like in percentage i am not sure how to find out.
Let's be precise here. They buy a nugget mass m for mp(g-adown) and sell it for mp(g+aup). Profit = mp(adown+aup). To get that as a percentage of what they paid, you divide by what they paid and multiply by 100%.
Do that.
 
  • #24
haruspex said:
Let's be precise here. They buy a nugget mass m for mp(g-adown) and sell it for mp(g+aup). Profit = mp(adown+aup). To get that as a percentage of what they paid, you divide by what they paid and multiply by 100%.
Do that.

umm, i understand what you wrote but like if u want me to be hundred percent accurate i cannot be because i don't know how much the price is or the mass.
 
  • #25
Please Wait said:
umm, i understand what you wrote but like if u want me to be hundred percent accurate i cannot be because i don't know how much the price is or the mass.
You don't need to. Just follow what I said and those things will cancel out.
 
  • #26
haruspex said:
You don't need to. Just follow what I said and those things will cancel out.

okay then Profit % = mp(adown+aup)/mp x 100%, is this correct?
 
  • #27
Please Wait said:
okay then Profit % = mp(adown+aup)/mp x 100%, is this correct?
I said to divide by what they paid. Did they pay mp?
 
  • #28
haruspex said:
I said to divide by what they paid. Did they pay mp?

if u mean just the price then
Profit % = mp(adown+aup)/mp x 100%

and for what they paid in acceleration wise
Profit % = mp(adown+aup)/mp(g+aup) x 100%
 
  • #29
Please Wait said:
if u mean just the price then
Profit % = mp(adown+aup)/mp x 100%

and for what they paid in acceleration wise
Profit % = mp(adown+aup)/mp(g+aup) x 100%

Neither. What did the commissioners pay for the nugget? Refer back to post #20.
 
  • #30
haruspex said:
Neither. What did the commissioners pay for the nugget? Refer back to post #20.

Profit % = mp(adown+aup)/p(mg-ma) x 100%
 
  • #31
Yes! Now do all the cancellations that are possible in that expression.
 
  • #32
Profit % = mp(adown+aup)/p(mg-ma) x 100%
Profit % = mp(adown+aup)/mp (g-a) x 100%
Profit % = (adown+aup)/(g-a) x 100%
Profit % = (2 m/s^2 + 2.5 m/s^2) / (9.8m/s^2 - 2 m/s^2) x 100%
Profit % = 4.5 m/s^2 / 7.8 m/s^2 x 100%
Profit % = 57.7 %
 
  • #34
haruspex said:
Bingo.

I would like to thank you for all your help.
You have been extremely patient and once again thank you.
 
  • #35
Well i was thinking of the orientation shouldn't it be like -6.4% see:

Profit % = mp(adown(down)+aup(up))/p(mg(down)-ma(down)) x 100%
Profit % = mp(adown+aup)/mp (g-a) x 100%
Profit % = (adown+aup)/(g-a) x 100%
Profit % = (2 m/s^2(down) - 2.5 m/s^2(down)) / (9.8m/s^2 (down)- 2 m/s^2(down)) x 100%
Profit % = -0.5m/s^2/ 7.8 m/s^2 x 100%
Profit % = -6.4%
 

1. What is spring constant and how does it affect acceleration?

Spring constant, also known as the force constant, is a measure of the stiffness of a spring. It is represented by the letter k and is measured in units of newtons per meter (N/m). The higher the spring constant, the stiffer the spring and the greater the force required to stretch or compress it. This affects acceleration because the greater the force applied to a spring, the greater the acceleration it will produce.

2. How does Hooke's Law relate to spring and acceleration?

Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed. This means that the more a spring is stretched or compressed, the greater the force it will produce. This relationship between force and distance is important in understanding the acceleration of a spring, as the greater the force applied to a spring, the greater the acceleration it will produce.

3. Can a spring have negative acceleration?

Yes, a spring can have negative acceleration. Negative acceleration, also known as deceleration, occurs when an object is slowing down. This can happen when a spring is compressed, as the force applied to the spring is in the opposite direction of the motion. In this case, the acceleration of the spring would be negative, but it is still accelerating, just in the opposite direction.

4. How does the mass attached to a spring affect its acceleration?

The mass attached to a spring does not directly affect its acceleration. According to Newton's Second Law, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. This means that the greater the mass attached to a spring, the greater the force needed to produce the same acceleration. However, the mass attached to a spring can affect the spring's natural frequency, which can impact its acceleration in a system with multiple springs.

5. What is the relationship between spring acceleration and frequency?

The relationship between spring acceleration and frequency is inverse. This means that as the frequency of a spring increases, its acceleration decreases, and vice versa. This is because the natural frequency of a spring is determined by its stiffness and mass, and a higher frequency means a higher natural frequency, which results in a lower acceleration. However, this relationship only applies to a single spring and not a system with multiple springs.

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