Question
The following picture shows a roller coaster car travelling down a track.
1
m = 100 kg
v1=0 m/s
h = 25 m
2
v2 = ?
h = 10 m
The mass of the roller coaster car is 100 kg.
The velocity of the car at point 1 is v = 0 m/s.
Using the Law of Conservation of Mechanical Energy, calculate the velocity of the car at point 2.
Assume no frictional forces. Use 9.81 m/s^2 as the acceleration due to gravity. You may use your
equation sheet to answer this problem.
answers:
17.2 m/s
25.1 m/s
9.89 m/s
12.4 m/s
1
m = 100 kg
v1=0 m/s
h = 25 m
2
v2 = ?
h = 10 m
The mass of the roller coaster car is 100 kg.
The velocity of the car at point 1 is v = 0 m/s.
Using the Law of Conservation of Mechanical Energy, calculate the velocity of the car at point 2.
Assume no frictional forces. Use 9.81 m/s^2 as the acceleration due to gravity. You may use your
equation sheet to answer this problem.
answers:
17.2 m/s
25.1 m/s
9.89 m/s
12.4 m/s
Answers
GPT-4o mini
To find the velocity of the roller coaster car at point 2 using the Law of Conservation of Mechanical Energy, we start by noting that the total mechanical energy is conserved when there are no non-conservative forces like friction.
At point 1 (the top of the track where the height is 25 m), the mechanical energy consists of gravitational potential energy (PE) since the velocity is 0 m/s (kinetic energy (KE) is 0). The gravitational potential energy is given by the equation:
\[
PE = mgh
\]
Where:
- \( m = 100 \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( h = 25 \, \text{m} \)
Calculating the potential energy at point 1:
\[
PE_1 = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 25 \, \text{m} = 24525 \, \text{J}
\]
At point 2 (at height 10 m), the mechanical energy will consist of both potential energy and kinetic energy.
Calculating the potential energy at point 2:
\[
PE_2 = mgh = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{m} = 9810 \, \text{J}
\]
Now, using conservation of mechanical energy:
\[
PE_1 = PE_2 + KE_2
\]
Where \( KE_2 \) (kinetic energy at point 2) is given by:
\[
KE = \frac{1}{2} mv^2
\]
So we can write:
\[
24525 \, \text{J} = 9810 \, \text{J} + \frac{1}{2} \cdot 100 \, \text{kg} \cdot v_2^2
\]
Now, rearranging the equation to solve for \( v_2 \):
\[
24525 \, \text{J} - 9810 \, \text{J} = \frac{1}{2} \cdot 100 \, \text{kg} \cdot v_2^2
\]
Calculating the left side:
\[
14715 \, \text{J} = \frac{100}{2} v_2^2
\]
\[
14715 \, \text{J} = 50 v_2^2
\]
Now, divide both sides by 50:
\[
v_2^2 = \frac{14715 \, \text{J}}{50} = 294.3
\]
Finally, take the square root to find \( v_2 \):
\[
v_2 = \sqrt{294.3} \approx 17.2 \, \text{m/s}
\]
Thus, the velocity of the car at point 2 is **17.2 m/s**.
At point 1 (the top of the track where the height is 25 m), the mechanical energy consists of gravitational potential energy (PE) since the velocity is 0 m/s (kinetic energy (KE) is 0). The gravitational potential energy is given by the equation:
\[
PE = mgh
\]
Where:
- \( m = 100 \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( h = 25 \, \text{m} \)
Calculating the potential energy at point 1:
\[
PE_1 = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 25 \, \text{m} = 24525 \, \text{J}
\]
At point 2 (at height 10 m), the mechanical energy will consist of both potential energy and kinetic energy.
Calculating the potential energy at point 2:
\[
PE_2 = mgh = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{m} = 9810 \, \text{J}
\]
Now, using conservation of mechanical energy:
\[
PE_1 = PE_2 + KE_2
\]
Where \( KE_2 \) (kinetic energy at point 2) is given by:
\[
KE = \frac{1}{2} mv^2
\]
So we can write:
\[
24525 \, \text{J} = 9810 \, \text{J} + \frac{1}{2} \cdot 100 \, \text{kg} \cdot v_2^2
\]
Now, rearranging the equation to solve for \( v_2 \):
\[
24525 \, \text{J} - 9810 \, \text{J} = \frac{1}{2} \cdot 100 \, \text{kg} \cdot v_2^2
\]
Calculating the left side:
\[
14715 \, \text{J} = \frac{100}{2} v_2^2
\]
\[
14715 \, \text{J} = 50 v_2^2
\]
Now, divide both sides by 50:
\[
v_2^2 = \frac{14715 \, \text{J}}{50} = 294.3
\]
Finally, take the square root to find \( v_2 \):
\[
v_2 = \sqrt{294.3} \approx 17.2 \, \text{m/s}
\]
Thus, the velocity of the car at point 2 is **17.2 m/s**.