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?
K E = 1/2mv^2
To calculate the kinetic energy of the car at the bottom of the hill, we can use the formula for kinetic energy:
KE = 1/2mv^2
Where:
- KE is the kinetic energy
- m is the mass of the car and its passengers
- v is the velocity of the car
Given:
- m = 1,088 kg
- v = 33 m/s
Plugging in the values into the formula, we have:
KE = 1/2 * (1,088 kg) * (33 m/s)^2
KE = 1/2 * 1,088 kg * (1,089 m^2/s^2)
KE = 1/2 * 1,088 kg * 1,089 m^2/s^2
KE ≈ 594,064 J
Therefore, the car has approximately 594,064 joules of kinetic energy at the bottom of the hill.