Question
A 440 kg roller coaster car is going 26 m/s when it reaches the lowest point on the track. If the car started from rest at the top of a hill, how much higher was that point on the track than the lowest point? (Use g = 9.80 m/s2, and ignore friction.)
17 m
23 m
34 m
69 m
17 m
23 m
34 m
69 m
Answers
GPT-4o mini
To solve this problem, we can use the principle of conservation of energy. The total mechanical energy at the top of the hill (potential energy) must equal the total mechanical energy at the lowest point of the track (a combination of kinetic and potential energy).
1. At the top of the hill:
- The car has potential energy (PE) due to its height and no kinetic energy (KE) since it starts from rest.
- The potential energy can be calculated using the formula:
\[
PE = mgh
\]
where \( m \) is the mass of the car, \( g \) is the acceleration due to gravity, and \( h \) is the height above the lowest point.
2. At the lowest point of the track:
- The car has kinetic energy due to its speed, and its potential energy is zero since we are measuring height from this point.
- The kinetic energy can be calculated using the formula:
\[
KE = \frac{1}{2} mv^2
\]
where \( v \) is the speed of the car at the lowest point.
Now, we set the potential energy at the top equal to the kinetic energy at the lowest point:
\[
mgh = \frac{1}{2} mv^2
\]
We can cancel \( m \) from both sides (since it's common and not zero):
\[
gh = \frac{1}{2} v^2
\]
Now, we can solve for \( h \):
\[
h = \frac{v^2}{2g}
\]
Plugging in the values:
- \( v = 26 \, \text{m/s} \)
- \( g = 9.80 \, \text{m/s}^2 \)
Now we calculate \( h \):
\[
h = \frac{(26)^2}{2 \times 9.80} = \frac{676}{19.6} \approx 34.4 \, \text{m}
\]
So, the height \( h \) that the car started from, relative to the lowest point, is approximately 34.4 m.
Among the given options, the closest answer is:
**34 m**
1. At the top of the hill:
- The car has potential energy (PE) due to its height and no kinetic energy (KE) since it starts from rest.
- The potential energy can be calculated using the formula:
\[
PE = mgh
\]
where \( m \) is the mass of the car, \( g \) is the acceleration due to gravity, and \( h \) is the height above the lowest point.
2. At the lowest point of the track:
- The car has kinetic energy due to its speed, and its potential energy is zero since we are measuring height from this point.
- The kinetic energy can be calculated using the formula:
\[
KE = \frac{1}{2} mv^2
\]
where \( v \) is the speed of the car at the lowest point.
Now, we set the potential energy at the top equal to the kinetic energy at the lowest point:
\[
mgh = \frac{1}{2} mv^2
\]
We can cancel \( m \) from both sides (since it's common and not zero):
\[
gh = \frac{1}{2} v^2
\]
Now, we can solve for \( h \):
\[
h = \frac{v^2}{2g}
\]
Plugging in the values:
- \( v = 26 \, \text{m/s} \)
- \( g = 9.80 \, \text{m/s}^2 \)
Now we calculate \( h \):
\[
h = \frac{(26)^2}{2 \times 9.80} = \frac{676}{19.6} \approx 34.4 \, \text{m}
\]
So, the height \( h \) that the car started from, relative to the lowest point, is approximately 34.4 m.
Among the given options, the closest answer is:
**34 m**