2. Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F (Fig.(1)).Suppose that F=68.0N, M1=12.0kg, M2=18.0kg, and the coefficient of kinetic friction between each block and and the surface is 0.20.

Question

A, Draw a free body diagram for each block.
B, Determine the tension T and the magnitude of the acceleration of the system .
M1-----T-----M2----F
Figure(1)

Answer 1

To solve this problem, we'll go through the steps systematically. We'll start with part A (the free body diagrams) and then move on to part B (determining the tension and acceleration).

A. Free Body Diagrams

1. Block M1 (mass = 12.0 kg):

   • Forces acting on M1:

     • Weight (W1) acting downward: \( W_1 = M_1 \cdot g = 12.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 117.72 , \text{N} \)
     • Normal force (N1) acting upward: \( N_1 = W_1 = 117.72 , \text{N} \) (since the surface is horizontal)
     • Tension (T) acting to the right
     • Friction force (f1) acting to the left: \[ f_1 = \mu_k \cdot N_1 = 0.20 \cdot 117.72 , \text{N} = 23.544 , \text{N} \]

   • Free Body Diagram for M1:

       ↑ N1
       |
       |
    ---|---  T → 
       |    ← f1
       ↓ W1
   

2. Block M2 (mass = 18.0 kg):

   • Forces acting on M2:

     • Weight (W2) acting downward: \( W_2 = M_2 \cdot g = 18.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 176.58 , \text{N} \)
     • Normal force (N2) acting upward: \( N_2 = W_2 = 176.58 , \text{N} \)
     • tension (T) acting to the left
     • Friction force (f2) acting to the left: \[ f_2 = \mu_k \cdot N_2 = 0.20 \cdot 176.58 , \text{N} = 35.316 , \text{N} \]

   • Free Body Diagram for M2:

       ↑ N2
       |
       |
    ← T |               
       |   
       |   ← f2
       ↓ W2
   

B. Determine Tension T and Magnitude of the Acceleration

1. Net force on the system:

   • The total horizontal force applied is \( F = 68.0 , \text{N} \).
   • The total frictional force opposing the motion is the sum of frictional forces acting on both blocks: \[ f_{\text{total}} = f_1 + f_2 = 23.544 , \text{N} + 35.316 , \text{N} = 58.86 , \text{N} \]

2. Net force exerted on the system: \[ F_{\text{net}} = F - f_{\text{total}} = 68.0 , \text{N} - 58.86 , \text{N} = 9.14 , \text{N} \]

3. Total mass of the system: \[ M_{\text{total}} = M_1 + M_2 = 12.0 , \text{kg} + 18.0 , \text{kg} = 30.0 , \text{kg} \]

4. Calculate the acceleration (a): Using Newton's second law, \( F = m \cdot a \): \[ a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{9.14 , \text{N}}{30.0 , \text{kg}} \approx 0.3047 , \text{m/s}^2 \]

5. Determine the tension (T): We can analyze either block to find the tension. Let’s consider Block M2: \[ F_{\text{net on M2}} = T - f_2 = M_2 \cdot a \] \[ T - 35.316 , \text{N} = 18.0 , \text{kg} \cdot 0.3047 , \text{m/s}^2 \] \[ T - 35.316 , \text{N} = 5.48 , \text{N} \] \[ T = 5.48 , \text{N} + 35.316 , \text{N} = 40.796 , \text{N} \approx 40.8 , \text{N} \]

Final Results:

• The magnitude of the acceleration of the system is approximately \( 0.30 , \text{m/s}^2 \).
• The tension in the rope (T) is approximately \( 40.8 , \text{N} \).