Answer :
To find the acceleration of gravity when a 15 kg falling rock produces a force of 147 N, we can use Newton's second law of motion. This law tells us that force equals mass times acceleration, written as the formula:
[tex]\[ F = m \cdot a \][/tex]
Where:
- [tex]\( F \)[/tex] is the force in newtons (N),
- [tex]\( m \)[/tex] is the mass in kilograms (kg),
- [tex]\( a \)[/tex] is the acceleration in meters per second squared (m/s²).
In this situation:
- The force [tex]\( F \)[/tex] exerted by the falling rock is 147 N.
- The mass [tex]\( m \)[/tex] of the rock is 15 kg.
We need to solve for the acceleration [tex]\( a \)[/tex]. Rearrange the formula to isolate [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F}{m} \][/tex]
Plug in the given values:
[tex]\[ a = \frac{147 \, \text{N}}{15 \, \text{kg}} \][/tex]
When we perform the division, we find:
[tex]\[ a = 9.8 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of gravity is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]. This is a typical value for the acceleration due to gravity on Earth's surface.
[tex]\[ F = m \cdot a \][/tex]
Where:
- [tex]\( F \)[/tex] is the force in newtons (N),
- [tex]\( m \)[/tex] is the mass in kilograms (kg),
- [tex]\( a \)[/tex] is the acceleration in meters per second squared (m/s²).
In this situation:
- The force [tex]\( F \)[/tex] exerted by the falling rock is 147 N.
- The mass [tex]\( m \)[/tex] of the rock is 15 kg.
We need to solve for the acceleration [tex]\( a \)[/tex]. Rearrange the formula to isolate [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F}{m} \][/tex]
Plug in the given values:
[tex]\[ a = \frac{147 \, \text{N}}{15 \, \text{kg}} \][/tex]
When we perform the division, we find:
[tex]\[ a = 9.8 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of gravity is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]. This is a typical value for the acceleration due to gravity on Earth's surface.
