250 kg roller coaster car has 20000 J of potential energy at the top hill what is velocity bottom hill?

Explanation:

The conservation of energy principle states that the total mechanical energy of a closed system remains constant, regardless of the changes that occur within the system. In this case, the closed system is the roller coaster car and the track.

At the top of the hill, the roller coaster car has only potential energy, given by:

$$PE = mgh$$

where:

- PE is the potential energy in joules (J)

- m is the mass of the car in kilograms (kg)

- g is the acceleration due to gravity (9.8 m/s²)

- h is the height of the hill in meters (m)

At the bottom of the hill, the roller coaster car has only kinetic energy, given by:

$$KE = (1/2)mv^2$$

where:

- KE is the kinetic energy in joules (J)

- m is the mass of the car in kilograms (kg)

- v is the velocity of the car in meters per second (m/s)

Since the total mechanical energy of the system remains constant, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill:

$$mgh = (1/2)mv^2$$

Solving for v, we get:

$$v = \sqrt{2gh}$$

Substituting the given values, we get:

$$v = \sqrt{2(9.8 m/s²)(20 m)} = 22.4 m/s$$

Therefore, the velocity of the roller coaster car at the bottom of the hill is 22.4 m/s.