https://file.aiquickdraw.com/mathbot/file/17268480781476d4r611s.jpg

Let's analyze the problem shown in the image.

The block has a mass of \( 10 \, \text{kg} \) and is being pulled along a flat surface. The forces acting on the block are:
- A pulling force of \( 30 \, \text{N} \) to the right.
- A frictional force of \( 10 \, \text{N} \) opposing the motion to the left.

To find the acceleration of the block, we will use **Newton's second law of motion**:

\[
F_{\text{net}} = m \cdot a
\]

Where:
- \( F_{\text{net}} \) is the net force acting on the block,
- \( m \) is the mass of the block, and
- \( a \) is the acceleration.

### Step 1: Calculate the net force
The net force is the difference between the applied force and the frictional force:

\[
F_{\text{net}} = F_{\text{pull}} - F_{\text{friction}} = 30 \, \text{N} - 10 \, \text{N} = 20 \, \text{N}
\]

### Step 2: Apply Newton's second law
Now, we can solve for acceleration:

\[
a = \frac{F_{\text{net}}}{m} = \frac{20 \, \text{N}}{10 \, \text{kg}} = 2.0 \, \text{m/s}^2
\]

### Final Answer:
The acceleration of the block is \( \boxed{2.0 \, \text{m/s}^2} \), which corresponds to option D.

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Would you like more details on any part of the solution?

Here are 5 related questions for further practice:
1. How would the acceleration change if the pulling force increased to 40 N?
2. What would happen to the acceleration if the frictional force increased to 15 N?
3. How would the problem change if the surface was inclined?
4. How is the acceleration affected if the mass of the block is doubled?
5. What would the acceleration be if there were no friction?

**Tip:** Always ensure that the forces are balanced correctly by paying attention to the direction when calculating net force.

The diagram below shows a block that is being pulled along a flat surface. What is the acceleration of the block?

Learn how to apply Newton's Second Law of Motion to calculate the acceleration of a block with a mass of 10 kg. Given a pulling force of 30 N and a frictional force of 10 N, this problem provides a step-by-step explanation, perfect for students in Grades 8-10.

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