Question

Conservative Forces and Potential Energy?

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 67 J and each mass has moved a distance of 15 m. Determine the values of m₁ and m₂

Answer 1

Let mass m1 goes up with acceleration a and mass m2 goes down with acceleration a

If T is tension in strings then

T - m1g = m1a

m2g-T = m2a

Solving above equations,

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

Given, Kinetic energy of system is 67 J after 5s

0.5 m1v^2 + 0.5 m2v^2 = 67

(m1+ m2)v^2 = 2 x 67 = 134

At this time v = 6 m/s

(m1+m2)(6)^2 = 134

m1+m2 = 3.72

Masses move 15 m each and acquire speed 6 m/s at end of 5 s, so

v=vo +at

6 = 0 +a(5)

a = 6/5 m/s^2

Using value of a and m1 +m2 in

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

6/5 = (m2-m1)(9.81)/3.72

m2-m1 = 0.455

Also m1 +m2 = 3.72

Therefore m1 = 1.63 kg and m2 = 2.09 kg