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



Responses

592,416 J

1,184,832 J

17,952 J

2,978,944 J

1 answer

The formula for kinetic energy (KE) is given by:

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

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, we can substitute the values into the formula:

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

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

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

Now plug that 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 the product:

\[ KE = 544 , \text{kg} \times 1,089 , \text{m}^2/\text{s}^2 = 592,416 , \text{J} \]

Thus, the kinetic energy of the car at the bottom of the hill is:

\[ \text{KE} = 592,416 , \text{J} \]

The correct response is:

592,416 J