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
592,416 J
592,416 J
17,952 J
17,952 J
2,978,944 J
2,978,944 J
1,184,832 J
To calculate the kinetic energy of the car at the bottom of the hill, we can use the formula:
KE = 1/2 * m * v^2
Where:
- KE represents the kinetic energy,
- m represents the mass of the car and passengers, and
- v represents the speed of the car.
Plugging in the values given:
m = 1,088 kg
v = 33 m/s
KE = 1/2 * 1,088 kg * (33 m/s)^2
KE = 1/2 * 1,088 kg * 1,089 m^2/s^2
KE = 593,792 J
Rounded to the nearest whole number, the kinetic energy of the car at the bottom of the hill is 592,416 J.
To calculate the kinetic energy of the car at the bottom of the hill, we can use the formula:
KE = 1/2 * m * v^2
Where:
KE = Kinetic energy
m = Mass of the car and passengers
v = Velocity (speed) of the car
Given:
Mass (m) = 1,088 kg
Velocity (v) = 33 m/s
Now, let's substitute the given values into the formula:
KE = 1/2 * 1,088 kg * (33 m/s)^2
Using the formula, we can calculate the kinetic energy:
KE = 1/2 * 1,088 kg * 1,089 m^2/s^2
KE = 596,896 Joules (J)
Therefore, the kinetic energy of the car at the bottom of the hill is approximately 596,896 J.
To calculate the kinetic energy of the roller-coaster at the bottom of the hill, we can use the formula:
KE = 1/2 * m * v^2
Where:
m = total mass of the car and passengers = 1,088 kg
v = velocity of the car = 33 m/s
Plugging in the values, we have:
KE = 1/2 * 1088 * (33)^2
= 1/2 * 1088 * 1089
= 592,416 J
Therefore, the kinetic energy of the car at the bottom of the hill is 592,416 Joules.