A 250.0 kg roller coaster car has 20,000.0 J of potential energy at the top of the hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?
The kinetic energy (1/2) M V^2 will equal the lost potential energy (20,000 J) at the bottom of the hill. This assumes that it had nearly zero kinetic energy at the top of the hill.
You know M, so solve for V.
V = sqrt(2*20,000/M)
To find the velocity of the roller coaster car at the bottom of the hill, you can use the principle of conservation of energy. The potential energy (PE) at the top of the hill will be converted into kinetic energy (KE) at the bottom of the hill.
The formula for potential energy is PE = mgh, where m is the mass of the roller coaster car, g is the gravitational acceleration (approximately 9.8 m/s^2), and h is the height of the hill.
The formula for kinetic energy is KE = 0.5mv^2, where v is the velocity of the roller coaster car.
Equating the potential energy at the top to the kinetic energy at the bottom, we can set up the equation:
PE = KE
mgh = 0.5mv^2
In this case, mass (m) and the height of the hill (h) are given, so we can substitute the given values:
250.0 kg * 9.8 m/s^2 * h = 0.5 * 250.0 kg * v^2
Since the height of the hill is not provided, we need additional information to solve the problem. However, if we assume that the height of the hill is 10.0 m (for demonstration purposes), we can proceed with the calculation.
Substituting the given values:
250.0 kg * 9.8 m/s^2 * 10.0 m = 0.5 * 250.0 kg * v^2
Now, solve for v:
(250.0 kg * 9.8 m/s^2 * 10.0 m) / (0.5 * 250.0 kg) = v^2
(24500 J) / (125 kg) = v^2
196 m^2/s^2 = v^2
Taking the square root of both sides:
v = √196 m^2/s^2
v ≈ 14 m/s
Therefore, assuming a hill height of 10.0 m, the velocity of the roller coaster car at the bottom of the hill is approximately 14 m/s.