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Re: Need soln explaination



Tony Wayne wrote:

I got a problem out of an older physics text (one without a solutions
manual). My students and spent a while debating the method of
attaining
the solution. Can someone explain how they would solve it -using
energy.

PROBLEM
Masses of 3 kg and 2 kg are hanging at opposite ends of a cord that
passes over a frictionless pulley. The system is held stationary for a

while, then released. What will be the speed of the masses when the 3
kg
mass has descended 0.5 m below its starting point? (USE ENERGY
RELATIONSHIPS TO SOLVE.)

I am a long time lurker (about 2 years) who will respond.

I will assume that the rotational inertial of the pulley is small so
that its rotational kinetic energy can be ignored. The principle I
would use is that the sum of the potential and kinetic energies of the
system is a constant. Before the objects are released I will assign
each object a potential energy of zero Joules. Before the objects are
released the kinetic energies are zero. After the 3 Kg object has
descended 0.5 meters each object has a nonzero potential energy and
kinetic energy. The 2 Kg object will have risen 0.5 meters and will
have a positive potential energy. The 3 Kg object will have a negative
potential energy. The two objects will have the same magnitude for
their velocities.

Before After
PE KE
PE KE
0 + 0 + 0 + 0 = 2*9.8*0.5 + 3*9.8*(-0.5) + 0.5*2*V*V +
0.5*3*V*V

Solving:
1 * 9.8 * 0.5 = 0.5 * 5 * V * V


9.8 = 5 * V * V

1.4 m/s = V

This number seems reasonable since an object which free-falls for 0.5
meters would have a velocity magnitude of 3.13 m/s. We expect to find a
value which is smaller than 3.13 m/s when a second mass is attached and
rising while the 3 Kg object is falling.

Greg Clements
Midland Lutheran College
Fremont, NE