To solve the problem step-by-step, we will follow these steps:
### Step 1: Identify the masses and the external force
We have three blocks:
- Block A: mass \( m_A = 5 \, \text{kg} \)
- Block B: mass \( m_B = 10 \, \text{kg} \)
- Block C: mass \( m_C = 15 \, \text{kg} \)
An external force \( F = 30 \, \text{N} \) is applied to block A.
### Step 2: Calculate the total mass of the system
The total mass \( M \) of the system is the sum of the masses of the three blocks:
\[
M = m_A + m_B + m_C = 5 \, \text{kg} + 10 \, \text{kg} + 15 \, \text{kg} = 30 \, \text{kg}
\]
### Step 3: Calculate the acceleration of the system
Using Newton's second law, the acceleration \( a \) of the system can be calculated as:
\[
F = M \cdot a \implies a = \frac{F}{M} = \frac{30 \, \text{N}}{30 \, \text{kg}} = 1 \, \text{m/s}^2
\]
### Step 4: Analyze the forces on block C
Since block C is only connected to block B by a string, the tension \( T_2 \) in the string connecting B and C can be calculated using the equation:
\[
T_2 = m_C \cdot a = 15 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 15 \, \text{N}
\]
### Step 5: Analyze the forces on block B
Now, we will analyze block B, which is connected to block A by another string. The tension \( T_1 \) in the string connecting A and B can be calculated by considering the forces acting on block B:
\[
T_1 - T_2 = m_B \cdot a
\]
Substituting the known values:
\[
T_1 - 15 \, \text{N} = 10 \, \text{kg} \cdot 1 \, \text{m/s}^2
\]
\[
T_1 - 15 \, \text{N} = 10 \, \text{N}
\]
\[
T_1 = 10 \, \text{N} + 15 \, \text{N} = 25 \, \text{N}
\]
### Step 6: Find the difference in tensions
Now, we can find the difference in the tensions \( T_1 \) and \( T_2 \):
\[
\text{Difference} = T_1 - T_2 = 25 \, \text{N} - 15 \, \text{N} = 10 \, \text{N}
\]
### Final Answer
The difference in the tensions in the strings connecting A and B and B and C is \( 10 \, \text{N} \).
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