# Atwood machine

by physics_geek
Tags: atwood, machine
 P: 84 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
HW Helper
P: 2,249
Hi physics_geek,

 Quote by physics_geek 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 $\sum F = m a$ for each of the diagrams then gives two equations with two unknowns.

(You can also use an energy appoach here.)
 P: 1 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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