A 250.0 kg roller coaster has 2.000 x 10^4 Joule of potential energy at the top of the hill , if the track is frictionless , what is the coasters velocity at the bottom of the hill?

To find the coaster's velocity at the bottom of the hill, we can apply the principle of conservation of energy. According to this principle, the total mechanical energy of an object remains constant as long as no external forces are acting on it.

In this case, the roller coaster only has potential energy at the top of the hill. As it moves down the hill, this potential energy is converted into kinetic energy. So, the equation we can use to solve this problem is:

Potential Energy (PE) = Kinetic Energy (KE)

The potential energy at the top of the hill is given as 2.000 x 10^4 Joules.

The kinetic energy at the bottom of the hill can be calculated using the formula:

KE = 1/2 * m * v^2

Where:
KE = Kinetic energy
m = Mass of the roller coaster
v = Velocity of the roller coaster

Since the roller coaster has a mass of 250.0 kg, we can substitute these values into the equation:

2.000 x 10^4 J = 1/2 * 250.0 kg * v^2

Now, we can solve for v:

First, multiply both sides of the equation by 2 to cancel out the 1/2 factor:

2 * 2.000 x 10^4 J = 250.0 kg * v^2

4.000 x 10^4 J = 250.0 kg * v^2

Next, divide both sides of the equation by 250.0 kg to isolate v^2:

(4.000 x 10^4 J) / (250.0 kg) = v^2

Simplify the right side of the equation:

160 J/kg = v^2

Finally, take the square root of both sides to solve for v:

v = √(160 J/kg)

v ≈ 12.65 m/s

Therefore, the coaster's velocity at the bottom of the hill is approximately 12.65 m/s.