# Help with simple work / nrg conservation question

1. Nov 15, 2004

### Mr. Hiyasaki

The question is
"Two masses comprise an inertial system: the first, a 2 Kg mass which rests on a rough table top; then second, 5 Kg mass which hangs freely 0.4 m above the floor from an ideal pulley at the edge of the table top. The two masses are connected by an inextensible string run over the pulley. The coefficient of kinetic friction between the table top and the first mass is .40."

first I tried
$$m_{1} = 2 kg$$
$$T-F_{k}=m_{1}a \ where\ F_{k}= \mu N \ and \ N = m_{1}g$$
so
$$T=m_{1}a+ \mu m{1}g$$

Then mass 2
$$m_{2} = 5 kg$$

$$F= MA$$
$$m_{2}g -T = m_{2}a$$

substituting T for what T = in part 1 gives me
$$m_{2}g -m_{1}a - \mu m{1}g = m_{2}a$$

so a = $$(m_{2}g - \mu m_{1}g)/(m_{1} + m_{2})$$

pluging in 5 kg for m2 98. m/s^2 for g .4 for mu and 2 kg for m1 gives me an acceleration of 5.88 m/S^2

then to solve for velocity I would go $$W_{nc} = E_{f} - E_{i}$$
so $$-\mu m_{1}gd = 1/2m_{1}v^2 + 1/2 m_{2}v^2-m_{2}gd$$

then solving using the same numbers above and using .4 for d v comes out to be 2.55 m/s

could I also solve for T in equation 1 to get a conservative work force of 19.6 J?

someone also showed me that

delta K = work total = work gravity + work friction
which is 1/2(m2 + m1)(v^2 -0) = -m2g(-d -0) + (- mu m1g)d
then it comes down to v = $$\sqrt{\frac{2gd(m_{2}- \mu m_{1}}{m_{2}+m_{1}}}$$

but that gives me a v = 4.704 m/s which is not 2.55 m/s

and someone told me the anwser to the velocity component is 2.88 m/s

I am confused

moreover, how could i figure out how far it traveled once the free hanging block hit the ground?

Last edited: Nov 15, 2004
2. Nov 15, 2004

### Diane_

I presume you're trying to figure the acceleration of the system in your first part.

I get the same acceleration for the system, 5.88 m/s^2. (Note - in your text, you mention m1 as 3 kg, but you seem to have used the proper number in your calculations.)

In your energy calculations, though, I'm a little confused by your signs. Try it this way: think of the energy in the system as "gozinta's" and "gozouta's". Energy comes into the system due to the change in potential energy of block 1, at least until it hits the ground. Energy goes "out" of the system, appearing as kinetic energy in the blocks and work done by friction. So:

change in PE for block 1 = change in KE for block 1 + change in KE for block 2 - work by friction

m1 gd = 1/2m1v^2 + 1/2 m2v^2 - (mu)m1gd

Note a simplification:

m1 gd = 1/2(m1 + m2)v^2 - (mu)m1 gd

In other words, it's the total kinetic energy of block 1 and block 2 that matters - they act as a unit. This should make a lot of sense.

Solving for v, we get

v = sqrt(2gd(m1 + (mu)m2)/(m1+m2))

Note the difference in sign from your version.

This results in a velocity of 2.55 m/s, which you indicate is the correct answer.

For the rest of it, find the kinetic energy of the block on the table when the other one hits the floor and figure out how far it would have to slide for friction to do exactly that amount of work on it.

3. Nov 15, 2004

### Mr. Hiyasaki

i was checking over this and noticed that

$$v_{f}^2 = v_{o}^2 + 2a delta x$$
2.55(m/s)^2 = 0 + 2 (5.88m/s^2) (.4m)
is not valid

so does that mean it is wrong? or am i just confused?