Questions: A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a 14 kg mass on the left and a 2.4 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2.9 m apart. The acceleration of gravity is 9.8 m / s^2. At what rate are the two masses accelerating when they pass each other? Answer in units of m / s^2.

A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a 14 kg mass on the left and a 2.4 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2.9 m apart.

The acceleration of gravity is 9.8 m / s^2.

At what rate are the two masses accelerating when they pass each other?

Answer in units of m / s^2.

Transcript text: A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a 14 kg mass on the left and a 2.4 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2.9 m apart. The acceleration of gravity is $9.8 \mathrm{~m} / \mathrm{s}^{2}$. At what rate are the two masses accelerating when they pass each other? Answer in units of $\mathrm{m} / \mathrm{s}^{2}$.

Solution

Solution Steps

Step 1: Identify the forces acting on the masses

The forces acting on the 14 kg mass are its weight (14 kg * 9.8 m/s²) and the tension in the cord (T).
The forces acting on the 2.4 kg mass are its weight (2.4 kg * 9.8 m/s²) and the tension in the cord (T).

Step 2: Write the equations of motion for each mass

For the 14 kg mass (m1): \[ m_1 g - T = m_1 a \] \[ 14 \times 9.8 - T = 14a \] \[ 137.2 - T = 14a \]
For the 2.4 kg mass (m2): \[ T - m_2 g = m_2 a \] \[ T - 2.4 \times 9.8 = 2.4a \] \[ T - 23.52 = 2.4a \]

Step 3: Solve the system of equations

Add the two equations to eliminate T: \[ 137.2 - T + T - 23.52 = 14a + 2.4a \] \[ 113.68 = 16.4a \] \[ a = \frac{113.68}{16.4} \] \[ a \approx 6.93 \, \text{m/s}^2 \]