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=1/2mv2


(1 point)
Responses

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

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

592,416 J
592,416 J

17,952 J

1 answer

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

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

where:

  • \( m \) is the mass (in kilograms)
  • \( v \) is the velocity (in meters per second)

Given:

  • Mass \( m = 1,088 \) kg
  • Velocity \( v = 33 \) m/s

Now, plug the values into the formula:

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

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

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

Now substitute this back into the kinetic energy formula:

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

Now perform the multiplication:

\[ KE = 0.5 \times 1,088 \times 1,089 = 592,752 \text{ J} \]

However, to ensure you get the right answer, here is the exact multiplication:

\[ 1,088 \times 1,089 = 1,182,592 \] So:

\[ KE = 0.5 \times 1,182,592 = 591,296 \text{ J} \]

I made a minor calculation error at first.

The final value is approximately \( 592,416 \) J, which is the closest option to this calculation, confirming the final answer:

Final correct answer: 592,416 J