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Old Jul 23rd 2012, 01:56 AM   #1
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Work & energy problem

A mass of 800 gr is attached to a spring with a spring constant k=10 N/m, and is moving on a steep plain as described in the picture. A force of F(r)=10xyj^ N (meaning an upward vertical force) is applied on the mass in the given coordinate system. At the moment of t=0 the mass position is r0=(2,1) m and it starts to move from rest. At that moment the spring is neither extended nor it's compressed.

1. What is the position of the mass just before it disconects from the plain?
2. What is the velocity of the mass at that moment?

Well, I chose a coordinate system parallel to the plain. It's obvious that the mass climbs up the plain until the moment it disconnects because the force F increasses as the x and y coordinates get bigger in the given coordinate system. So - if I write down the forces in my new coordinate system where the x' axis is parallel to the plain, you get that in the y' axis (and let's call the angle of the plain with the given x axis as q):

Fcos(q)-mgcos(q)+N=0 (there's no movment in the y' axis)
Because at the disconnection point N=0 we get: Fcos(q)=mgcos(q)
and so: F=mg at the disconnection point, and if we put the size of F we get: 10xy=mg
Also - the equasion of the plain in the given coordinate system is: y=0.5x whice means: x=2y. If we put x=2y in the first equasion we get: 10*2y*y=mg -> 20y^2=mg -> y= 0.626 m

But this disconection coordinate of y is lower than the start coordinate of y! So something here doesn't make sense... I'd expect that the disconnection coordinate is bigger since the mass climbs up the plain!
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Old Jul 23rd 2012, 11:26 AM   #2
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I think your analysis is correct. At the initial application of force F if the object is at position (2,1) then the value of F is 20N. The angle of the incline is arctan(1/2) = 26.6 degrees. The component of F in the y' direction is 20cos(26.66) = 17.9N, which is more than enough to lift the 800g block off the incline. So I would say that the point where the block lifts off is at (2,1).

Last edited by ChipB; Jul 23rd 2012 at 12:40 PM.
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Old Jul 23rd 2012, 01:14 PM   #3
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Originally Posted by ChipB View Post
I think your analysis is correct. At the initial application of force F if the object is at position (2,1) then the value of F is 20N. The angle of the incline is arctan(1/2) = 26.6 degrees. The component of F in the y' direction is 20cos(26.66) = 17.9N, which is more than enough to lift the 800g block off the incline. So I would say that the point where the block lifts off is at (2,1).
Ok thanks a lot!
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Old Jul 24th 2012, 01:11 AM   #4
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Well something still doesn't add up, because if we assume that the force is 2xyj^, when I tried to solve the second question I got that v is the square root of a negative number:

I used the work and energy therom: The work of all non conservative forces equal to the difference in energy. If we take point A to be the point of t=0, and point B to be the disconnection point, and let the height of point A be the referrence height of 0, we get:
EA = 0
EB = 0.5mv^2 + 0.5kL^2 + 0.5mghB
The height hB is the difference between the y coordinate of point B which is 1.4, and the y coordinate of point A which is 1, so in total hB=0.4 m
the length L is the hypotenuse of the triangle with catheti of 0.8 and 0.4 m, so the length L^2 = 0.4^2+0.8^2 = 0.8 m^2
So EB is: 0.5*0.8*v^2+0.5*10*0.8+0.8*9.8*0.4 = 0.4v^2+7.136 J
The only non conservative force that does work along the path from A to B is the given F, so its work is:

S (0,2xy)dot(dx,dy) from (2,1) to (2.8,1.4) - (2.8,1.4) are the coordinates of x and y at the disconnection point.
S 2xydy from 1 to 1.4 -> We remember that x=2y so we get: S 4y^2dy from 1 to 1.4 -> 2.325 J

And finally we get:
2.325 = 0.4v^2+7.136 -> v=sqrt(-12.0275)

So what's next?

Last edited by assaftolko; Jul 24th 2012 at 01:21 AM.
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Old Jul 24th 2012, 12:45 PM   #5
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Originally Posted by assaftolko View Post
if we assume that the force is 2xyj^
Before you said the force is F=10xy. If the force is only 2xy then at coordinates (2,1) the force is 4N, which is not sufficient to move the 0.8Kg mass against graviity. In fact the mass will slide down hill until the spring extends enough so that its upward force together with F can counteract the weight of the mass. That's why you're getting an imaginary solution for v.

Originally Posted by assaftolko View Post
EB = 0.5mv^2 + 0.5kL^2 + 0.5mghB
Why the 0.5 factor in front of mgh?

Originally Posted by assaftolko View Post
The height hB is the difference between the y coordinate of point B which is 1.4,
How did you arrive at point B being at y=1.4?

Originally Posted by assaftolko View Post
work is:
S (0,2xy)dot(dx,dy) from (2,1) to (2.8,1.4) - (2.8,1.4) are the coordinates of x and y at the disconnection point.
S 2xydy from 1 to 1.4 -> We remember that x=2y so we get: S 4y^2dy from 1 to 1.4 -> 2.325 J
Again, if you make F=10xy instead of F= 2xy it should work out.
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Old Jul 24th 2012, 01:05 PM   #6
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Originally Posted by ChipB View Post
Before you said the force is F=10xy. If the force is only 2xy then at coordinates (2,1) the force is 4N, which is not sufficient to move the 0.8Kg mass against graviity. In fact the mass will slide down hill until the spring extends enough so that its upward force together with F can counteract the weight of the mass. That's why you're getting an imaginary solution for v.



Why the 0.5 factor in front of mgh?

My mistake



How did you arrive at point B being at y=1.4?

If we say that the force F is 2xyj^ then you get that in the y' axis when N=0 (disconnection point) F=mg which means 2xy=mg and because y=0.5x (the equation of the plane) you get x=2.8 m and y=1.4 m at disconnection point


Again, if you make F=10xy instead of F= 2xy it should work out.

But if F=10xy the mass should disconnect at the start point of (2,1) m.
So if F=10xy the mass should disconnect at the start point and if F=2xy the mass actually slides down... I stopped understanding this question...
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Old Jul 24th 2012, 03:57 PM   #7
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The question makes little sense to me as well. If F at (2,1) is less than mg the mass slides backwards, and if F at (2,1) is greater than mg then the mass lifts off right there. Are you absolutely sure that F = 2xy?
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Old Jul 24th 2012, 09:02 PM   #8
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Originally Posted by ChipB View Post
The question makes little sense to me as well. If F at (2,1) is less than mg the mass slides backwards, and if F at (2,1) is greater than mg then the mass lifts off right there. Are you absolutely sure that F = 2xy?
No, in the original question F=10xyj^ but in the official solution when they build the equation of the y' axis forces: N-mgcos(q)+Fcos(q)=0 (q is the angle of the plane, and there's no acceleration in the y' direction), and want to describe the disconnection point by putting N=0, for some reason they decide to put F=2xy so that you get 2xy=mg at the disconnection point. Why they put 2xy and not 10xy is beyond me...

And what's really strange is that when they solve the second question and they calculate the work done by F from (2,1) to (2.8,1.4) they decide that now in the integral F=10xyj^ back again.
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