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

1 answer

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

\[ KE = \frac{1}{2} m v^2 \]

where:

  • \( m \) is the mass of the car and passengers, and
  • \( v \) is the velocity of the car at the bottom of the hill.

Given:

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

Now, we can substitute the values into the kinetic energy formula:

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

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

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

Now substitute back into the kinetic energy formula:

\[ KE = \frac{1}{2} \times 1,088 , \text{kg} \times 1,089 , \text{m}^2/\text{s}^2 \]

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

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

Now multiply:

\[ KE = 544 , \text{kg} \times 1,089 , \text{m}^2/\text{s}^2 \]

Calculating the final result:

\[ KE = 592,176 , \text{kg m}^2/\text{s}^2 = 592,176 , \text{J} \]

Therefore, the kinetic energy of the car at the bottom of the hill is approximately:

\[ \boxed{592,176 , \text{J}} \]