1. The problem statement, all variables and given/known data 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. 2. Relevant 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? 3. 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 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!