Newton’s Second Law and Tension in Connected Masses (Physics with Calculus)

Q1. Two crates, one with mass 5.0 kg and the other with mass 10.0 kg, are connected by a light string and pulled with a force of 25 N on a frictionless surface. What is the acceleration of the system? What is the tension in the string connecting the crates?

Background

Topic: Newton’s Second Law, Tension, and Systems of Connected Objects

This question tests your understanding of how to apply Newton’s Second Law ($\sum F = ma$) to a system of connected masses, and how to analyze the forces acting on each mass to find acceleration and tension.

Key Terms and Formulas

• Newton’s Second Law: $\sum F = ma$

• Tension (T): The force transmitted through a string or rope connecting objects.

• System Acceleration: For objects connected and moving together, they share the same acceleration.

• Total Mass: $m_{\text{total}} = m_1 + m_2$

Step-by-Step Guidance

1. Draw a free-body diagram for the system, showing all forces acting on both crates (the applied force, tension, and any other relevant forces).

2. Write Newton’s Second Law for the entire system (both crates together):

   $\sum F = (m_1 + m_2)a$

   Here, $\sum F$ is the net external force (the pulling force), and $a$ is the acceleration of the system.

3. Plug in the values for the masses and the applied force to solve for the acceleration $a$ (but do not calculate the final value yet):

   $a = \frac{F}{m_1 + m_2}$

4. To find the tension $T$ in the string, focus on just one crate (for example, the 5.0 kg crate). Write Newton’s Second Law for that crate:

   $T = m_1 a$

   Substitute the value of $a$ you found in the previous step into this equation (but do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer:

Acceleration: $a = \frac{25}{5.0 + 10.0} = \frac{25}{15} = 1.67\ \text{m/s}^2$

Tension in the string: $T = (5.0)(1.67) = 8.3\ \text{N}$

We used Newton’s Second Law for the system to find the acceleration, then applied it to one crate to find the tension.