# M3 elasticity energy question!

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#1
One end of a natural string length 1.5m and mod of elasticity 49n is attached to a hook on a ceiling. A particle of mass 2kg is attached at the other end so the string hangs at an equilibrium point. The particle is then pulled down an additional 0.4m. When released find how much the particle rises by.

So I know before release it is 2.5m below the ceiling, since it hangs in equilibrium at 2.1m, before being pulled the additional 0.4m

So I thought, since the string has been extended by 1m in total, the elastic energy would be 49/3 J since Elastic energy = (1/2)kx^2 where k is the stiffness. But this gets the wrong answer.

Also, am I correct in thinking that from release point to the highest point, the loss of elastic potential energy = gain in gravitational potential energy?

I think this is wrong also - but anyway, thanks for any help! 0
7 years ago
#2
(Original post by Student10011)
One end of a natural string length 1.5m and mod of elasticity 49n is attached to a hook on a ceiling. A particle of mass 2kg is attached at the other end so the string hangs at an equilibrium point. The particle is then pulled down an additional 0.4m. When released find how much the particle rises by.

So I know before release it is 2.5m below the ceiling, since it hangs in equilibrium at 2.1m, before being pulled the additional 0.4m

So I thought, since the string has been extended by 1m in total, the elastic energy would be 49/3 J since Elastic energy = (1/2)kx^2 where k is the stiffness. But this gets the wrong answer.
Seems reasonable to me.

Also, am I correct in thinking that from release point to the highest point, the loss of elastic potential energy = gain in gravitational potential energy?
Again, seems fine.

I get 4/5 m.

What does the book say?
0
7 years ago
#3
(Original post by Student10011)
So I thought, since the string has been extended by 1m in total, the elastic energy would be 49/3 J since Elastic energy = (1/2)kx^2 where k is the stiffness. But this gets the wrong answer.
I haven't worked this through, but the only elastic energy available to accelerate the particle upwards is that added by the final 0.4 m extension. Once the particle reaches the equilibrium position on its upward journey, the net force on it is 0. Thereafter it decelerates.

Imagine instead that the particle was pulled down only an extra 1 mm from the equilibrium position then released; it would move with SHM until it was 1 mm above the equilibrium position and the energy stored in the string due to the 0.6 m extension due to its weight would not be converted back into PE.
0
#4
(Original post by ghostwalker)
Seems reasonable to me.

Again, seems fine.

I get 4/5 m.

What does the book say?
The book says 0.8 also. But if gain in gpe = loss in elastic energy, then surely 49/3 (loss in elastic) = 2gh (Gain in GPE), which gives h = 5/6, 0.8333. The book gives 0.8 exactly 0
7 years ago
#5
(Original post by Student10011)
The book says 0.8 also. But if gain in gpe = loss in elastic energy, then surely 49/3 (loss in elastic) = 2gh (Gain in GPE), which gives h = 5/6, 0.8333. The book gives 0.8 exactly At its new position, the string is still under tension. It doesn't lose all of its EPE.

Let the extension be "x", then apply your equation - in bold - and you'll get a quadratic.
0
#6
(Original post by atsruser)
I haven't worked this through, but the only elastic energy available to accelerate the particle upwards is that added by the final 0.4 m extension. Once the particle reaches the equilibrium position on its upward journey, the net force on it is 0. Thereafter it decelerates.

Imagine instead that the particle was pulled down only an extra 1 mm from the equilibrium position then released; it would move with SHM until it was 1 mm above the equilibrium position and the energy stored in the string due to the 0.6 m extension due to its weight would not be converted back into PE.
Okay, so I said from the point to which it is pulled down to, upto its equilibrium position, the loss in epe = (49(0.4)^2)/3 and this equals the gain in kinetic energy. Then the loss in KE from equillibrium point to peak point equals the gain in GPE. This gives height risen above equilibrium point as 2/15. Add 2/15 to the 0.4 extension and you get 0.533..which again is incorrect! 0
#7
(Original post by ghostwalker)
At its new position, the string is still under tension. It doesn't lose all of its EPE.

Let the extension be "x", then apply your equation - in bold - and you'll get a quadratic.
But I would have thought, once the particle attached rises above the equillibrium position, the string is no longer taut, so tension isn't a factor
0
7 years ago
#8
(Original post by Student10011)
But I would have thought, once the particle attached rises above the equillibrium position, the string is no longer taut, so tension isn't a factor
The string is indeed taut while the particle is above the equilibrium position, since it is still extended beyond its natural length.

To do this correctly by energy considerations (rather than the purely physical argument that I was hinting at), do as follows:

1. At its lowest point, the particle is 2.5 m below the point of attachment. The string has m. What is the extension of the string? Hence what is the EPE in the string?

Suppose that the particle reaches a point above the equilibrium position. At this point, by conservation of energy, some of the EPE that you found in 1. has been converted into PE of the particle. (Draw a picture now to make all of these distances clear in your mind)

2. Write down an expression, in terms of , for the gain in PE of the particle from its lowest point.

3. At its highest point, what, in terms of , is the extension of the string? Hence write down an expression for the EPE in the string at this point

4. Apply conservation of energy to write down an equation relating the values found in 1), 2), and 3)

5. Solve this equation to find 0
#9
(Original post by atsruser)
The string is indeed taut while the particle is above the equilibrium position, since it is still extended beyond its natural length.

To do this correctly by energy considerations (rather than the purely physical argument that I was hinting at), do as follows:

1. At its lowest point, the particle is 2.5 m below the point of attachment. The string has m. What is the extension of the string? Hence what is the EPE in the string?

Suppose that the particle reaches a point above the equilibrium position. At this point, by conservation of energy, some of the EPE that you found in 1. has been converted into PE of the particle. (Draw a picture now to make all of these distances clear in your mind)

2. Write down an expression, in terms of , for the gain in PE of the particle from its lowest point.

3. At its highest point, what, in terms of , is the extension of the string? Hence write down an expression for the EPE in the string at this point

4. Apply conservation of energy to write down an equation relating the values found in 1), 2), and 3)

5. Solve this equation to find Ah yes cheers man, got the correct answer doing that and considering the string still has tension above the equilibrium position. Thanks for the help!
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