An Atwood machine with m1 = 5 kg and m2 = 2 kg, frictionless pulley. Find acceleration and rope tension (g = 9.8 m/s^2).

An Atwood machine with m1 = 5 kg and m2 = 2 kg, frictionless pulley. Find acceleration and rope tension (g = 9.8 m/s^2).

• A. a ≈ 1.0 m/s^2; T ≈ 14 N
• B. a ≈ 4.2 m/s^2; T ≈ 28 N
• C. a ≈ 9.8 m/s^2; T ≈ 0 N
• D. a ≈ 0.5 m/s^2; T ≈ 11 N

Answer: (B)

When two masses share a rope over a frictionless pulley, the same tension acts on both sides and the heavier mass accelerates downward while the lighter one goes up. The acceleration is set by the difference in weights relative to the total mass: a = (m1 − m2) g / (m1 + m2). Plugging in m1 = 5 kg, m2 = 2 kg, and g = 9.8 m/s^2 gives a = (3)(9.8)/7 ≈ 4.2 m/s^2. To find the tension, use either mass equation. For the lighter mass, T − m2 g = m2 a, so T = m2(g + a) = 2(9.8 + 4.2) = 28 N. The same value arises from the heavier mass equation: T = m1(g − a) = 5(9.8 − 4.2) = 28 N. So the acceleration is about 4.2 m/s^2 and the rope tension about 28 N.

When two masses share a rope over a frictionless pulley, the same tension acts on both sides and the heavier mass accelerates downward while the lighter one goes up. The acceleration is set by the difference in weights relative to the total mass: a = (m1 − m2) g / (m1 + m2).

Plugging in m1 = 5 kg, m2 = 2 kg, and g = 9.8 m/s^2 gives a = (3)(9.8)/7 ≈ 4.2 m/s^2.

To find the tension, use either mass equation. For the lighter mass, T − m2 g = m2 a, so T = m2(g + a) = 2(9.8 + 4.2) = 28 N. The same value arises from the heavier mass equation: T = m1(g − a) = 5(9.8 − 4.2) = 28 N.

So the acceleration is about 4.2 m/s^2 and the rope tension about 28 N.