Two common sense questions (solutions provided)

[flyingpig]

Homework Statement

1) Please forget the concept of center of mass for now

[PLAIN]http://img97.imageshack.us/img97/1773/unledurk.png [Broken]

OKay my question is, what is the height of the square on the incline? Is it the green line or the blue line? If this was any physics problem, which is the height? ALl measurements are made with respect to the invisible ground which I forgot to draw and the blue line is not supposed to cross extend, I couldn't delete it without deleting the incline

2) The masses shown in Fig are connected by a
massless string over a frictionless, massless pulley
and are released from rest. Use energy conserva-
tion to find the maximum height
reached by the 4.0-kg mass,

[PLAIN]http://img716.imageshack.us/img716/2043/unledeqt.png [Broken]


Solutions given

The 4-kg mass, with an upward velocity of
5.17 m/s, will rise (as a projectile) an additional
v²/2g = (5.17 m/s)2/(2×9.8 m/s2) = 1.36 m to a
maximum height of 5 m + 1.36 m = 6.36 m off the
floor. (Note: the mechanical energy of both masses is
no longer conserved after the 7-kg mass hits the floor,
because a non-conservative contact force acts to stop
it.)

Question about solution

Does that mean when it reaches h = 5 and then h = 6.36 for brief moment and then it will come back down to h = 5 again for the m = 4kg block because the rope can't actually extend (not elastic)? Is the nonconservative force in the context meaning the heat produced (friction) between the contact forces?

Thank you

 

[Delphi51]

Technically the center of mass position should be used at the top and also at the bottom, but using the bottom of the mass in both cases would be exactly the same. Neither blue nor green is exactly right; half way between (average) would be perfect.

The non-conservative force is when the 7 kg block hits the ground and its remaining kinetic energy is lost (converted to heat, I suppose). Yes, the other block will end up at height 5.