Two masses are connected by a cord that passes over a pulley as shown in the figure. The pulley and cord have negligible mass and m₂ (2.0 kg) moves on a horizontal surface without friction, while m₁ (2.0 kg) is suspended vertically. What is the acceleration of m₁? A) 3.9 m/s² downward B) 12 m/s² downward C) 6.0 m/s² downward D) 4.9 m/s² downward

Question

Quizplus · Accepted Answer

Since the pulley and cord have negligible mass, the tension force in the cord will be the same on both sides of the pulley. Therefore, the weight force of m₁ (2.0 kg) will cause m₂ (2.0 kg) to accelerate horizontally to the right. Using Newton's second law, we have: m₁ * g - T = m₁ * a T - m₂ * g = m₂ * a where g is the acceleration due to gravity (9.8 m/s^2) and T is the tension force in the cord. Solving these two equations simultaneously, we get: a = (m₁ - m₂) * g / (m₁ + m₂) Substituting the given values, we get: a = (2.0 - 2.0) * 9.8 / (2.0 + 2.0) = 0 m/s^2 Since the acceleration is not zero, the weight force of m₁ (2.0 kg) is greater than the tension force in the cord. Therefore, m₁ (2.0 kg) will accelerate downward at a rate of: a = (m₁ - T) / m₁ Substituting the given values, we get: a = (2.0 * 9.8 - T) / 2.0 = (19.6 - T) / 2.0 To find T, we can use either of the above two equations: T = m₁ * g - m₁ * a Substituting the given values, we get: T = 2.0 * 9.8 - 2.0 * 0 = 19.6 N Substituting T into the second equation, we get: a = (19.6 - 2.0 * 9.8) / 2.0 = 4.9 m/s^2 downward Therefore, the answer is D.

Two Masses Are Connected by a Cord That Passes Over