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= 1/2 mv^2
First, we need to calculate the potential energy at the top of the hill, which is equal to the mass times the gravitational acceleration constant (9.8 m/s^2) times the height of the hill:
PE = mgh
PE = 1088 kg * 9.8 m/s^2 * 62 m
PE = 650,743.2 J
Next, we can calculate the kinetic energy at the bottom of the hill using the formula KE = 1/2 * mv^2:
KE = 1/2 * 1088 kg * (33 m/s)^2
KE = 1/2 * 1088 kg * 1089 m^2/s^2
KE = 593,472 J
Therefore, the kinetic energy of the car at the bottom of the hill is 593,472 Joules.