In an Atwood machine, two masses are connected by a string over a pulley. Let's approach this step-by-step to solve for the following:

Given:

• $m_1 = 12 \, \text{kg}$
• $m_2 = 13 \, \text{kg}$
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$

(a) Acceleration of the system

The net force acting on the system is due to the difference in the gravitational forces on $m_1$ and $m_2$, as they are connected by a string and move in opposite directions.

The force difference is: $F_{\text{net}} = m_2 g - m_1 g$

The total mass of the system is: $M = m_1 + m_2$

Using Newton's Second Law, the acceleration $a$ is given by: $a = \frac{F_{\text{net}}}{M} = \frac{(m_2 - m_1) g}{m_1 + m_2}$

Substitute the values:

= \frac{9.8}{25} = 0.392 \, \text{m/s}^2$$ ### (b) Tension in the string The tension $$T$$ in the string can be found by analyzing the forces acting on either mass. Let’s consider mass $$m_1$$ (similar calculations can be done for $$m_2$$): For $$m_1$$, the forces are: - The gravitational force: $$m_1 g$$ - The tension force $$T$$ acting upwards - The net force on $$m_1$$ is $$m_1 a$$ Using Newton's Second Law for $$m_1$$: $$T - m_1 g = -m_1 a$$ $$T = m_1 g - m_1 a$$ Substitute the values: $$T = 12 \cdot 9.8 - 12 \cdot 0.392 = 117.6 - 4.704 = 112.896 \, \text{N}$$ ### (c) Speed of either mass after moving a distance of 2.4 m The speed $$v$$ can be found using the kinematic equation, since the system starts from rest: $$v^2 = u^2 + 2 a s$$ where: - $$u = 0$$ (initial velocity), - $$a = 0.392 \, \text{m/s}^2$$ (acceleration), - $$s = 2.4 \, \text{m}$$ (distance). Thus: $$v^2 = 0 + 2 \cdot 0.392 \cdot 2.4$$ $$v^2 = 1.8816$$ $$v = \sqrt{1.8816} \approx 1.37 \, \text{m/s}$$ ### Summary of results: - (a) The acceleration of the system is $$0.392 \, \text{m/s}^2$$. - (b) The tension in the string is $$112.9 \, \text{N}$$. - (c) The speed of either mass after moving 2.4 m is $$1.37 \, \text{m/s}$$. Would you like more detailed explanations for any step? ### 5 Related Questions: 1. How would the results change if both masses were equal in the Atwood machine? 2. What effect would a non-negligible pulley mass have on the system’s dynamics? 3. How do you calculate the tension if there’s friction in the pulley system? 4. Can the Atwood machine setup be used to measure gravitational acceleration $$g$$? 5. How does the acceleration compare if the pulley is no longer frictionless? ### Tip: Always check whether the pulley is ideal (frictionless and massless) in Atwood machine problems; otherwise, it complicates the tension and acceleration calculations.