A 4.4-cm-diameter, 24 g plastic ball is attached to a 1.2-m-long string and swung in a vertical circle. The ball's speed is 5.3 m/s at the point where it is moving straight up. What is the magnitude of the net force on the ball? Air resistance is not negligible.

To solve this problem, we need to find the net force acting on the ball when it's at the point where it is moving straight up.

### Step 1: Define the forces acting on the ball
At the top of the vertical circle, there are two forces acting on the ball:
1. **Gravitational force (weight)**: This is the force due to gravity, directed downward.
   \[
   F_{\text{gravity}} = mg
   \]
   where:
   - \( m \) is the mass of the ball (in kg),
   - \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).

2. **Tension in the string**: The string applies a force that acts along the string, which at this point is directed upward.

3. **Centripetal force**: The net force on the ball must also provide the centripetal force to keep the ball moving in a circular path. This force is directed towards the center of the circle and is given by:
   \[
   F_{\text{centripetal}} = \frac{mv^2}{r}
   \]
   where:
   - \( v \) is the speed of the ball at the top (5.3 m/s),
   - \( r \) is the radius of the circle (the length of the string, 1.2 m).

### Step 2: Calculate the gravitational force
First, convert the mass of the ball to kilograms:
\[
m = 24 \, \text{g} = 0.024 \, \text{kg}
\]
Now, calculate the gravitational force:
\[
F_{\text{gravity}} = mg = 0.024 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.2352 \, \text{N}
\]

### Step 3: Calculate the centripetal force
Now calculate the centripetal force needed to keep the ball moving in a circle:
\[
F_{\text{centripetal}} = \frac{mv^2}{r} = \frac{0.024 \, \text{kg} \times (5.3 \, \text{m/s})^2}{1.2 \, \text{m}}
\]
\[
F_{\text{centripetal}} = \frac{0.024 \times 28.09}{1.2} = \frac{0.67416}{1.2} = 0.5618 \, \text{N}
\]

### Step 4: Calculate the net force
The net force is the vector sum of the forces acting on the ball. At the top of the circle, the tension in the string must counteract the gravitational force and provide enough additional force to create the centripetal acceleration. However, the direction of gravity and centripetal acceleration are the same (both downward), so we can add the two forces directly:

\[
F_{\text{net}} = F_{\text{gravity}} + F_{\text{centripetal}} = 0.2352 \, \text{N} + 0.5618 \, \text{N} = 0.797 \, \text{N}
\]

Therefore, the magnitude of the net force on the ball is approximately **0.797 N**.

### Final answer:
The magnitude of the net force on the ball is **0.797 N**.

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Would you like further clarification on any step, or a more detailed explanation? Here are some related questions to expand your understanding:

1. How does air resistance affect the net force in a vertical circle?
2. What is the role of tension at different points in the circular motion?
3. How would the net force change if the ball were moving faster at the top?
4. What happens to the net force at the bottom of the circular motion?
5. How do we calculate the tension in the string at any point during the circular motion?

**Tip:** The direction of the net force in circular motion is always towards the center of the circle, and its magnitude is influenced by both gravitational and centripetal forces.

Discover how to calculate the net force acting on a plastic ball (24 g) attached to a 1.2-m-long string and swung in a vertical circle with a speed of 5.3 m/s. This problem covers key physics concepts like centripetal force, gravitational force, and circular motion. Suitable for Grades 10-12.

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