=3D=3D>PROBLEM
=3D=3D>Masses of 3 kg and 2 kg are hanging at opposite ends of a cord tha=
t
=3D=3D>passes over a frictionless pulley. The system is held stationary f=
or a
=3D=3D>while, then released. What will be the speed of the masses when th=
e 3 kg
=3D=3D>mass has descended 0.5 m below its starting point? (USE ENERGY
=3D=3D>RELATIONSHIPS TO SOLVE.)
Oh! This has a right answer. All others are wrong. A good =
opportunity to embarrass myself in front of my peers. =
Here goes.
I assume that the cord is not stretched at any time during the =
experiment, i.e. there is no Hooke's Law potential energy =
stored in it. I use KE for kinetic energy & PE for gravitational =
potential energy. I use g =3D 10m/s^2 for simplicity. I ignore =
air friction. I use three spaces to indicate multiplication.
"^2" means "squared."
At the start the experiment KE =3D 0, I assign PE =3D 0.
At the end of the experiment the PE =3D mgh
=3D 3kg 10m/s^2 (-0.5m) =
+2kg 10m/s^2 (+0.5m) =
=3D -5 J
Since the 2kg mass rose it gained PE & vice versa.
But energy is conserved, so there must be an INcrease =