Hi mates, I have problems solving the third part of this exercise, I've already done all the previous calculations.
Given the following circuit, where the switch S is open, the power supply = 50 volts and:
The initial charge in the C capacitor: QC = 0 coulombs
I mean, the 0,15 meters is what the 2nd box raises by the action of the spring and the 0.52 is what the 2nd box raises by releasing the system. other value for velocity I calculated by making an energy balance from the height -0,2m (before releasing the system to the higest ponit the second box...
I redid it. Now what I've done is calculate what heigt the 2nd box will reach (gave equals to 0.15 meters (the height obtained was the same using 2 different reference systems).
Then calculated de acceleration of the system via free-body diagrams (Acceleration= 11,29m/s^2)
The sexond box is compressing the spring before the system is released.
The first box descends o,6 meters parallel to the plane.
But the descending of the first box (the left one) means the rising of the second one (the one tied directly to the spring).
Whats is the final speed of the second box(mass = 1kg) when the first one (mass = 5 kg) has descended a distance equal to 0.6 meters in the rough ramp inclined 60° respect the x axis. Consider the spring is compressed a length x = 0.2 meters. The second box is tied to the...