Two masses, frictionless pulley

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SUMMARY

The discussion focuses on a physics problem involving two masses connected by a light string over a frictionless pulley. The 8.7-kg mass (m1) falls a vertical distance of 1.09 m, and the goal is to determine the speed of the 3.9-kg mass (m2) just before m1 hits the ground and the maximum height attained by m2. Utilizing the conservation of mechanical energy, the relevant equation is established as m1*g*h1i = (1/2)m1*V^2 + (1/2)m2*V^2 + m2*g*h2f, leading to a straightforward algebraic solution for V. The approach for part (b) involves using the velocity from part (a) to calculate the height attained by m2.

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  • Familiarity with kinetic energy (KE) and potential energy (PE) equations
  • Basic algebra skills for solving equations
  • Knowledge of gravitational force (g) and its application in physics problems
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  • Learn how to derive and apply kinetic and potential energy equations
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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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