# Increase in mass when raised?!

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#1
Calculate the mass increase of a 10kg object when it is raised through a height of 2m?

(So.... I hink this is meant to be simple and it gains a certain amount each 10m it is raised, also, my calculator is rubbish as 10+ (2.18x10^-15) comes back as 10... obv it thinks its too small to notice :/ )

I've done this at school but I just can't remember how. Thanks x
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10 years ago
#2
(Original post by Mangafairy)
Calculate the mass increase of a 10kg object when it is raised through a height of 2m?
Try working out the change in gravitational potential energy.
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10 years ago
#3

Inertial and gravitational mass will not change. It should be possible to calculate is using Newton's law of universal gravitation.
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#4
I worked the potential energy out at 196.2J But how does this help?
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#5
(Original post by AnonymousPenguin)

Inertial and gravitational mass will not change. It should be possible to calculate is using Newton's law of universal gravitation.

The question just says in the textbook

'calculate the mass increse of:
a) a 10kg object when it is raised through a height of 2.0m'
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10 years ago
#6
(Original post by Mangafairy)
The question just says in the textbook

'calculate the mass increse of:
a) a 10kg object when it is raised through a height of 2.0m'
When the object is raised the weight should decrease due to the reduced strength of the gravitational field and mass should stay the same. Perhaps it's a trick question, what do the solutions say?

I don't see how potential energy would have anything to do with mass. You need to calculate the change in the gravitational field. Have you been given data for mass of Earth and its radius?
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10 years ago
#7
(Original post by Mangafairy)
I worked the potential energy out at 196.2J But how does this help?
Use with that PE

Edit : This gives the correct answer, but the above posters may be correct and the question might be wrong. I don't know very much about Physics.
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#8
I do have data for those in the back on the formulae and I will try it. The radius I assume is Radius of Earth plus 2m?

Which formula were you thinking I should use, F=Gx (mm)/r^2
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#9
Use with that PE

Edit : This gives the correct answer, but the above posters may be correct and the question might be wrong. I don't know very much about Physics.

Why would I use E=mc^2? And what should I do with PE if I did use this?
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10 years ago
#10
(Original post by Mangafairy)
Why would I use E=mc^2? And what should I do with PE if I did use this?
Increase in PE does not increase mass
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10 years ago
#11
Yes I was thinking of that formula. You'll need to remove one m to get field strength. Just calculate it with the radius you have and then with radius + 2m and you should get the difference in field strength. Then multiply that with mass and see if it's right.

My teacher would answer 0 to this question, but the IBO used a different interpretation of "mass" a couple of years ago. The question was, I am told, disqualified in the end.

Sliced bread: Umm, is relativity taught for what this kid is doing? I've never seen that formula used in that way and I don't really see why potential energy would add to standing energy :/ .
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#12
I work out the Pe. which is 196J and using E=mc^2, I find that m is 2.18x10^-15.

But I still don't understnd how the mass increases as it is raised :/ Which I assume means that hey did want me to use E=mc^2 O.o
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10 years ago
#13
(Original post by Mangafairy)
I work out the Pe. which is 196J and using E=mc^2, I find that m is 2.18x10^-15.

But I still don't understnd how the mass increases as it is raised :/
That formula tells you how the mass of objects changes if they give away or receive energy. It is generally used in exercises like "the sun gives away xxx energy in radiation per hour. How much mass does it lose?" This formula is generally taught together with special relativity, are you doing that now?

I've never seen it used in this context and I really can't wrap my head around the fact that a change in potential energy would add to the energy of an object. Potential energy isn't really a property of an object the same way as kinetic or internal energy is :/ . So I guess that's where my help ends and I'm looking forward to someone explaining this.
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#14
It was in my topic on Energy and Mass, which does spend ages talking about E=mc^2. But I still don't see why it's used here. If I try it using g=(Gm)/r^2 do I need the mass of the eart or the object as m?
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10 years ago
#15
(Original post by Mangafairy)
I work out the Pe. which is 196J and using E=mc^2, I find that m is 2.18x10^-15.

But I still don't understnd how the mass increases as it is raised :/ Which I assume means that hey did want me to use E=mc^2 O.o
You know that the object has gained 196.2 J of energy
and you know that the change in energy is = change in mass *c^2

c is the speed of light in vacuum 3x10^8 approximately.
You can rearrange for the change in mass.

As I said before, I don't really know if change in PE increases the mass/the theory behind it, but doing this will give you the answer you stated at the top.
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10 years ago
#16
Well I guess that means I don't fully understand the theory behind it. Sorry if I've mislead you and wasted your time :/ .
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#17
Yea, I suppose I'll have to accept it and not understand fully :/ But I wouldn't naturally think to do that in an exam, I'm gonna sit messing around with GPE and stuff and still get it wrong :P

A Change in Energy= Change in Mass x C^2...................nice....
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#18
And thanks for your help x
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10 years ago
#19
you do the E=mc^2 thing. i thought.
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10 years ago
#20
(Original post by AnonymousPenguin)
Yes I was thinking of that formula. You'll need to remove one m to get field strength. Just calculate it with the radius you have and then with radius + 2m and you should get the difference in field strength. Then multiply that with mass and see if it's right.
The change in the field strength would affect the weight, not the mass.

(I'm also in the "not sure that changing the PE would affect the relativistic mass" camp, but it's pretty clear that is what the question is expecting you to do).
0
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