a roller coaster weighing 500kg starts at rest from a point of 45m above the ground. what is its speed when it reaches point b at ground level?
mgh=1/2 m v^2
v= sqrt(2gh) where h is the distance of the fall.
To find the speed of the roller coaster when it reaches point B at ground level, we can use the principle of conservation of energy.
First, let's find the potential energy at the starting point (point A) and the kinetic energy at point B.
The potential energy (PE) at point A is given by the formula:
PE = m * g * h
Where:
m = mass of the roller coaster (500 kg)
g = acceleration due to gravity (9.8 m/s²)
h = height above the ground at point A (45 m)
PE(A) = 500 kg * 9.8 m/s² * 45 m
Next, the kinetic energy (KE) at point B is given by the formula:
KE = 0.5 * m * v²
Where:
m = mass of the roller coaster (500 kg)
v = velocity of the roller coaster at point B (unknown)
We can equate the potential energy at point A to the kinetic energy at point B, since energy is conserved:
PE(A) = KE(B)
500 kg * 9.8 m/s² * 45 m = 0.5 * 500 kg * v²
Now, let's solve for v:
(500 kg * 9.8 m/s² * 45 m) / (0.5 * 500 kg) = v²
v² = (500 kg * 9.8 m/s² * 45 m) / (0.5 * 500 kg)
v² = 9.8 m/s² * 45 m
v² = 441 m²/s²
Taking the square root of both sides:
v = √441 m²/s²
v ≈ 21 m/s
Therefore, the speed of the roller coaster when it reaches point B at ground level is approximately 21 m/s.