Conservation Of Mechanical Energy

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Homework Help Overview

The problem involves two blocks of different masses connected by a rope over a frictionless pulley. The scenario describes the motion of the blocks after being released from rest, with one block descending and the other ascending. The task is to determine the mass of the more massive block given specific parameters such as distance fallen and speed attained.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of mechanical energy, questioning the initial conditions and the setup of potential and kinetic energy equations. There is confusion regarding the signs used in the equations and the interpretation of energy terms.

Discussion Status

Participants are actively engaging with the problem, offering corrections and alternative perspectives on the equations used. Some guidance has been provided regarding the correct formulation of kinetic and potential energy, but no consensus has been reached on the correct approach to find the mass of the more massive block.

Contextual Notes

There is an ongoing discussion about the initial mechanical energy being considered as zero, which some participants question. The problem also involves assumptions about the direction of motion and the signs of the variables in the equations.

tizzful
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Homework Statement


Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.30 , its speed is 1.00 .
If the total mass of the two blocks is 18.0 , what is the mass of the more massive block?
Take free fall acceleration to be 9.80 .
I set the heavier block as m1.

Homework Equations


m1gh1+0.5m1v1^2=m2gh2+0.5m2v2^2

The Attempt at a Solution


They both start at height 0 and velocity 0 and so the initial PE and KE is going to be 0, and so the initial Mechanical energy is also 0 (I'm pretty sure this is wrong but don't know how to fix it). Then m1 drops 1.30m so that's h1 and m2 goes up -1.30m=h2. v should be equal between both, m1=1, m2=-1.

m1(gh1+0.5v^2)=m2(gh2+0.5v^2)
m1/m2=(gh2+0.5v^2)/(gh1+0.5v^2)
=(9.8*-1.30+1/2*-1^2)/(9.8*1.30+1/2*1^2)
Therefore
m1=-m2

So its wrong ahah I was wondering if someone could help me?
Thanks in advance!:shy:
 
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tizzful said:
m1/m2=(gh2+0.5v^2)/(gh1+0.5v^2)
=(9.8*-1.30+1/2*-1^2)/(9.8*1.30+1/2*1^2)

Hi tizzful! :smile:

erm … it's not "-1^2" … :redface:

No wonder they came out minus each other! :rolleyes:
 
tiny-tim said:
Hi tizzful! :smile:

erm … it's not "-1^2" … :redface:

No wonder they came out minus each other! :rolleyes:

actually the -1 gets squared and so it becomes one.. Its negative because the height is negative because down is positive and up is negative.. But from what you're saying why isn't it -1? It's also in the opposite direction...
 
tizzful said:
… why isn't it -1? It's also in the opposite direction...

Noooooo … the kinetic energy mv^2/2 is always positive!

It depends only on speed, not direction!

You have too much imagination! :smile:
 
ahahah thank you! But I know KE is always positive because if velocity is negative it gets squared making it positive.. And that's what happened in this case.. But i still can't figure out the answer.. I think there's something wrong with me saying initial ME = 0...
 
tizzful said:
m1(gh1+0.5v^2)=m2(gh2+0.5v^2)
m1/m2=(gh2+0.5v^2)/(gh1+0.5v^2)

ah … I see now … your basic equation is wrong …

KE gained is (m1 + m2)v^2/2

PE gained is (m1 - m2)gh. :smile:
 
When you have gotten the answer using that method (which is probably the easiest), you can also try doing it by finding the acceleration on the big mass, and then use s=(at^2)/2. You will end up with the exactly same equation. :)
 

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