How fast will a roller coaster, mass 1200 kg, be moving at the bottom of an 18 m tall hill?
To determine the speed of the roller coaster at the bottom of the hill, we can use the principle of conservation of energy.
The roller coaster's potential energy at the top of the hill is converted to kinetic energy at the bottom. The potential energy (PE) is given by the formula PE = mgh, where m is the mass (1200 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (18 m).
PE = mgh = (1200 kg)(9.8 m/s^2)(18 m) = 211,680 J
At the bottom of the hill, all of the potential energy is converted to kinetic energy, which is given by the formula KE = (1/2)mv^2, where v is the velocity of the roller coaster.
Setting the potential energy equal to the kinetic energy, we have:
211,680 J = (1/2)(1200 kg)v^2
Rearranging the equation to solve for v:
v^2 = (2 * 211,680 J) / (1200 kg)
v^2 = 352.8 m^2/s^2
Taking the square root of both sides, we find:
v = sqrt(352.8 m^2/s^2) ≈ 18.77 m/s
Therefore, the roller coaster will be moving at approximately 18.77 m/s at the bottom of the 18 m tall hill.