Answer :
To find out how quickly the roller coaster cart will be moving at the bottom of the hill, we can use the principle of conservation of energy. This principle tells us that the total energy at the beginning is equal to the total energy at the end.
1. Identify Energy Types:
At the top of the hill, the cart has:
- Potential Energy (because it's elevated)
At the bottom of the hill, the cart has:
- Kinetic Energy (because it's moving)
2. Set Up the Equation:
The potential energy at the top is converted entirely to kinetic energy at the bottom. Mathematically, this can be expressed as:
[tex]\[
\text{Potential Energy (top)} = \text{Kinetic Energy (bottom)}
\][/tex]
3. Express Each Energy Type:
- Potential Energy (PE) = [tex]\( m \times g \times h \)[/tex]
- [tex]\( m \)[/tex] = mass of the cart = 694 kg
- [tex]\( g \)[/tex] = acceleration due to gravity = 9.81 m/s²
- [tex]\( h \)[/tex] = height of the hill = 49.9 meters
- Kinetic Energy (KE) = [tex]\( \frac{1}{2} \times m \times v^2 \)[/tex]
- [tex]\( v \)[/tex] = velocity at the bottom of the hill, which we are trying to find
4. Substitute and Solve for [tex]\( v \)[/tex]:
[tex]\[
m \times g \times h = \frac{1}{2} \times m \times v^2
\][/tex]
Cancel out the mass ([tex]\( m \)[/tex]) from both sides, since it's non-zero:
[tex]\[
g \times h = \frac{1}{2} \times v^2
\][/tex]
Rearrange to solve for [tex]\( v^2 \)[/tex]:
[tex]\[
v^2 = 2 \times g \times h
\][/tex]
Take the square root to find [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h}
\][/tex]
5. Calculate the Velocity:
Plug in the values:
[tex]\[
v = \sqrt{2 \times 9.81 \times 49.9}
\][/tex]
When you calculate this, you'll find:
[tex]\[
v \approx 31.29 \, \text{m/s}
\][/tex]
6. Choose the Closest Option:
The closest answer to this calculation is option:
d. [tex]\(31.3 \, \text{m/s}\)[/tex]
So, the roller coaster cart will be moving at approximately [tex]\(31.3 \, \text{m/s}\)[/tex] when it reaches the bottom of the hill.
1. Identify Energy Types:
At the top of the hill, the cart has:
- Potential Energy (because it's elevated)
At the bottom of the hill, the cart has:
- Kinetic Energy (because it's moving)
2. Set Up the Equation:
The potential energy at the top is converted entirely to kinetic energy at the bottom. Mathematically, this can be expressed as:
[tex]\[
\text{Potential Energy (top)} = \text{Kinetic Energy (bottom)}
\][/tex]
3. Express Each Energy Type:
- Potential Energy (PE) = [tex]\( m \times g \times h \)[/tex]
- [tex]\( m \)[/tex] = mass of the cart = 694 kg
- [tex]\( g \)[/tex] = acceleration due to gravity = 9.81 m/s²
- [tex]\( h \)[/tex] = height of the hill = 49.9 meters
- Kinetic Energy (KE) = [tex]\( \frac{1}{2} \times m \times v^2 \)[/tex]
- [tex]\( v \)[/tex] = velocity at the bottom of the hill, which we are trying to find
4. Substitute and Solve for [tex]\( v \)[/tex]:
[tex]\[
m \times g \times h = \frac{1}{2} \times m \times v^2
\][/tex]
Cancel out the mass ([tex]\( m \)[/tex]) from both sides, since it's non-zero:
[tex]\[
g \times h = \frac{1}{2} \times v^2
\][/tex]
Rearrange to solve for [tex]\( v^2 \)[/tex]:
[tex]\[
v^2 = 2 \times g \times h
\][/tex]
Take the square root to find [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h}
\][/tex]
5. Calculate the Velocity:
Plug in the values:
[tex]\[
v = \sqrt{2 \times 9.81 \times 49.9}
\][/tex]
When you calculate this, you'll find:
[tex]\[
v \approx 31.29 \, \text{m/s}
\][/tex]
6. Choose the Closest Option:
The closest answer to this calculation is option:
d. [tex]\(31.3 \, \text{m/s}\)[/tex]
So, the roller coaster cart will be moving at approximately [tex]\(31.3 \, \text{m/s}\)[/tex] when it reaches the bottom of the hill.