# Homework Help: Redemption - I Challenge Dextercioby

1. Jan 7, 2005

### apchemstudent

Please DON'T delete this thread. I want to make a point here...
I want other experts in physics to determine who’s wrong in this situation…

This was the problem taken from:

“A 5 kg block is fastened to a vertical spring that has a spring constant of 1000 newtons per meter. A 3 kg block rests on top of the 5 kg block. The blocks are pushed down and released so that they oscillate.

Determine the magnitude of the max acceleration that the blocks can attain and still remain in contact at all times.”

Dextercioby agrees that the answer should be 9.8 m/s^2.

(I know I did something wrong in the other posts, but I’ve fixed it up)

Us + Up = Us(o) + Up(o)

kx^2/2 + 0 = mg2(x-.078) + 0 <- ------ It’s 2 times (x-0.078) due to the oscillation and 0.078 is the equilibrium point.

500x^2 – 156.8 x + 12.2304 = 0

x = 0.168 or x = 0.1456, but it’s asking for max acceleration so x = 0.168

Fs = 1000*0.168 = 168

Fnet = Fs – Fg
= 168 – 8*9.8
= 89.6 N

Amax = 89.6/8 = 11.2 m/s^2

I am pretty sure this is the correct answer. Everyone other then Dextercioby, please check if my method is right or wrong and can you explain why it might be wrong or right... thnx

Last edited: Jan 7, 2005
2. Jan 8, 2005

### Andrew Mason

At first glance I would say that the answer has to be g. The downward acceleration on the top mass cannot exceed g. The acceleration of the lower mass is due to the spring extension/compression from the equilibrium point, 200x-8g/8, and can be greater than g. If the downward acceleration on the lower mass (ie. it is max. at the top) is greater than g at ANY point, the bottom one accelerates away from the top one. If it is any less, they stay together.

AM

Last edited: Jan 8, 2005
3. Jan 8, 2005

### Gokul43201

Staff Emeritus
I second AM.

4. Jan 8, 2005

### Staff: Mentor

I agree with AM and Gokul43201.

Since the top mass only rests upon the bottom mass (it's not attached), the greatest downward acceleration it can have is g.

The more interesting question is how far down can you push the system so that it oscillates without the top mass losing contact. (My guess is that this is the real question. Else why give all the extraneous information?)

As long as the acceleration never exceeds g the two masses stay together and can be treated as a single system. The angular frequency is $\omega = \sqrt{k/(M + m)}$, where M and m are the masses of the bottom and top masses. Thus the maximum acceleration in terms of the amplitude (A) of the SHM is: $a_{max} = A\omega^2$. Set that equal to g to find that $A = (M + m)g/k$.

Another way to look at this that will make that answer "obvious" is this: Call the unstretched postion of the spring y = 0. The equilibrium position is where the net force on the masses equals zero, or $k \Delta y = (M + m)g$, thus $y_{eq} = -(M + m)g/k$. The system will oscillate about that equilibrium point. Note that this means that the maximum amplitude (calculated above) is just enough to bring the system up to the y = 0 point (the unstretched position). This should make sense, since the only time the masses could have an acceleration of g downward is when the only force acting on them is gravity--and that's only true when the spring is unstretched.

5. Jan 8, 2005

### marlon

I also agree with AM, Gokul and Doc Al...

marlon

6. Jan 8, 2005

### Sirus

Believe it or not, I also agree with AM, Gokul and Doc Al...

7. Jan 8, 2005

### apchemstudent

boohoo ... looks like dextercioby won once again... heh it was fun while it lasted...

8. Jan 8, 2005

### Gokul43201

Staff Emeritus
Sirus, you forgot to include marlon ...do you own a horse ? :yuck:

9. Jan 8, 2005

### Sirus

That's because marlon said "I also agree with AM, Gokul and Doc Al...", so it would be pointless for me to say that I agree that marlon agrees with those posters.

um...what?

10. Jan 17, 2005

### Gokul43201

Staff Emeritus
Think 'Godfather'...

11. Jan 17, 2005

### Curious3141

If in doubt, just draw a force diagram. Only two forces act on the 3 kg block, its weight (downward) and the normal reaction force (upward). The resultant of these two forces results in the block's acceleration.

So we have $$F_N - mg = ma$$ When the blocks just lose contact, $$F_N = 0$$ giving $$a = -g$$ (meaning an acceleration of $g$ in a downward direction. Quite simple, right ?