To calculate the gravitational potential energy (GPE) of the rollercoaster cart, you can use the formula:
\[ \text{GPE} = m \cdot g \cdot h \]
Where:
- \( m \) is the mass in kilograms (kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)),
- \( h \) is the height in meters (m).
Given:
- \( m = 125 , \text{kg} \)
- \( g = 9.81 , \text{m/s}^2 \)
- \( h = 90 , \text{m} \)
Now, plug the values into the formula:
\[ \text{GPE} = 125 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot 90 , \text{m} \]
\[ \text{GPE} = 125 \cdot 9.81 \cdot 90 \]
\[ \text{GPE} = 125 \cdot 882.9 \]
\[ \text{GPE} \approx 110361.25 , \text{J} \]
So, the gravitational potential energy (GPE) of the cart at a height of 90 meters is approximately 110,361.25 Joules.
Now, for finding the speed (m/s) of the rollercoaster cart as it descends from that height, assuming all the GPE is converted to kinetic energy (K.E.), you can use the kinetic energy formula:
\[ \text{K.E.} = \frac{1}{2} m v^2 \]
Where \( v \) is the velocity in meters per second (m/s).
Setting GPE equal to K.E.:
\[ 110361.25 , \text{J} = \frac{1}{2} \cdot 125 \cdot v^2 \]
Rearranging for \( v \):
\[ 110361.25 = 62.5 \cdot v^2 \]
\[ v^2 = \frac{110361.25}{62.5} \]
\[ v^2 \approx 1765.8 \]
\[ v \approx \sqrt{1765.8} \]
\[ v \approx 42.0 , \text{m/s} \]
Thus, the speed of the rollercoaster cart as it descends from a height of 90 meters is approximately 42.0 m/s.