Two masses, frictionless pulley

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Homework Help Overview

The problem involves two masses connected by a light string over a frictionless pulley, focusing on the conservation of mechanical energy as one mass falls and the other rises. The specific scenario includes a mass of 8.7 kg falling a distance of 1.09 m and a mass of 3.9 kg moving as a result.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of energy principles and the relevant equations. Some express uncertainty about where to begin, while others attempt to set up the equations for mechanical energy before and after the fall.

Discussion Status

Some participants have provided guidance on setting up the conservation of energy equation, noting the need to account for both masses and their respective kinetic and potential energies. There is acknowledgment of the initial conditions and relationships between the velocities of the two masses.

Contextual Notes

Participants highlight the importance of considering both masses in the energy equations and the initial conditions, such as starting from rest and the heights involved. There is an emphasis on the need for clarity regarding the definitions of variables and terms used in the equations.

cassienoelle
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Homework Statement


Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 8.7-kg mass m1 is released and falls through a vertical distance of h = 1.09 m before hitting the ground. Use conservation of mechanical energy to determine:
*picture attached*

a) how fast the 3.9-kg mass m2 is moving just before the 8.7-kg mass hits the ground; and
b) the maximum height attained by the 3.9-kg mass.



Homework Equations





The Attempt at a Solution

 

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So what exactly are you having trouble with? If you haven't tried to solve the problem or you don't tell us where you're stuck, you won't get much help.

What are the relevant equations for conservation of energy problems?
 
i'm stuck. period.
i have NO idea where to even start.
 
Okay so:
MEi = MEf
m1=8.7 hi=1.09 hf = 0
m2 = 3.9 hi = 0 hf = ?
KEi + PEi = KEf + PEf
1/2mv^2 + mgh = 1/2mv^2 + mgh
(1/2)(m??1 or 2??) V^2 + m?g0 = 1/2 m? 0 + mgh
 
cassienoelle said:
Okay so:
MEi = MEf
m1=8.7 hi=1.09 hf = 0
m2 = 3.9 hi = 0 hf = ?
KEi + PEi = KEf + PEf
1/2mv^2 + mgh = 1/2mv^2 + mgh
(1/2)(m??1 or 2??) V^2 + m?g0 = 1/2 m? 0 + mgh

Yeah you're on the right path. Just remember that you have two masses, so you need KE and PE for both masses, before and after (8 terms). If you want to write out the complete equation, it's like this:

(KE of m1 before) + (KE of m2 before) + (PE of m1 before) + (PE of m2 before) = (KE of m1 after) + (KE of m2 after) + (PE of m1 after) + (PE of m2 after)

(1/2)m1*v1i^2 + (1/2)m2*v2i^2 + m1*g*h1i + m2*g*h2i = (1/2)m1*v1f^2 + (1/2)m2*v2f^2 + m1*g*h1f + m2*g*h2f

And like you said, h2i = 0, h1f = 0, and you also have v1i = 0 and v2i = 0 (they both start from rest). Moreover, you know that both velocities are the same since they're connected by a string, so v1f = v2f = V (we'll call both final velocites V). Now your equation reduces to:

m1*g*h1i = (1/2)m1*V^2 + (1/2)m2*V^2 + m2*g*h2f

Now that's not so bad. The rest is algebra, solving for V.

Edit: That does it for part (a). For part (b), simply use the V you got from part (a) as an initial upward velocity against gravity (ignore the other mass and string and everything else).
 

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