Let's solve the problem step by step: ### **Given:** - Mass of upper block (\(m_1\)) = 5.00 kg (tied to the wall) - Mass of lower block (\(m_2\)) = 10.0 kg (pulled by force) - Applied horizontal force (\(F\)) = 45.0 N - Coefficient of kinetic friction (\(\mu_k\)) = 0.200 --- ## **(a) Free-body diagrams & Action-Reaction Forces** ### **For the 5.00-kg Block (upper block):** - **Forces acting:** - Weight (\(m_1g\)) downward - Normal force (\(N_1\)) upward from the 10 kg block - Tension (\(T\)) in the string to the left - Friction force (\(f_{k,1}\)) to the right, due to movement of the 10 kg block under it ### **For the 10.0-kg Block (lower block):** - **Forces acting:** - Weight (\(m_2g\)) downward - Normal force from the ground (\(N_2\)) upward - Normal force (\(N_1\)) downward (from 5-kg block) - Applied force (\(F\)) to the right - Friction force (\(f_{k,2}\)) to the left (from ground) - Friction force (\(f_{k,1}\)) to the left (from 5-kg block) - Action-reaction pair: the friction force \(f_{k,1}\) (on 10 kg block left, on 5 kg block right) --- ## **(b) Tension in the String and Acceleration of 10-kg Block** 1. **Find the friction forces:** - **Between 5-kg and 10-kg block**: \[ f_{k,1} = \mu_k N_1 \] \(N_1\) is simply the weight of the upper block: \[ N_1 = m_1 g = 5.00 \times 9.8 = 49.0\ \text{N} \] \[ f_{k,1} = 0.200 \times 49.0 = \boxed{9.80~\text{N}} \] - **Between 10-kg block and ground**: \[ N_2 = m_2 g + m_1 g = (10.0 + 5.00) \times 9.8 = 147~\text{N} \] \[ f_{k,2} = \mu_k N_2 = 0.200 \times 147 = \boxed{29.4~\text{N}} \] 2. **Forces on 5.00-kg block (horizontal direction):** The block is **tied to the wall** and does not accelerate, so \[ T = f_{k,1} \] \[ \boxed{T = 9.80~\text{N}} \] (Tension balances the friction force.) 3. **Forces on 10.0-kg block (horizontal direction):** Net force: \[ F_{\text{net}} = F - f_{k,1} - f_{k,2} \] \[ F_{\text{net}} = 45.0 - 9.80 - 29.4 = 5.80~\text{N} \] Acceleration: \[ a = \frac{F_{\text{net}}}{m_2} \] \[ a = \frac{5.80}{10.0} = \boxed{0.580~\text{m/s}^2} \] --- ## **Final Answers:** **Tension in the string:** \[ \boxed{9.80~\text{N}} \] **Acceleration of the 10-kg block:** \[ \boxed{0.580~\text{m/s}^2} \] --- If you need the free-body diagrams drawn, let me know!

Let's carefully analyze and compute the problem step by step with detailed calculations. --- ### **Given Data:** - Mass of upper block, \(m_1 = 5.00\,kg\) - Mass of lower block, \(m_2 = 10.0\,kg\) - Applied force, \(F = 45.0\,N\) - Coefficient of kinetic friction, \(\mu_k = 0.200\) - Acceleration due to gravity, \(g = 9.8\,m/s^2\) --- ### **Step 1: Free-Body Diagrams and Action-Reaction Forces** - **Upper block (5 kg):** - **Forces:** - Tension in the string, \(T\), pulling to the left. - Normal force from the lower block, \(N_1\), upward. - Weight, \(W_1 = m_1 g = 5.00 \times 9.8 = 49.0\,N\), downward. - Friction force between blocks, \(f_{k,1}\), acting to oppose relative motion, which in this case acts to the right on the upper block if the lower block moves right. - **Lower block (10 kg):** - **Forces:** - Applied force, \(F = 45.0\,N\), to the right. - Friction force from the ground, \(f_{k,2}\), to the left. - Friction force from the upper block, \(f_{k,1}\), to the left. - Normal force from the ground, \(N_2\), upward. - Weight, \(W_2 = m_2 g = 10.0 \times 9.8 = 98.0\,N\), downward. **Note:** The upper block is tied to the wall, so it does not accelerate; it only experiences tension and friction. --- ### **Step 2: Calculate Normal and Friction Forces** **a) Normal force from lower block on upper block:** Since the upper block rests on the lower block, the normal force exerted by the lower block on the upper block (and vice versa) is equal to the weight of the upper block: \[ N_1 = W_1 = 49.0\,N \] **b) Friction force between the two blocks:** Using the coefficient of kinetic friction: \[ f_{k,1} = \mu_k N_1 = 0.200 \times 49.0 = 9.80\,N \] **c) Normal force from ground on the lower block:** The total normal force balances the total weight: \[ N_2 = W_1 + W_2 = 49.0 + 98.0 = 147.0\,N \] **d) Friction force between ground and lower block:** \[ f_{k,2} = \mu_k N_2 = 0.200 \times 147.0 = 29.4\,N \] --- ### **Step 3: Find Tension \(T\) in the String** Since the **upper block is tied to the wall** and **does not accelerate**, the only horizontal force on it is the tension \(T\). The friction between the upper and lower blocks opposes the motion of the lower block, so the tension must balance the friction force: \[ T = f_{k,1} = 9.80\,N \] **Final tension:** \[ \boxed{T = 9.80\,N} \] --- ### **Step 4: Determine the Acceleration of the Lower Block** Now, analyze the **lower block**: - **Net force in the horizontal direction:** \[ F_{\text{net}} = F - f_{k,2} - T \] - Plugging in the values: \[ F_{\text{net}} = 45.0 - 29.4 - 9.80 = 5.80\,N \] - **Calculate acceleration:** \[ a = \frac{F_{\text{net}}}{m_2} = \frac{5.80\,N}{10.0\,kg} = 0.580\,m/s^2 \] --- ### **### Final Answers:** | Quantity | Value | Units | | --- | --- | --- | | **Tension in the string** | \(\boxed{9.80}\) | N | | **Acceleration of the 10-kg block** | \(\boxed{0.580}\) | m/s² | --- **Summary:** - The tension in the string is approximately **9.80 N**. - The lower block accelerates to the right at approximately **0.580 m/s²**. Let me know if you'd like further clarification or diagrams!

