Two masses, frictionless pulley

4 replies · 12K views · last reply
cassienoelle
Messages
63
Reaction score
0

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

 
Attachments
  • Diagram of an Atwood machine: a light pulley hangs from the ceiling with a rope over it, one end holding mass m1 (taller cylinder) at height h above the floor and the other end holding mass m2 resting on the floor.
    84b257235efec188243b3de648ddec440256f024bc927fa5d01e0d019a4d3abcaf0c2311e181d84806bf7496ccf632d3.png
    7.9 KB · Views: 1,312
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).