Calculating Maximum Height of a Hanging Block on a Pulley System

[jphillip]

two blocks (2.2kg and 3.2kg) both hanging 1.80 m from the ground on a pulley(massless / frictionless) that is 4.80 m from the ground.

What is the maximum height the smaller block (2.2kg) reaches, when the system is released?

It appears that I need to find its velocity first of the system as the 3.2kg falls 1.8 m to the ground.
(3.2kg-2.2kg) / (3.2kg+2.2kg) = .185g * (9.8m/s²) = 1.81 m/s²

Then once the big block hits the ground, calc the additional height of the little block will rise additionally from the velocity created.

v2 - v²_i / 2 (g)
1.81² / 2(9.8 m/s²) = .168 m + 1.8m (initial drop / increase)

Is this correct?

 

[tiny-tim]

welcome to pf!

hi jphillip! welcome to pf!

(3.2kg-2.2kg) / (3.2kg+2.2kg) = .185g * (9.8m/s²) = 1.81 m/s²

where did you get this formula from?

(it's only partly correct )

(your second formula is fine)

 

[jphillip]

I got the first forumla from an example in my book that appeared to be calculating what I thought I needed. An elevator question, what did I miss or am I missing

 

[tiny-tim]

thought so!

always work out the formula yourself …

what should it be? ​

 

[jphillip]

I believe F = ma which worked out a = F/m
So
3.2kg - 2.2kg = 1kg * 9.8 = F
(3.2 + 2.2) = m ? (not sure here)

I think

 

[tiny-tim]

I believe F = ma …

well, that certainly would work, but you want v not a …

so it's quicker to use conservation of energy ​

 

[jphillip]

O.K.

V² = 2(1.81m/s²)(1.80m)

 

[tiny-tim]

O.K.

V² = 2(1.81m/s²)(1.80m)

yes!

 

[jphillip]

So that gets me the velocity² I would then have to solve for the additional distance correct

 

[tiny-tim]

yes

(same method as you used at the start)

 

[jphillip]

\sqrt{2*1.81*1.80} = v = 2.55

2.55² / (2*9.8m/s²) = y = .332 + 3.6 (1.80(starting height)+1.80 accl height)

total height 3.93m

 

[tiny-tim]

i'm finding it difficult to follow this without any formulas

(and why did you bother to square-root that, when you were only going to square it again immediately after? )

 

[jphillip]

Here are the formulas that I have used... initial height = 1.80m
N - Newtons
kg - kilograms
(Block 1 31.31N - Block 2 21.56N) / (block 1 3.3kg + block 2 2.2kg) = a

So
(31.36N - 21.56N) / (3.2kg + 2.2kg) = 1.8148 m/s²

Then solve for Velocity using v² = v²_i + 2a(y)

V²= 0 + 2(1.8148m/s²)(1.80) = 6.5338 which is velocity not squared

Now solve for the actual distance using the same formula on
v = 6.5338 (not squared from above)
a = 9.8 gravity
v / 2(a) = y

6.5338 / 2(9.8) = .332 which should be the height of the lighter block 2.2kg with momentum after the 3.2 kg block hits the ground.

 

[tiny-tim]

hi jphillip!

(just got up :zzz: …)

i think you have the correct result, but what do you mean by "not squared"?

that 6.53338 is v²

have you done conservation of energy in class (KE + PE = constant)?

you can avoid finding the acceleration by using mgh - Mgh = (m + M)v²/2
​

 

[jphillip]

I have not made it to conservation of energy in Class. Thanks for the help I am sure I will be back for more. I am really enjoying this class, just not the summer version of I believe LOL.

 

[tiny-tim]

I have not made it to conservation of energy in Class.

ok, that's something to look forward to!

just to recap, your equations (without conservation of energy) should have been

(F = ma): mg - T = ma, Mg - T = -Ma, so (subtracting) (m-M)g = (m+M)a, or a = (m-M)/(m+M)

v² = 2as₁

v² = 2gs₂