The force of gravity between the moon and an object near its surface is much smaller than the force of gravity between Earth and the same object near Earth’s surface. A bowling ball with a mass of 7.51 kg is pulled downward with an unbalanced force of 12.2 N. What is the acceleration of the falling bowling ball on the moon?

Answer 1

To find the acceleration of the bowling ball on the Moon, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

\[ F = m \cdot a \]

Where:

• \( F \) is the net force acting on the object (in Newtons),
• \( m \) is the mass of the object (in kilograms),
• \( a \) is the acceleration (in meters per second squared).

In this case, we know:

• The net force \( F = 12.2 , \text{N} \)
• The mass of the bowling ball \( m = 7.51 , \text{kg} \)

We can rearrange the formula to solve for acceleration:

\[ a = \frac{F}{m} \]

Substituting the known values:

\[ a = \frac{12.2 , \text{N}}{7.51 , \text{kg}} \approx 1.63 , \text{m/s}^2 \]

Therefore, the acceleration of the falling bowling ball on the Moon is approximately 1.63 m/s².