In the Atwood machine, two masses m1 = 12 kg and m2 = 13 kg are connected by a light inextensible string passing over a frictionless pulley. Find the (a) acceleration of the system when released from rest, (b) the tension in the string, and (c) the speed of either mass after moving a distance of 2.4 m.

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:
\[
a = \frac{(13 - 12) \cdot 9.8}{12 + 13}
= \frac{9.8}{25}
= 0.392 \, \text{m/s}^2
\]

### (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.

Solve an Atwood machine problem involving two masses (12 kg and 13 kg) connected by a string over a frictionless pulley. Learn how to calculate the system's acceleration, the tension in the string, and the speed of the masses after moving 2.4 m. This problem applies Newton's Second Law of Motion and kinematic equations. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation