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
2,978,944 J
2,978,944 J
1,184,832 J
1,184,832 J
592,416 J
592,416 J
17,952 J
To find the kinetic energy of the roller-coaster at the bottom of the hill, we can use the equation:
KE = 1/2 * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
Given:
Mass (m) = 1088 kg
Velocity (v) = 33 m/s
Plugging in these values into the equation, we have:
KE = 1/2 * 1088 kg * (33 m/s)^2
KE ≈ 592,416 J
Therefore, the correct answer is 592,416 J.