Asimple atwood’s machine uses a massless
pulley and two masses m1 and m2. starting
from rest, the speed of the two masses is
4.2 m/s at the end of 4.7 s. at that time, the
kinetic energy of the system is 94 j and each
mass has moved a distance of 9.87 m.
9.87 m
find the value of heavier mass. the accel-
eration due to gravity is 9.81 m/s2 .
answer in units of kg
002 (part 2 of 2) 10.0 points
find the value of lighter mass.
answer in units of kg

The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t). 
Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) 
then a = 6.7/1.9 = 3.526 m/s². 
The Atwood's Machine also has the formula d= distance = 1/2a(t²) 
distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², 
a = 3.526 m/s² the same acceleration. 
a= g(m1-m2) / m1+m2) 
m1a + m2a = m1g - m2g 
m1a - m1g = -m2g - m2a 
3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 
-6.28 m1 = -13.34 m2 
0.47 m1= m2 
if 24J = 1/2mv² 
then 24J = 1/2 m1 ( 6.7)² 
48/ 44.89 = m1 
1.069 kg = m1 , then 
0.47(1.069) = m2 
0.503 kg = m2

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