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 = 1/2 mv^2

Answer 1

To calculate the kinetic energy (KE) of the roller coaster at the bottom of the hill, you can use the formula:

\[
KE = \frac{1}{2} mv^2
\]

Where:
- \( m \) is the mass of the roller coaster (800 kg)
- \( v \) is the speed of the roller coaster (18 m/s)

Now, substituting the values into the formula:

\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times (18 \, \text{m/s})^2
\]

First, calculate \( (18 \, \text{m/s})^2 \):

\[
(18)^2 = 324 \, \text{m}^2/\text{s}^2
\]

Now substitute it back into the kinetic energy formula:

\[
KE = \frac{1}{2} \times 800 \, \text{kg} \times 324 \, \text{m}^2/\text{s}^2
\]

Calculating further:

\[
KE = 400 \times 324
\]
\[
KE = 129600 \, \text{J}
\]

Thus, the kinetic energy of the roller coaster at the bottom of the hill is **129,600 Joules**.