Certainly! Here's a detailed description of the free-body diagrams for both blocks, which you can sketch based on the following instructions: --- ### **Diagram for the 5 kg Block (Upper Block)** - **Position:** Placed on top of the 10 kg block, tied to the wall. - **Forces:** - **Weight:** \(W_1 = 49.0\,N\) downward, labeled as \(m_1g\). - **Normal force from the lower block:** \(N_1\) upward, equal in magnitude to \(W_1\), so \(N_1 = 49\,N\). - **Tension in the string:** \(T\), acting horizontally to the left (since tension pulls towards the wall). - **Friction force from the lower block:** \(f_{k,1} = 9.80\,N\), acting horizontally to the right (opposes the tension if the lower block moves right). **Note:** The block is tied to the wall, so it does not move, and the tension balances the friction force. --- ### **Diagram for the 10 kg Block (Lower Block)** - **Position:** The block is pulled to the right by the applied force \(F = 45\,N\). - **Forces:** - **Applied force:** \(F = 45\,N\), directed to the right. - **Friction force from the ground:** \(f_{k,2} = 29.4\,N\), acting to the left. - **Friction force from the upper block:** \(f_{k,1} = 9.80\,N\), acting to the left. - **Normal force from ground:** \(N_2 = 147\,N\), upward, balancing the total weight. - **Weight:** \(W_2 = 98\,N\), downward. - **Normal force from the upper block:** \(N_1 = 49\,N\), acting downward on the lower block (equal to the weight of the upper block). **Additional notes:** - The friction between the lower and upper blocks acts to oppose relative motion, so it acts to the left on the lower block. - The net force causes acceleration \(a = 0.580\,m/s^2\) to the right. --- ### **Sketching Tips:** - Draw two boxes stacked vertically. - For the upper box: - Arrow downward labeled \(49\,N\). - Arrow upward labeled \(49\,N\). - Arrow to the left labeled \(T = 9.80\,N\). - Arrow to the right labeled \(f_{k,1} = 9.80\,N\). - For the lower box: - Arrow to the right labeled \(45\,N\). - Arrow to the left labeled \(f_{k,2} = 29.4\,N\). - Arrow to the left labeled \(f_{k,1} = 9.80\,N\). - Arrow upward labeled \(147\,N\). - Arrow downward labeled \(98\,N\). --- If you'd like, I can generate a visual diagram image for you, but this description should help you create an accurate free-body diagram manually.

Certainly! Here's a visual representation of the free-body diagrams for both blocks based on the problem description: --- ### Free-Body Diagram for the 5 kg Block (Upper Block): ``` ↑ N₁ = 49 N | | (Normal force from lower block) | [5 kg Block] ← T = 9.80 N (Tension) → fₖ₁ = 9.80 N (Friction from lower block) ↓ W₁ = 49 N ``` ### Free-Body Diagram for the 10 kg Block (Lower Block): ``` → F = 45 N (Applied force) ↓ W₂ = 98 N ↑ N₂ = 147 N (Normal force from ground) ← fₖ₂ = 29.4 N (Friction from ground) ← fₖ₁ = 9.80 N (Friction from upper block) ``` --- ### Explanation: - The upper block experiences tension pulling it to the left while the friction from the lower block opposes this tension. - The lower block experiences the applied force to the right, friction from the ground to the left, and friction from the upper block also to the left. - The normal forces balance the weights vertically. - The friction forces are calculated using \(\mu_k N\). --- If you'd like a more formal, graphical diagram image, I recommend using drawing tools or software like GeoGebra or a physics diagram generator. But the above sketches should help you visualize the forces clearly. Let me know if you'd like me to create an actual image for you!