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Need soln explaination



=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 =

in the KE of the system of 5 J. =

KE =3D 5 J
=3D 1/2 3kg v^2
+ 1/2 2kg v^2
=3D 5/2 v^2

So 5J =3D 5/2kg v^2
v =3D 1.4 m/s

Notice that the velocities are in opposite directions, but since =

KE is a scalar (or because there's no such thing as negative KE), =

this does not matter, you just add them.

I'd guess a large fraction of us think we can solve this.
There will be lots of answers. For the sake of our egos, I hope
they are all the same.

Cheers,
Bill (no tests left to write or give, 5 left to grade) Larson
Geneva