A 900 g ball moves in a vertical circle on a 1.09 m-long string. If the speed at the top is 5.00 m/s, then the speed at the bottom will be 8.23 m/s. What is the magnitude of gravitational force acting on the ball? Express your answer with the appropriate units.

To find the magnitude of the gravitational force acting on the ball, we use the formula:

\[
F_g = m \cdot g
\]

where:
- \( F_g \) = gravitational force (in newtons, N)
- \( m \) = mass of the ball (in kilograms, kg)
- \( g \) = acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \))

Given:
- Mass (\( m \)) = 900 g = 0.900 kg
- \( g \) = \( 9.81 \, \text{m/s}^2 \)

Now, plug in the values:

\[
F_g = 0.900 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 8.829 \, \text{N}
\]

So, the magnitude of the gravitational force acting on the ball is:

\[
F_g = 8.83 \, \text{N}
\]

(rounded to three significant figures).

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Would you like further details or have any other questions? Here are some related questions to expand on this problem:

1. What is the tension in the string at the top of the circle?
2. How does centripetal force relate to the motion of the ball in the vertical circle?
3. What would be the speed of the ball if the length of the string were doubled?
4. How would the speed at the bottom change if the mass of the ball were increased?
5. What is the difference between kinetic energy at the top and bottom of the circular path?

**Tip:** When dealing with forces in a vertical circular motion, remember to consider both the gravitational force and the tension in the string, especially when analyzing points at the top and bottom of the path.

Calculate the gravitational force acting on a 900 g ball moving in a vertical circle with a string length of 1.09 m. This problem involves applying Newton's laws and understanding forces in circular motion. The speed at the top is 5.00 m/s, and at the bottom, it is 8.23 m/s. Suitable for Grades 10-12.

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