Calculating Elastic Potential Energy with Varying Masses

[flyingpig]

Homework Statement

A 2.6-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 1.3 J. What is the elastic potential energy of the system when the 2.6-kg block is replaced by a 4.8-kg block?

The Attempt at a Solution

[tex]E_i = mgh + \frac{1}{2}kx^2[/tex]

[tex]E_f = 2mgh' + \frac{1}{2}kx'^2[/tex]

Now I am stuck...

 

[tiny-tim]

hi flyingpig!

hint: the mgh doesn't matter …

you're only interested in the elastic https://www.physicsforums.com/library.php?do=view_item&itemid=269"

(not the gravitational potential energy)​

 

[flyingpig]

But there is a mass...

 

[tiny-tim]

hi flyingpig!

you can only use E_i and E_f (conservation of energy) when there's some process in which energy is conserved

in this case, the mass is placed carefully in the equilibrium position …

(or it's allowed to oscillate until it loses enough energy to achieve equilibrium)

there's no energy conservation process to analyse!

use a forces equation (not an energy one) to find a relation between x and m ​

 

[PeterO]

Homework Statement

A 2.6-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 1.3 J. What is the elastic potential energy of the system when the 2.6-kg block is replaced by a 4.8-kg block?

/QUOTE]

The heading says it all - use ratios with energy and mass.

The 2.6 kg block extends the spring a certain distance.

The energy stored in the spring is given by 1/2 k x^2 and has a value of 1.3J

We are using the same spring so k doesn't change.

if you double x, you get 4 times the energy stored (2x)^2 = 4 x^2
If instead you triple x you get 9 times the energy (3x)^2 = 9 x^2


So when you put the new mass on, by what factor do you increase x?

Peter

 

[flyingpig]

How do we know the object was placed at x = 0?

 

[tiny-tim]

in this case, the mass is placed carefully in the equilibrium position …

How do we know the object was placed at x = 0?

it wasn't …

by "equilibrium", i meant the position in which the weight balances the tension

 

[flyingpig]

I am still stuck...

 

[PeterO]

I am still stuck...

hanging a 2.6 kg mass on the spring will extend the spring.

the F = kx formula for a spring shows the connection between force applied, extension x and the spring constant k.

You only know one of these, F , since that is the weight of the block.

However, you also know how much energy was stored by that extension

E = 1/2 k x^2 = 1.3 J I think it was? [it was given]

now 1/2 kx^2 equals x/2 x kx - but you know the value of kx [see above] so you can work out what x is, and thus get out k.

Once you know that you can find what happens with the new mass.

Of course if you are good at variation, you can find the answer more directly, but most people aren't good at variation, and if you were you would have the answer by now.

 

[flyingpig]

Are you suggesting [tex]\frac{1}{2}kx^2 = E[/tex], so [tex]F = kx = \frac{2E}{x}[/tex]?

 

[PeterO]

I am still stuck...

Using Variation - not sure you will Follow.

Energy in a first spring spring is given by E = ½kx²

Energy in second spring can be shown as E' = ½kx'² =

ratio of second to first means E'/E = x'²/x² = (x'/x)²

Now the spring is extended each time according to F = kx, and since it is a weight attached this means mg =kx

By the same method, the new situation is m'g = kx'

with ratios we have x'/x = m'/m

In the energy ratio above that means E'/E = (m'/m)² or E' = E(m'/m)²

so E' = 1.3 x (4.8/2.6)² which you can now evaluate to 4.4 J

 

[PeterO]

Are you suggesting [tex]\frac{1}{2}kx^2 = E[/tex], so [tex]F = kx = \frac{2E}{x}[/tex]?

Yes.

 

[flyingpig]

Wow so this question combines dyanmics + energy

 

[flyingpig]

Yes.

But that didn't get me anywhere

[tex]F = kx = \frac{2E}{x} = mg[/tex]

[tex]\frac{4E}{x} = 2mg[/tex]

 

[PeterO]

But that didn't get me anywhere

[tex]F = kx = \frac{2E}{x} = mg[/tex]

[tex]\frac{4E}{x} = 2mg[/tex]

it won't get you anywhere is you just multiply each side by 2 ?

What made you do that?

 

[flyingpig]

Oh right...

for some reason i thought 2.6 * 2 = 4.8

So I should multiply by 4.8/2.6

 

[PeterO]

Oh right...

for some reason i thought 2.6 * 2 = 4.8

So I should multiply by 4.8/2.6

I wouldn't. What were you trying to achieve.

I would use the original data to find x, use that value to then find K.

then use the new weight to find the new x, then the new x and the k value to find the new energy.

[actually I would use variation, like the solution I gave, but the sequence outlined above shows the sequence you have to follow otherwise]