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=12mv2


(1 point)
Responses

592,416 J
592,416 J

1,184,832 J
1,184,832 J

17,952 J
17,952 J

2,978,944 J
2,978,944 J

Answer 1

To calculate 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:
- \( m = 1,088 \) kg (mass of the car and passengers)
- \( v = 33 \) m/s (speed at the bottom of the hill)

Now, substituting the values into the formula:

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

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

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

Now substitute this back into the kinetic energy formula:

\[
KE = \frac{1}{2} \times 1,088 \times 1,089
\]

Calculating further:

\[
KE = 544 \times 1,089
\]
\[
KE = 592,416 \, \text{J}
\]

So the kinetic energy of the car at the bottom of the hill is:

**592,416 J**