Question

A simple Atwood’s machine uses a massless
pulley and two masses m1 and m2. Starting
from rest, the speed of the two masses is
4.1 m/s at the end of 7.6 s. At that time, the
kinetic energy of the system is 86 J and each
mass has moved a distance of 15.58 m. Find the value of heavier mass. The acceleration due to gravity is 9.81 m/s2 .
Answer in units of kg

I know it seems like I'm just looking for quick fix, but I really did attempt numerous times, and still don't get it. PLease Help!

Answer 1

The acceleration rate is
a = 4.1/7.6 = 0.5395 m/s^2

For an Atwood's machine (with no friction or pulley inertia)

a = [(M1 - M2)/(M1 + M2]*g

You don't need to use the 15.58 meter motion information. You know the accleration already.

Total KE = (M1 +M2)Vfinal^2)/2

Use that to solve for M1 + M2

Then use the a value to solve for M1 - M2, and then, from that, the heavier mass M1

Answer 2

dfa