Finding the Position and Velocity of a Mass on a Slope Using Work and Energy

[assaftolko]

Homework Statement

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.

Homework Equations

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?

The Attempt at a Solution

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 equation of the plain in the given coordinate system is: y=0.5x whice means: x=2y. If we put x=2y in the first equation 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!

 

[tiny-tim]

welcome to pf!

hi assaftolko! welcome to pf!

(btw, a plain is usually horizontal … you mean a plane )

… 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!

yes, there must be a misprint: the initial F is 20 N, but mg = 8 N

hmm … judging from answer א, the initial F needs to be 4 N,

so i'd suggest you try it with F = 2xyj ​

 

[assaftolko]

hi assaftolko! welcome to pf!

(btw, a plain is usually horizontal … you mean a plane )yes, there must be a misprint: the initial F is 20 N, but mg = 8 N

hmm … judging from answer א, the initial F needs to be 4 N,

so i'd suggest you try it with F = 2xyj ​

Yes that's exactly what they did! For some reason when they wrote down the equation F=mg for the disconnection point, they put 2xy=mg and not 10xy=mg. But the question says clearly that F=10xyj^! but something still doesn't make sense to me:

Logicly - I would expect that a force of 10xy would make the mass go up the plane and disconnect from it much more easily than a force of 2xy... So I really don't see how it makes sense that for a force of 2xy the mass disconnects at a higher point than it started, but for a bigger force of 10xy it disconnects lower than the point it started?

 

[tiny-tim]

Logicly - I would expect that a force of 10xy would make the mass go up the plane and disconnect from it much more easily than a force of 2xy... So I really don't see how it makes sense that for a force of 2xy the mass disconnects at a higher point than it started, but for a bigger force of 10xy it disconnects lower than the point it started?

i don't understand your objection …

the greater force field, 10xy, is greater at each point

so in particular it is greater at (2,1), and will lift the mass immediately …

what's wrong with that? ​

 

[assaftolko]

i don't understand your objection …

the greater force field, 10xy, is greater at each point

so in particular it is greater at (2,1), and will lift the mass immediately …

what's wrong with that? ​

Because that's not what happens... The mass first drops down the plane and only after it drops some distance does the force F is big enough to overcome gravity? How does that make sense considering the force F gets smaller as the mass drops down the plane and gravity doesn't change during this time? Is it maybe because the spring applies force in the upward x' axis as it streaches while the mass moves downwards?

Sorry if I seem to be nagging I just really don't understand why is it possible that the mass will go downward before disconnecting...

 

[tiny-tim]

Because that's not what happens... The mass first drops down the plane …

no it doesn't!

it lifts off immediately! (because F > mg) ​

 

[assaftolko]

no it doesn't!

it lifts off immediately! (because F > mg) ​

Well I see your point but how do you explain what I got in my inital calculations? I wrote down my calculations In the first message, and showed how the y coordinate at the disconnection point is 0.626 m which is smaller than 1 m - the inital y coordinate of the mass. If you'll calculate the x coordinate for the disconnection point you'll see that it's also smaller than the inital x coordinate. Why are my calculations wrong?

 

[tiny-tim]

but the equation you were using was really an inequation (an inequality) …

it was satisfied by (2,1)

 

[assaftolko]

but the equation you were using was really an inequation (an inequality) …

it was satisfied by (2,1)

Ha? I'm not a native English speaker so I don't really know what you meant

 

[tiny-tim]

i'm saying that this isn't correct …

Fcos(q)-mgcos(q)+N=0 (there's no movment in the y' axis)
Because at the disconnection point N=0

… because, at the disconnection point, Fcos(q)-mgcos(q) can be greater than 0

you can only rely on v_y' being 0 if the mass is constrained to stay on the plane …

it isn't! ​

 

[assaftolko]

i'm saying that this isn't correct …


… because, at the disconnection point, Fcos(q)-mgcos(q) can be greater than 0

you can only rely on v_y' being 0 if the mass is constrained to stay on the plane …

it isn't! ​

Got you, thanks mate!

 

[assaftolko]

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?

 

[tiny-tim]

your equations are very difficult to follow ,

but if you're correct, and v² is negative at the point where it would lift off,

then that means that the mass never gets that high​

 

[assaftolko]

your equations are very difficult to follow ,

but if you're correct, and v² is negative at the point where it would lift off,

then that means that the mass never gets that high​

Sorry, I tried to be as clear as possible.

I don't understand what does it mean the mass never gets that high... If the force F is 2xyj^ then don't we agree the mass goes up the plane and disconnects at (2.8,1.4) m? I mean why shouldn't it? The calculations I did in the first question are correct for a force of 2xyj^ are they not?

 

[tiny-tim]

I don't understand what does it mean the mass never gets that high... If the force F is 2xyj^ then don't we agree the mass goes up the plane and disconnects at (2.8,1.4) m? I mean why shouldn't it?

there are two equations, one for x' and one for y'

the y' equation tells us where the mass would lift off (if it does) however strong the spring is (or indeed whether it's above or below the mass)

the y' equation doesn't include the spring at all

it certainly doesn't tell us anything about how far the spring allow the mass to climb

 

[assaftolko]

there are two equations, one for x' and one for y'

the y' equation tells us where the mass would lift off (if it does) however strong the spring is (or indeed whether it's above or below the mass)

the y' equation doesn't include the spring at all

it certainly doesn't tell us anything about how far the spring allow the mass to climb

Hmm that's interesting... I have to say I didn't think about that at all... this is very confusing and worrying... how could I know how much the spring really contracts?

 

[tiny-tim]

how could I know how much the spring really contracts?

you use the energy equation to find when v = 0 (assuming the mass stays on the slope)

 

[assaftolko]

you use the energy equation to find when v = 0 (assuming the mass stays on the slope)

Yeah but how can I know what is the work the force F do until that point where v=0? Clearly the force didn't act on the mass all the way to y=1.4 m...