Energy Conservation for Connected Objects with Friction

Solution

Method 1: Individual Free-Body Diagrams

For Block A (on the table), the forces are:

• Tension $T$ (rightward, along the string)
• Kinetic friction $f_k = \mu_k N$ (leftward, opposing motion)
• Normal force $N$ (upward)
• Weight $m_A g$ (downward)

Since Block A moves horizontally without vertical acceleration:

$$N = m_A g$$

Thus, the friction force is:

$$f_k = \mu_k m_A g$$

Applying Newton's second law in the horizontal direction (taking rightward as positive):

$$T - \mu_k m_A g = m_A a$$

For Block B (hanging vertically), the forces are:

• Weight $m_B g$ (downward)
• Tension $T$ (upward)

Applying Newton's second law in the vertical direction (taking downward as positive):

$$m_B g - T = m_B a$$

We now have two equations:

$$T - \mu_k m_A g = m_A a$$ $$m_B g - T = m_B a$$

Adding these equations eliminates $T$:

$$m_B g - \mu_k m_A g = m_A a + m_B a$$

Solving for $a$:

$$a = \frac{m_B g - \mu_k m_A g}{m_A + m_B} = \frac{(m_B - \mu_k m_A) g}{m_A + m_B}$$

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Method 2: System Approach

Treat both blocks as a single system. The string tension $T$ is an internal force and cancels out.

The net external force causing the system to accelerate is:

• Driving force: weight of Block B, $m_B g$ (pulls the system)
• Opposing force: kinetic friction on Block A, $f_k = \mu_k m_A g$

The total mass of the system is $m_A + m_B$.

Applying Newton's second law to the entire system:

$$F_{\text{net}} = (m_A + m_B) \cdot a$$ $$m_B g - \mu_k m_A g = (m_A + m_B) \cdot a$$

Solving directly for $a$:

$$a = \frac{(m_B - \mu_k m_A) g}{m_A + m_B}$$