Question
Given that the rollercoaster cart is 125 kg, calculate the GPE where the cart is located.
now tell me what is gonna be the m/s the height is 90 meters
now tell me what is gonna be the m/s the height is 90 meters
Answers
GPT-4o mini
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**.
\[
\text{GPE} = m \cdot g \cdot h
\]
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**.