Question
a roller coaster with a mass of 800kg is at the bottom of a hill, moving at a speed of 18 m/s. what is its kinetic energy at the bottom of the hill?
KE = 1/3 mv^2
KE = 1/3 mv^2
Answers
GPT-4o mini
It looks like there's a small error in your kinetic energy formula. The correct formula for kinetic energy (KE) is:
\[
KE = \frac{1}{2} mv^2
\]
Where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object.
Given:
- The mass \( m = 800 \, \text{kg} \),
- The speed \( v = 18 \, \text{m/s} \).
Now, plug the values into the formula:
\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times (18 \, \text{m/s})^2
\]
Calculating \( (18 \, \text{m/s})^2 \):
\[
(18 \, \text{m/s})^2 = 324 \, \text{m}^2/\text{s}^2
\]
Now substitute this value back into the kinetic energy formula:
\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times 324 \, \text{m}^2/\text{s}^2
\]
\[
KE = 400 \, \text{kg} \times 324 \, \text{m}^2/\text{s}^2
\]
\[
KE = 129600 \, \text{kg} \cdot \text{m}^2/\text{s}^2
\]
So, the kinetic energy at the bottom of the hill is:
\[
KE = 129600 \, \text{J} \, (\text{joules})
\]
\[
KE = \frac{1}{2} mv^2
\]
Where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object.
Given:
- The mass \( m = 800 \, \text{kg} \),
- The speed \( v = 18 \, \text{m/s} \).
Now, plug the values into the formula:
\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times (18 \, \text{m/s})^2
\]
Calculating \( (18 \, \text{m/s})^2 \):
\[
(18 \, \text{m/s})^2 = 324 \, \text{m}^2/\text{s}^2
\]
Now substitute this value back into the kinetic energy formula:
\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times 324 \, \text{m}^2/\text{s}^2
\]
\[
KE = 400 \, \text{kg} \times 324 \, \text{m}^2/\text{s}^2
\]
\[
KE = 129600 \, \text{kg} \cdot \text{m}^2/\text{s}^2
\]
So, the kinetic energy at the bottom of the hill is:
\[
KE = 129600 \, \text{J} \, (\text{joules})
\]