Problem 6: (17% of Assignment Value)
 Two children push horizontally on a third child on a sled. The first child pushes north with a force of magnitude 37.3N
, while the second child pushes south with a force of magnitude 123.5N
. The sled is moving south, and there is a frictional force of magnitude 17.2N
. The total mass of the child and the sled is 28.35kg
.

This problem includes a video solution that is accessible once the instructor has made problem solutions available for the problem set that includes this one. The video is incorporated at the end of the solution for the last part of the problem.




-   Part (a)
What is the magnitude, in newtons, of the net force on the child-sled system? 
Fnet
 = - 103.4
Fnet
 = −103.4
N
     ✘ Attempts Remain	

      Part (b)
What is the magnitude, in meters per squared second, of the acceleration of the child-sled system? 
a
 = 	
3.6|
m/s2

Let's solve the problem step by step.

### Part (a): Finding the net force \( F_{\text{net}} \)

We are given:
- Force by the first child (north): \( F_1 = 37.3 \, \text{N} \)
- Force by the second child (south): \( F_2 = 123.5 \, \text{N} \)
- Frictional force (south): \( F_f = 17.2 \, \text{N} \)
- The system is moving south, so we assign the south direction as negative.

#### Step 1: Calculate the total forces
Since one force is pushing north and the other is pushing south, the forces will combine considering their directions. The friction force also acts south (negative).

The net force \( F_{\text{net}} \) is calculated as:
\[
F_{\text{net}} = F_2 - F_1 - F_f
\]
Substitute the values:
\[
F_{\text{net}} = 123.5 \, \text{N} - 37.3 \, \text{N} - 17.2 \, \text{N}
\]
\[
F_{\text{net}} = 69.0 \, \text{N}
\]

So, the magnitude of the net force is \( F_{\text{net}} = 69.0 \, \text{N} \), and it acts in the south direction.

---

### Part (b): Finding the acceleration \( a \)

We are given:
- The total mass of the child and sled system: \( m = 28.35 \, \text{kg} \)
- The net force from part (a): \( F_{\text{net}} = 69.0 \, \text{N} \)

We can use Newton's second law to calculate the acceleration:
\[
F_{\text{net}} = m \cdot a
\]
Rearranging for \( a \):
\[
a = \frac{F_{\text{net}}}{m}
\]
Substitute the known values:
\[
a = \frac{69.0 \, \text{N}}{28.35 \, \text{kg}}
\]
\[
a \approx 2.43 \, \text{m/s}^2
\]

Thus, the magnitude of the acceleration is approximately \( a = 2.43 \, \text{m/s}^2 \).

---

### Summary:
- **Part (a)**: The net force on the system is \( 69.0 \, \text{N} \) towards the south.
- **Part (b)**: The magnitude of the acceleration is \( 2.43 \, \text{m/s}^2 \).

Would you like more details on any part of the solution, or clarification on any step?

Here are 5 related questions for further exploration:
1. What happens to the net force if the friction increases?
2. How would the acceleration change if the mass of the sled increases?
3. What is the direction of acceleration if the forces are perfectly balanced?
4. How does friction affect the motion of the sled in this scenario?
5. If the second child stops pushing, what would be the new net force?

**Tip:** When combining forces, always ensure you account for their directions accurately, using positive or negative signs.

Two children push horizontally on a third child on a sled. The first child pushes north with a force of magnitude 37.3N, while the second child pushes south with a force of magnitude 123.5N. The sled is moving south, and there is a frictional force of magnitude 17.2N. The total mass of the child and the sled is 28.35kg. Part (a) What is the magnitude, in newtons, of the net force on the child-sled system? Part (b) What is the magnitude, in meters per squared second, of the acceleration of the child-sled system?

Solve a physics problem involving two children pushing on a third child on a sled. Learn how to calculate the net force and acceleration of the system using Newton's second law and force vector addition. Suitable for Grades 9-12 students.

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