A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is μ1, and in region 2, the coefficient is 2. The positive direction is shown in the figure. (Figure 1) Figure Before: region 1 After: f PI + M₂ 12 < 1 of 1 region 2 > constant velocity. Express the net work in terms of M, g, L, ₁, and 2. View Available Hint(s) W = Submit Part B ✓ Correct This answer makes sense because it is as if the board spent half its time in region 1 average, it in fact did. (+₂) MgL 2 Wext = Previous Answers What is the total work done by the external force in pulling the board from region 1 to region moves at constant velocity) Express your answer in terms of M, g, L, ₁, and μ2. View Available Hint(s) Submit 5 ΑΣΦ ?

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**Dragging a Board**

A uniform board of length \( L \) and mass \( M \) lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is \( \mu_1 \), and in region 2, the coefficient is \( \mu_2 \). The positive direction is shown in the figure.

**Figure 1:**

- **Before:** The board is positioned with part of it in region 1 (\( \mu_1 \)) and the other part in region 2 (\( \mu_2 \)). The length of the board in region 1 is less than the total length \( L \).
- **After:** The board is fully within region 2 (\( \mu_2 \)), having moved a positive distance such that it is no longer occupying region 1 (\( \mu_1 \)).

---

**Constant Velocity**

**Question:** Express the net work in terms of \( M, g, L, \mu_1, \) and \( \mu_2 \).

**Solution:** 
\[ W = -\frac{(\mu_1 + \mu_2)MgL}{2} \]

**Explanation:** This answer makes sense because it is as if the board spent half its time in region 1 and half in region 2. On average, it in fact did.

---

**Part B**

**Question:** What is the total work done by the external force in pulling the board from region 1 to region 2 (assuming the board moves at constant velocity)?

Express your answer in terms of \( M, g, L, \mu_1, \) and \( \mu_2 \).

\[ W_{\text{ext}} = \]

**Note:** The exact expression for \( W_{\text{ext}} \) needs to be derived or provided.

--- 

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Transcribed Image Text:**Dragging a Board** A uniform board of length \( L \) and mass \( M \) lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is \( \mu_1 \), and in region 2, the coefficient is \( \mu_2 \). The positive direction is shown in the figure. **Figure 1:** - **Before:** The board is positioned with part of it in region 1 (\( \mu_1 \)) and the other part in region 2 (\( \mu_2 \)). The length of the board in region 1 is less than the total length \( L \). - **After:** The board is fully within region 2 (\( \mu_2 \)), having moved a positive distance such that it is no longer occupying region 1 (\( \mu_1 \)). --- **Constant Velocity** **Question:** Express the net work in terms of \( M, g, L, \mu_1, \) and \( \mu_2 \). **Solution:** \[ W = -\frac{(\mu_1 + \mu_2)MgL}{2} \] **Explanation:** This answer makes sense because it is as if the board spent half its time in region 1 and half in region 2. On average, it in fact did. --- **Part B** **Question:** What is the total work done by the external force in pulling the board from region 1 to region 2 (assuming the board moves at constant velocity)? Express your answer in terms of \( M, g, L, \mu_1, \) and \( \mu_2 \). \[ W_{\text{ext}} = \] **Note:** The exact expression for \( W_{\text{ext}} \) needs to be derived or provided. --- **Feedback Section** Readers can submit their responses or access further hints by interacting with the platform.
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