# Energy conservation( system of masses)

1. Oct 31, 2011

### murcy

Problem and Data
three masses conected m1= 800g, m2 =1100g m3=1200g. m2 rest on a table of coefficient μ=0.345. m1 and m3 hang vertically over frictionless/massless pulleys. The system is released from rest. Whats the velocity of m3 after falling 60.0cm

P.S "i know this system can be solve with finding tension and such, but i have to use energy conservation method."

2. Relevant equations

W(work)= ΔK+ΔUg(potential gravitational)+ΔUs(Spring energy)

3. The attempt at a solution

The velocity for m3 would be the same for all three masses because they are connected by a string. There's no k (initial) and spring energy involved.
This is what i have gotten so far;

I made a system of equations for each body:
Δy i defined as the change of height for potential energy.

m1;
W= (1/2)m1v^2+ m1gΔy

m2;
W= (1/2)m2v^2

m3;
W=(1/2)m3v^2-m3gΔy (the minus from negative displacement in height(Δy))

Im not sure about the following, i think im missing something;
"
*I also know that the system is losing energy by the interaction of M2 w/ the surface of the table...friction.
so friction(f)= N(normal)*μ(K).
In other words: f=m2*g*μ
and Work is also =f*r(displacement by the force)* cos (θ)
θ being the angle between the force and the displacement."
I'm just not sure how to related the energy lost W on all three bodies

Thanks for any input or help you guys can provide on my problem.

Last edited: Oct 31, 2011
2. Oct 31, 2011

### PhanthomJay

I think you have 3 masses total, M1, M2, and M3 (you shouldn't change from lower to upper case letters).
Each has the same velocity and each move the same distance.
The equation you have written for Work is for work done by non conservative forces (the work done by friction on M2 in this case). That's the part that goes on the left side of the equation. The change in KE and the change in PE go on the right side, as you have indicated in your relevant equation. Watch signage, and please show your work in arriving at a solution. And welcome to PF!

3. Oct 31, 2011

### murcy

Hi and thanks :),

sorry about the upper case M.(changed).

ok so the work on m2 would be;

w= (1/2)m2v^2
frcosθ= (1/2)m2v^2
N*μ(-1)=(1/2)m2v^2
-mg*μ=(1/2)m2v^2
-(1.1kg)(9.8m/s^2)*(0.345)=(1/2)(1.1kg)v^2
solve for V;
-3.719N=0.55kg v^2
v^2=-3.719N/0.55kg
v^2=-6.762m^2/s^2
v=sqrt(-6.762 m^2/s^2)......i know im wrong here since ill have a neg inside a sqrt :(, but if somehow the inside is pos ill have v= 2.60m/s...
im sure this is not right :/

Last edited: Oct 31, 2011
4. Oct 31, 2011

### murcy

I'm sorry but im still stuck in this problem :/

5. Oct 31, 2011

### PhanthomJay

You can't just isolate one block to determine its speed unless you include the work by the tension forces as well. Instead, look at the entire system of three blocks. Use your original equation, which is
$W_{friction} = (KE_{final} - KE _{initial}) + (PE_{final} - PE _{initial})$, where the initial and final KE's and PE's of each block must be determined and included in this equation.

6. Oct 31, 2011

### murcy

I understand what you mean now, this is what i have so far i hope is right;

Wfriction= (1/2)(m1+m2+m3)v^2+(m1-m3)gΔy

Wfriction=(1/2)(0.8kg+1.1kg+1.2kg)v^2+(0.8kg-1.2kg)(9.8m/s^2)(0.6m)

Wfriction=1.55kg*v^2-2.352joules

earlier i found;

Wfriction=fdcosθ and f=μk*N(m2g)

Wfriction=0.345*1.1kg*9.8m/s^2*0.6m*-1

Wfriction=-2.23 joules

so;

-2.23j=1.55kg*v^2-2.352j
i ended up with v= sqrt(0.07871m^2/s^2)
v=0.2806m/s i wish i could verify this answer but looks reasonable

7. Nov 1, 2011

### PhanthomJay

yes
Yeah, looks OK!

8. Nov 1, 2011

### murcy

Thanks a bunch Jay~