Question

A roller-coaster is at the top of a 62-meter hill. The car and its passengers have a total mass of 1,088 kilograms. By the time the car reaches the bottom of the hill, its speed is 74 miles per hour (33 meters per second). How much kinetic energy does the car have at the bottom of the hill? KE=1/2mv^2

Answer 1

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

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

where:
- \( m \) is the mass of the roller-coaster car and its passengers,
- \( v \) is the speed of the car at the bottom of the hill.

Given:
- \( m = 1,088 \) kilograms
- \( v = 33 \) meters per second

Now, we can substitute these values into the kinetic energy formula.

\[
KE = \frac{1}{2} \times 1,088 \, \text{kg} \times (33 \, \text{m/s})^2
\]

Calculating \( (33 \, \text{m/s})^2 \):

\[
33^2 = 1089 \, \text{m}^2/\text{s}^2
\]

Now substituting that back into the KE formula:

\[
KE = \frac{1}{2} \times 1,088 \times 1089
\]

Calculating \( \frac{1}{2} \times 1,088 \):

\[
\frac{1}{2} \times 1,088 = 544
\]

Now, we calculate:

\[
KE = 544 \, \text{kg} \times 1089 \, \text{m}^2/\text{s}^2
\]

\[
KE = 592,656 \, \text{kg} \cdot \text{m}^2/\text{s}^2
\]

The kinetic energy at the bottom of the hill is:

\[
KE = 592,656 \, \text{joules}
\]

So, the final answer is:

\[
\text{Kinetic Energy} = 592,656 \, \text{joules}
\]