Question

A simple Atwood's machine uses two masses, m1 and m2. Starting from rest, the speed of the two masses is 6 m/s at the end of 5 s. At that instant, the kinetic energy of the system is 60 J and each mass has moved a distance of 15 m. Determine the values of m1 and m2.

Answer 1

here,

from the statement of KE, we have that:

60 J = 1/2 * m1 * v^2 + 1/2 * m2 * v^2 = 1/2 * (6)^2(m1+m2) or that

60 = 36 * (m1+m2) or m1 + m2 = 1.67 kg .....(1)

we can write Newton's second law for each mass, where T is the tension in the rope:

assuming m1 moves up, we have:

T-m1g=m1a
T-m2g=-m2a

subtracting equations gives us:

(m2-m1)g=(m1+m2)a or a=(m2-m1)/(m2+m1) g

we can find the acceleration of the system also knowing that

the speed has changed by 6 m/s in 5 s for an accelration of 1.2, therefore:

1.2 = ((m2-m1)/(m1+m2)) g .....(2)

from (1) and (2)

m1 = 0.73 kg , m2 = 0.94 kg