M3 Elastic strings on inclined planes

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ThatPerson
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I'm getting slightly confused with energy conservation questions that involve elastic strings on inclined planes.

The question is from the old Heinemann M3 book (Ex2C Q8):

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A particle P of mass 2 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus 20N. One end of the string is fixed to a point A on a rough plane inclined at 45 degrees to the horizontal. P is held at rest on the plane below A with AP along a line of greatest slope of the plane and AP=0.6 m. The coefficient of friction between P and the plane is 0.2. P is released. Calculate the distance P moves down the plane before coming to instantaneous rest. Calculate also the speed when AP=0.8m.


My Working for the first part

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I'll take A as 0 GPE.

At P
EPE=0, KE=0, GPE = 2g(0.6)\sin(45)

Let x be the distance at which P moves before coming to instantaneous rest.

After moving distance x:
KE=0 GPE = 2g(0.6 + x)\sin(45) ,EPE = \dfrac{20 \times x^2}{2(0.6)}

Also Work Done against friction,  W= 0.2(2g\cos(45))x

Energy is conserved so

 W = \Delta{GPE} + \Delta{EPE}


After this I get x=-0.665m, which is correct, except I can't explain why it is negative [should be positive], when my positive direction was down the plane?

To calculate the change in GPE & EPE I'm doing (Final EPE/GPE - Start EPE/GPE) - should it be the other way round?

Also, I have a similar problem with the second part:
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AP=0.8,

EPE = \dfrac{20 \times 0.2^2}{1.2}, GPE = 2g(0.8)\sin(45), KE = \dfrac{1}{2}(2)v^2= v^2

Work done,W, against friction in moving from AP=0.6 to 0.8:

 W = \dfrac{\sqrt{2}}{25}g

After rearranging I get v=1.63ms^-1 when the correct answer is 1.25.
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ThatPerson
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Anyone?
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TeeEm
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(Original post by ThatPerson)
Anyone?
apologies that I cannot help at the moment
(very busy)

look at questions 6, 10 and 11 in this link while you are waiting for help

http://madasmaths.com/archive/maths_...gs_springs.pdf
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ghostwalker
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(Original post by ThatPerson)
To calculate the change in GPE & EPE I'm doing (Final EPE/GPE - Start EPE/GPE) - should it be the other way round?
I think of it as:

Starting energy - work done by system = final energy.

Which rearranges to "Work done = starting energy - final energy".

I note that you have the GPE as positive, although P is below your zero level.
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ThatPerson
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(Original post by ghostwalker)
I think of it as:

Starting energy - work done by system = final energy.

Which rearranges to "Work done = starting energy - final energy".

I note that you have the GPE as positive, although P is below your zero level.
I was going to put GPE as -ve but then I thought that energy is a scalar so it doesn't matter?
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ghostwalker
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(Original post by ThatPerson)
I was going to put GPE as -ve but then I thought that energy is a scalar so it doesn't matter?
It matters.

In the case of GPE, it arises from work done against a force, in this case gravity. As the amount of work done increases - i.e. you raise the object higher - so its energy increases.
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simonli2575
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(Original post by ThatPerson)
I was going to put GPE as -ve but then I thought that energy is a scalar so it doesn't matter?
It is scalar, but the change in the value can be negative.
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ThatPerson
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(Original post by ghostwalker)
It matters.

In the case of GPE, it arises from work done against a force, in this case gravity. As the amount of work done increases - i.e. you raise the object higher - so its energy increases.
Ah - That gives me the +ve answer. Thanks.

Can you see any error in the second part? The way I've interpreted the question, is that the system is still released at AP=0.6, but I need to work out the speed after P has moved to AP=0.8?
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ghostwalker
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(Original post by ThatPerson)
Ah - That gives me the +ve answer. Thanks.

Can you see any error in the second part? The way I've interpreted the question, is that the system is still released at AP=0.6, but I need to work out the speed after P has moved to AP=0.8?
Not got time just now (bed) - will look tomorrow.
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ghostwalker
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(Original post by ThatPerson)

Also, I have a similar problem with the second part:
[spoiler]
AP=0.8,

EPE = \dfrac{20 \times 0.2^2}{1.2}, GPE = 2g(0.8)\sin(45), KE = \dfrac{1}{2}(2)v^2= v^2

Work done,W, against friction in moving from AP=0.6 to 0.8:

 W = \dfrac{\sqrt{2}}{25}g

After rearranging I get v=1.63ms^-1 when the correct answer is 1.25.
For the GPE, the distance moved down the plane is 0.2, not 0.8
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ThatPerson
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(Original post by ghostwalker)
For the GPE, the distance moved down the plane is 0.2, not 0.8
But A is 0 GPE, and P is below A, so when P has moved down the plane by 0.2, isn't it then 0.8m from A?
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ghostwalker
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(Original post by ThatPerson)
But A is 0 GPE, and P is below A, so when P has moved down the plane by 0.2, isn't it then 0.8m from A?
True; my mistake. I was assuming that you had the change in GPE there, since you'd only posted one, rather than the initial and final GPEs.

Again, this would be negative relative to A.
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ThatPerson
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(Original post by ghostwalker)
True; my mistake. I was assuming that you had the change in GPE there, since you'd only posted one, rather than the initial and final GPEs.

Again, this would be negative relative to A.
I just tried the calculation again with the -ve signs, but I'm getting 1.43, when the answer in the back is 1.25.

I'm equating energy at P with energy at AP=0.8 + Work Done

 \dfrac{-1.2g}{\sqrt{2}} = \dfrac{20(0.2)^2}{1.2} - \dfrac{1.6g}{\sqrt{2}} + v^2 + \dfrac{\sqrt{2}g}{25}.

Edit: Never mind, I was being stupid - forgot the g. Thanks for your help.
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ghostwalker
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(Original post by ThatPerson)
Edit: Never mind, I was being stupid - forgot the g. Thanks for your help.
I was just about to say was the missing final g a typo. Good that you spotted it yourself.
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