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
K
E
=
1
2
m
v
2
(1 point)
Responses
2,978,944 J
2,978,944 J
592,416 J
592,416 J
1,184,832 J
1,184,832 J
17,952 J
To calculate the kinetic energy of the car 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 of the car and its passengers, and v is the velocity of the car.
Given:
Mass (m) = 1,088 kg
Velocity (v) = 33 m/s
Plugging in the values:
KE = 1/2 * 1088 kg * (33 m/s)^2
= 0.5 * 1088 kg * 1089 m^2/s^2
≈ 592,416 J
Therefore, the car has approximately 592,416 Joules of kinetic energy at the bottom of the hill.