The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Refer to the figure shown below.

Let m₁ and m₂ e the two masses.

Let a = the acceleration.

Let T = tension over the frictionless pulley.

Write the equations of motion.

m₂g - T = m₂a (1)

T - m₁g = m₁a (2)

Add equations (1) and (2).

m₂g - T + T - m₁g = (m₁ + m₂) a

(m₂ - m₁) g = (m₁ + m₂) a

Divide through by m₁.

(m₂/m₁ - 1) g = (1 + m₂/m₁) a

Define r = m₂/m₁ as the ratio of the two masses. Then

(r - 1) g = (1 + r) a

r (g-a) = a + g

r = (g - a) / (g + a)

With = 2 ft/s from rest, the acceleration is

a = 2/32.2 = 0.062 ft/s²

Therefore

r = (32.2 - 0.062) / (32.2 + 0.062) = 0.9962

Answer:

The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).