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atwood machine

 
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Nov8-08, 04:02 PM   #1
 

atwood machine


1. The problem statement, all variables and given/known data
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 7.50 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.80 m/s downward.
(a) How far will m1 descend below its initial level?

(b) Find the velocity of m1 after 1.80 s.

2. Relevant equations
f= ma
vf^2 = v^2 + 2ad

3. The attempt at a solution
i think for part a u use the equation i put above..but i dont know how to figure out acceleration...but i think its splitted between the two objects
 
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Nov9-08, 05:24 AM   #2
 
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Hi physics_geek,

Quote by physics_geek View Post
1. The problem statement, all variables and given/known data
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 7.50 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.80 m/s downward.
(a) How far will m1 descend below its initial level?

(b) Find the velocity of m1 after 1.80 s.

2. Relevant equations
f= ma
vf^2 = v^2 + 2ad

3. The attempt at a solution
i think for part a u use the equation i put above..but i dont know how to figure out acceleration...but i think its splitted between the two objects

To find the acceleration, start by drawing force diagrams for each of the objects. Using [itex]\sum F = m a[/itex] for each of the diagrams then gives two equations with two unknowns.

(You can also use an energy appoach here.)
 
Oct19-11, 06:33 PM   #3
 
Use the formula a= (gm1 + gm2)/(m2+m1) to find the acceleration then use the formula V^2=-2ad to find the distance. Bada bing bada boom!
Im still workin on the second part unfortunately...
 
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