Question

A simple Atwood's machine uses two masses, m₁ and m₂. Starting from rest, the speed of the two masses is 4.0 m/s at the end of 8.0 s. At that instant, the kinetic energy of the system is 90 J and each mass has moved a distance of 16.0 m. Determine the values of m₁ and m₂.

Answer 1

K.E.= 90J=  1/2m1v^2+1/2m2v^2 = 0.5*4^2(m1+m2)

So, m1+m2=11.25kg__________________ (A)

Now, assuming that m1 moves upwards and T is the tension in the rope,
T-m1g=m1a ______________________(B)
T-m2g=-m2a ________________(C)
subtracting (B)-(C) gives ;(m2-m1)g=(m1+m2)a

Or, a=(m2-m1)g / (m2+m1)

But, in t= 8s, velocity changes by4m/s. So, a= 4/8= 0.5m/s^2

So, (m2-m1) / (m2+m1) = 0.051

Or, (m2-m1)= 0.574 ___________(D)

So, (A) + (B) = 2m2= 11.82 ; So, m2= 5.9kg

m1= 11.25-5.9= 5.35kg