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GPT-3.5-Turbo ·
A roller coaster car with a mass of 500 kg is at the top of a hill that is 30 meters high. If the car is released, what will be its speed at the bottom of the hill? Assume that all the potential energy at the top of the hill is converted into kinetic energy at the bottom of the hill, and that there is no friction or air resistance.
The potential energy (PE) of the roller coaster at the top of the hill is given by the formula:
PE = mgh
where m is the mass of the roller coaster, g is the acceleration due to gravity (approximately 9.81 m/s^2), and h is the height of the hill. Substituting the given values:
PE = 500 kg x 9.81 m/s^2 x 30 m = 147,150 J
At the bottom of the hill, all of this potential energy is converted into kinetic energy (KE) according to the law of conservation of energy:
PE = KE
So, the kinetic energy of the roller coaster at the bottom of the hill is also 147,150 J.
The kinetic energy (KE) of an object is given by the formula:
KE = 1/2mv^2
where m is the mass of the object and v is its speed. Solving for v:
v = sqrt(2KE/m)
Substituting the given values:
v = sqrt(2 x 147,150 J / 500 kg) = 34.0 m/s
Therefore, the speed of the roller coaster at the bottom of the hill is approximately 34.0 m/s.
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