To solve this problem, we can use the principle of conservation of mechanical energy. At the top of the hill, the roller coaster has gravitational potential energy, and at the bottom of the hill, it has kinetic energy.
The potential energy (PE) at the top of the hill is given by the formula:
PE = mgh
Where:
m = mass of the roller coaster = 315 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the hill = 30 m
PE = 315 kg * 9.8 m/s^2 * 30 m
PE = 91,980 Joules
According to the conservation of mechanical energy, this potential energy is converted into kinetic energy at the bottom of the hill.
The kinetic energy (KE) at the bottom of the hill is given by the formula:
KE = 1/2 mv^2
Where:
m = mass of the roller coaster = 315 kg
v = velocity of the roller coaster at the bottom of the hill (what we need to find)
Setting the potential energy equal to the kinetic energy:
PE = KE
91,980 Joules = 1/2 * 315 kg * v^2
Rearranging the equation to solve for v:
v^2 = 2 * 91,980 Joules / 315 kg
v^2 = 2 * 291.61904761904765 m^2/s^2
v^2 = 583.2380952380952 m^2/s^2
v ≈ √(583.2380952380952) ≈ 24.16 m/s
Therefore, the speed of the roller coaster at the bottom of the hill is approximately 24.16 m/s.
A 315 kg roller coaster starts from rest at the top of a 30-m hill. Assuming the track is frictionless, what is roller coaster's speed at the bottom of the hill?
1 answer