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 1,184,832 J 1,184,832 J 17,952 J 17,952 J 2,978,944 J 2,978,944 J 592,416 J

1 answer

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

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

where:

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

Given:

  • \( m = 1088 , \text{kg} \)
  • \( v = 33 , \text{m/s} \)

Now, substitute the values into the equation:

\[ KE = \frac{1}{2} \times 1088 , \text{kg} \times (33 , \text{m/s})^2 \]

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

\[ (33 , \text{m/s})^2 = 1089 , \text{m}^2/\text{s}^2 \]

Now plug that back into the kinetic energy equation:

\[ KE = \frac{1}{2} \times 1088 , \text{kg} \times 1089 , \text{m}^2/\text{s}^2 \]

Calculating this step-by-step:

  1. Multiply \( 1088 \) kg by \( 1089 \) m²/s²:

\[ 1088 \times 1089 = 1,180,992 , \text{kg} \cdot \text{m}^2/\text{s}^2 \]

  1. Then multiply by \( \frac{1}{2} \):

\[ KE = \frac{1}{2} \times 1,180,992 , \text{kg} \cdot \text{m}^2/\text{s}^2 = 590,496 , \text{J} \]

Therefore, the correct answer is:

\[ KE = 590,496 , \text{J} \]

Among the options provided, none of them match this exact value. Please double-check the calculations or the available answer choices. However, based on the calculation above, the kinetic energy at the bottom is \( 590,496 , \text{J} \).