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The image you uploaded contains a physics problem involving two blocks, their masses, and forces acting on them. The questions specifically ask about the relationships between the tensions \( T_1 \) and \( T_2 \), and the friction forces \( f_1 \) and \( f_2 \).

Here’s a summary and analysis of the two questions:

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**Question 9:**

You are asked to determine the relationship between the tensions \( T_1 \) and \( T_2 \).

- \( M \) is the mass of the top block, which is pulled to the left with tension \( T_1 \).
- \( 3M \) is the mass of the bottom block, pulled to the right with tension \( T_2 \).
- The blocks are moving together, indicating they are in equilibrium relative to each other.

Given this setup, since the blocks are not accelerating relative to each other, the forces must be balanced. The correct answer appears to be:

\[
\boxed{A} \quad T_1 = T_2
\]

This means the tensions on both the blocks must be equal for them to move together at constant velocity.

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**Question 10:**

In **Question 10**, you are asked about the relationship between the friction forces \( f_1 \) (force of friction on the top block) and \( f_2 \) (force of friction on the bottom block). 

The key information includes:
- \( f_1 \) is the friction force experienced by the top block, mass \( M \), which is in contact with the bottom block.
- \( f_2 \) is the friction force experienced by the bottom block, mass \( 3M \), which is in contact with the ground.
- The coefficient of friction between the blocks is \( \mu \), and the surface below the bottom block is frictionless.

### Analysis:
- The friction force \( f_1 \) for the top block is proportional to the normal force between the two blocks. The normal force here is \( M \cdot g \), where \( g \) is the acceleration due to gravity.
  
- For the bottom block, the normal force is due to the combined weight of both blocks, so the total normal force acting on the bottom block is \( (M + 3M)g = 4Mg \).

Since friction \( f \) is proportional to the normal force (\( f = \mu N \)), we can express the friction forces as:
- \( f_1 = \mu M g \)
- \( f_2 = \mu (4M) g = 4 \mu M g \)

Thus, \( f_2 = 4 f_1 \), leading to the correct relationship:
\[
\boxed{C} \quad f_1 = \frac{1}{3} f_2
\]

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Would you like more details on any part of the solution or have other questions?

Here are 5 additional questions to extend this topic:
1. How would the solution change if the surface beneath the bottom block was not frictionless?
2. What would happen if the coefficient of friction \( \mu \) was zero?
3. How would the system behave if the mass of the top block was doubled?
4. How do you calculate the net force acting on both blocks if they were accelerating?
5. How does the direction of motion affect the tensions and friction forces?

**Tip:** In systems like this, always check whether the forces acting on all objects are balanced. This helps determine if the system is moving at constant speed or accelerating.

Two blocks are stacked on top of each other. The top block of mass M is attached to a rope that pulls to the left with tension T1. The bottom block of mass 3M is attached to a rope that pulls to the right with tension T2. The coefficient of friction between the blocks is μ. The surface beneath the bottom block is frictionless. You need to determine the relationship between the tensions T1 and T2 and the friction forces f1 and f2.

Solve for the relationships between tension forces T1 and T2, and friction forces f1 and f2, for two blocks on a frictionless surface. Learn how Newton's laws and friction laws apply. Suitable for students in Grades 11-12.

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