In an Atwood-style system with m1 = 2 kg on a table and m2 = 3 kg hanging, what is the acceleration and the tension?

In an Atwood-style system with m1 = 2 kg on a table and m2 = 3 kg hanging, what is the acceleration and the tension?

• A. a ≈ 9.8 m/s^2; T ≈ 39.2 N
• B. a ≈ 9.8 m/s^2; T ≈ 19.6 N
• C. a ≈ 5.88 m/s^2; T ≈ 11.76 N
• D. a ≈ 2.0 m/s^2; T ≈ 4.0 N

Answer: (C)

This question tests how to analyze a two-mass rope system with one mass on a frictionless table. Write Newton’s second law for both masses and use the fact the rope has the same tension throughout. For the mass on the table, the only horizontal force is the tension, so T = m1 a. For the hanging mass, the downward weight minus the tension equals its mass times acceleration: m2 g − T = m2 a. Substitute T = m1 a into the second equation: m2 g − m1 a = m2 a, which gives a = m2 g / (m1 + m2). Plug in m1 = 2 kg, m2 = 3 kg, and g ≈ 9.8 m/s^2: a = (3 × 9.8) / (2 + 3) = 29.4 / 5 ≈ 5.88 m/s^2. Tension follows from T = m1 a = 2 × 5.88 ≈ 11.76 N. So the acceleration is about 5.88 m/s^2 and the tension about 11.76 N. The hanging mass accelerates downward while the table mass moves horizontally to the right.

This question tests how to analyze a two-mass rope system with one mass on a frictionless table. Write Newton’s second law for both masses and use the fact the rope has the same tension throughout.

For the mass on the table, the only horizontal force is the tension, so T = m1 a.

For the hanging mass, the downward weight minus the tension equals its mass times acceleration: m2 g − T = m2 a.

Substitute T = m1 a into the second equation: m2 g − m1 a = m2 a, which gives a = m2 g / (m1 + m2).

Plug in m1 = 2 kg, m2 = 3 kg, and g ≈ 9.8 m/s^2:

a = (3 × 9.8) / (2 + 3) = 29.4 / 5 ≈ 5.88 m/s^2.

Tension follows from T = m1 a = 2 × 5.88 ≈ 11.76 N.

So the acceleration is about 5.88 m/s^2 and the tension about 11.76 N. The hanging mass accelerates downward while the table mass moves horizontally to the right.