M3 Question

A pretty standard Hooke's law problem taken from a book, which also states the answer, but the answer is counter intuitive and might be wrong, which is why I am asking you guys. So here it is:

A light elastic string is of natural length 0.5 m is stretched between two points P and Q on a smooth horizontal table. The distance PQ is 0.75 m and the modulus of elasticity is 30N.
A particle of mass 0.5 kg is attached to the midpoint of PQ. The particle is pulled 0.1m towards Q and released from rest.
Find the speed of the particle as it passes through the midpoint of PQ.

Thank you.

Scroll to see replies

Reply 1

Pangol

I make the answer 2.19 m s-1 (3 s.f.), found via a conservation of energy method. Does that agree with the book?

Reply 2

-Phoenix-

Original post by Pangol

I make the answer 2.19 m s-1 (3 s.f.), found via a conservation of energy method. Does that agree with the book?

I also used the energy conservation method but managed to get 1.549 m/s (I think I'm wrong...may I know how you did it?)

Reply 3

Pangol

Original post by -Phoenix-

I also used the energy conservation method but managed to get 1.549 m/s (I think I'm wrong...may I know how you did it?)

I did do it rather quickly, I may have some of the values wrong.

When the particle is released from rest, the system has EPE (which I made 3.075 J) and no KE.

When is passes through the midpoint, the system has EPE (which I made 1.875 J) and the particle has KE (which, if I have the other values correct, would be 1.2 J).

Since we know the mass of the particle, it's now easy to work out it's speed.

Reply 4

Pangol

(Actually, I think I have made some calculation errors, but the idea seems right.)

Reply 5

username1667697

Original post by Pangol

I make the answer 2.19 m s-1 (3 s.f.), found via a conservation of energy method. Does that agree with the book?

Yes that is the answer in the book.

EPE loss = KE gain (This is obvious)

EPE loss = (30/2x0.25)(0.225^2 + 0.025^2 - 0.125^2 - 0.125^2)
This EPE loss gives v = 2.19 m/s, which is what you and the book gives as the answer.
Now the part which confuses me is the EPE provided by each 0.25m of the string. Shouldn't it be
EPE loss = (30/2x0.25)(0.225^2 - 0.025^2 + 0.125^2 - 0.125^2) ?
Since the right part of the strings potential energy is to the right?

Reply 6

-Phoenix-

Original post by Pangol

I literally got half of all your energy values:
EPE at rest = 1.5375
EPE at midpoint = 0.9375
Did you use ((lambda)x^2)/2L for EPE?

Reply 7

Pangol

(Actually, I think I have made some calculation errors, but the idea seems right.)

Original post by -Phoenix-

I literally got half of all your energy values:
EPE at rest = 1.5375
EPE at midpoint = 0.9375
Did you use ((lambda)x^2)/2L for EPE?

I am now back to thinking that I was right again! I did indeed use that EPE formula. I think it is best to think of the system consisting of two identical strings of half the total original length, as the particle is fixed between them.

Reply 8

-Phoenix-

Original post by Pangol

(Actually, I think I have made some calculation errors, but the idea seems right.)

I am now back to thinking that I was right again! I did indeed use that EPE formula. I think it is best to think of the system consisting of two identical strings of half the total original length, as the particle is fixed between them.

Oh yes! you are right. I forgot to divide the natural length by two when doing the calculation even though I divided the extension by two.
Thank you :smile:

Reply 9

Pangol

Original post by utah

Now the part which confuses me is the EPE provided by each 0.25m of the string. Shouldn't it be
EPE loss = (30/2x0.25)(0.225^2 - 0.025^2 + 0.125^2 - 0.125^2) ?
Since the right part of the strings potential energy is to the right?

I think that you are mixing up the properties of tension and energy. Tension is a force, and so has a direction, but EPE is a form of energy, which does not have a direction. You simply add the two EPE values when released to get the total EPE at that point, and the same again when the particle passes through the middle of PQ.

Reply 10

Pangol

Incidentally, this exact question came up only a few weeks ago (I knew it felt familiar!).

https://www.thestudentroom.co.uk/showthread.php?p=70655634&highlight=

Reply 11

username1667697

Original post by Pangol

Can you explain the EPE values, and mainly the values of extension you chose and particularly the directions of the EPE's(i.e. signs)?

Reply 12

Pangol

Original post by utah

Can you explain the EPE values, and mainly the values of extension you chose and particularly the directions of the EPE's(i.e. signs)?

Probably worth looking in the other thread I linked to - there are diagrams there.

But the important thing is that EPE (like all energy) does not have a direction. Energy is a scalar, not a vector.

Reply 13

username1667697

Original post by Pangol

Probably worth looking in the other thread I linked to - there are diagrams there.

But the important thing is that EPE (like all energy) does not have a direction. Energy is a scalar, not a vector.

It just doesn't make sense intuitively.
I am trying to say that when the particle is in the stretched positions, the potential from the left string is to the left(since it has been stretched to the right), and the potential from the right string is to the right(since it has been stretched to the left). So should the EPE's be subtracted rather than added?

Reply 14

Pangol

Original post by utah

No. I understand why you would think this, but it is not correct. The tension in the strings have directions - they are forces, and one acts to the left, one to the right. But elastic potential energy, like kinetic energy and gravitational potential energy, does not have a direction. They are just numbers, and therefore the two EPEs should be added.

Reply 15

username1667697

Original post by Pangol

OK. I am going to bother you some more with weird thoughts.

Suppose you are hovering 1km above the earth. Now, your potential energy is mgh. But you only have the potential to fall towards earth. You don't have the potential to fall into space, do you?

Reply 16

Pangol

Original post by utah

It is the force acting on you (your weight) that points down, not the GPE. I do sympathise, but you're not going to win on this one! Check your text book for the definition of energy (any kind you like) - they should be clearly defined as scalars, not vectors.

Reply 17

Pangol

I've thought of an example that might help you to see that EPE has no direction.

If I stretch an elastic band between the index fingers of each hand, then it now has EPE stored in it.

If I slip it off my right index finger, it shoots off towards the left. If, instead, I slip it off my left index finger, it shoots off towards the right.

So if you are right that EPE has a direction, which direction was it pointing in before I decided how to let the band go?

Reply 18

username1667697

Original post by Pangol

How about this? There are two masses of the same mass, or two charges of the same charge, some distance apart. Now if you place some object with some mass or some charge, exactly in the middle of the two objects, then, is the object's potential, the sum of the potential's provided from the other two objects, or simply 0?