A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the masses are gently released, what is the resulting acceleration of the masses

acceleration = 2.4525‬ m/s²

Explanation:

dа ta: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 * 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a) ‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 * 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and (b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²