Find the maximum velocity of a roller coaster with a mass of 3,320 kg when it reaches the bottom of a 83 meter drop. (1 point)
To find the maximum velocity of the roller coaster, we can use the principle of conservation of energy.
At the top of the drop, the coaster only has potential energy. As it descends, this potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the top to the kinetic energy at the bottom.
Potential energy (PE) = mgh
where m = mass, g = acceleration due to gravity, and h = height
Kinetic energy (KE) = (1/2)mv^2
where v = velocity
Setting the two equations equal, we have:
mgh = (1/2)mv^2
Canceling out the mass:
gh = (1/2)v^2
Rearranging the equation:
v^2 = 2gh
Substituting in the values given in the question:
v^2 = 2 * 9.8 m/s^2 * 83 m
v^2 ≈ 1623.2 m^2/s^2
Taking the square root of both sides:
v ≈ √1623.2 m^2/s^2
v ≈ 40.3 m/s
Therefore, the maximum velocity of the roller coaster when it reaches the bottom of the 83-meter drop is approximately 40.3 m/s.