# Homework Help: Spring with pulley system

1. Nov 11, 2012

### eonden

1. The problem statement, all variables and given/known data
A block of 4 kg hangs from an ideal string which passes by a pulley and is attached at the other end to a block of 6kg, which is in turn lying on a rough table of friction coefficion μ=0.2. The block on the table is pushed towards a spring with elastic constant k = 60 N/m, compressing it 30cm, and ten released. Find out the speed of the blocks when the hanging one has descended 40cm.

2. Relevant equations
W (of non-conservative forces) = difference of the mechanical force
Potencial elastic energy = (k*x^2)/2

3. The attempt at a solution
Let us put the height "0" where the 4kg mass hangs. We will therefore have an ecuation such as:
Em1 (when the spring is compressed) = Em2 (when the mass has already fallen 0,4m) +WFF.
in such a way that
K/2 * L^2 = ((m1+m2)*v^2)/2 + mgh + FF * h (as the distance fallen will be the same the 6kg block passes in the rough surface)
as the height is negative (we consider the point 0 as the original position):
(m1+m2)*v^2 = K*L^2 +2mgh -2*FF*h
such that:
10*v^2 = 60*(0.3)^2 + 2*4*9.8*0.4 - 2*0.2*6*9.8*0.4
The problem is that the solution shall be 0.57 m/s.
Where is my mistake? Can someone explain me?

2. Nov 11, 2012

### Spinnor

I did your problem and assumed there was no spring and got a speed of 1.48m/s. The addition of the spring should make the speed greater. If all the information you gave is correct I am stopped.

3. Nov 11, 2012

### haruspex

You need to separate the movement into two stages. After 30cm, the spring will no longer be in contact with the block. It's not attached.

4. Nov 11, 2012

### Simon Bridge

Welcome to PF;
So far this sounds like the standard "mass hanging off the end of a table" situation - all the pulley does is change the direction of the motion.
So - when the spring is released, the resulting motion has the table-block move towards the pulley and the hanging block goes down?

I must ask: when the hanging one has decended 40mc from where? Equilibrium, or from the initial position?

... for the static problem or the compressed one? It makes a difference.
... so $m_1$ is on the table and $m_2$ is hanging...
... when you "prime" the rig, you are storing energy in the spring, and as gravitational PE in lifting $m_2$. When the spring is released, this energy goes into kinetic for both blocks and work against friction for $m_1$. How did you account for the situation that h>L (and the spring stores energy again)?
But when h > L in the above derivation, then the spring will be extended - slowing the speed.

I'd have picked x=0 for the mass positions in the static problem (i.e. before compressing the spring) and chosen $+x_2$ = down for $m_2$. That way a displacement from this equilibrium of x will be the same for both masses. But I can do it the above way too.

Last edited: Nov 11, 2012
5. Nov 11, 2012

### Simon Bridge

I don't see that in the description ... but then: I don't see anything saying the spring is attached either!

If the block is attached to the spring, then the spring will extend - taking care of Spinnor's objection that the model answer is slower than what you'd get without the spring.

6. Nov 11, 2012

### haruspex

"The block on the table is pushed towards a spring ..."
I read that as implying it is not attached.

7. Nov 12, 2012

### eonden

Yes, I also consider that the mass is not attached to the spring as you push the mass towards it.
Harupex, you are right, that is why I only consider the spring in the first stage, as it's pushing forwards.
Simon, I consider the height 0 when the spring is compressed, so that the potential energy in the moment 1 is 0.

8. Nov 12, 2012

### Simon Bridge

@eonden: in which case, the spring releases all it's energy to the motion.
want to check:
Code (Text):
octave:102> m1=6;
octave:103> m2=4;
octave:104> f=0.2;
octave:105> k=60;
octave:106> L=0.3;
octave:107> h=0.4;
octave:108> g=9.8;
octave:109> v=sqrt( (2*m2*g*h + k*L^2 - 2*f*m1*g*h)/(m1+m2) )
v =  1.6538
octave:110> v=sqrt( (2*m2*g*h + k*L^2 - 2*f*m1*g*h-k*(h-L)^2)/(m1+m2) )
v =  1.6356
... doing it both ways.
We see that allowing the spring to be attached does not account for the lower speed in the model answer: it only removes 0.3J from the energy.

This suggests that either the model answer is incorrect or there is a factor of 3 missing in the denominator (or something else going on that has yet to be mentioned). Did we miss a x3 or a x0.3 someplace?

9. Nov 12, 2012

### haruspex

I agree, the answer must be at least 1.6. Even if we remove the spring it gives about 1.5.
It did occur to me that we're not told the string is horizontal from pulley to block on table, but if it's not then we're missing other vital info.
Btw, in a proper general treatment, it would be important to check that the spring is not so powerful as to make the string go slack initially (it isn't). Were it so there'd be an impulse to consider.

10. Nov 12, 2012

### Simon Bridge

I suppose we could work it backwards and find out what the energy discrepancy is ... but it is probably faster for OP to take this result to others doing the same problem - and the teacher.