Question

A roller-coaster is at the top of a 54-meter hill. The car and its passengers have a total mass of 1,063 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? (1 point)
Responses
KE=1/2mv^2
592,416.0 J
592,416.0 J

578,803.5 J
578,803.5 J

29,403.0 J
29,403.0 J

2,910,494.0 J

Answer 1

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

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

where:
- \( m \) is the mass (in kilograms),
- \( v \) is the velocity (in meters per second).

Given:
- Mass, \( m = 1,063 \) kg
- Velocity, \( v = 33 \) m/s

Now, plug in the values:

\[
KE = \frac{1}{2} \times 1,063 \, \text{kg} \times (33 \, \text{m/s})^2
\]

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

\[
(33)^2 = 1,089 \, \text{m}^2/\text{s}^2
\]

Now, substitute this back into the equation:

\[
KE = \frac{1}{2} \times 1,063 \times 1,089
\]

Calculating \( \frac{1}{2} \times 1,063 = 531.5 \):

Now, multiply:

\[
KE = 531.5 \times 1,089
\]

This equals:

\[
KE \approx 578,803.5 \, \text{J}
\]

So, the kinetic energy at the bottom of the hill is approximately **578,803.5 J**.

Therefore, the correct response is:

**578,803.5 J**.