A roller coaster cart of mass m = 293 kg starts stationary at point A, where h1 = 23.3 m and a while later is at B, where h2 = 8.4 m. The acceleration of gravity is 9.8 m/s^2.
What is the speed of the cart at B, ignoring the effect of friction?
To find the speed of the cart at point B, we can use the conservation of energy principle.
The potential energy at point A is equal to the sum of the kinetic energy and potential energy at point B.
Potential energy at point A (Ua) = mass (m) × acceleration due to gravity (g) × height at A (h1)
Potential energy at point B (Ub) = mass (m) × acceleration due to gravity (g) × height at B (h2)
Kinetic energy at point B (Kb) = 0.5 × mass (m) × velocity at B squared (v^2)
According to the conservation of energy principle:
Ua = Ub + Kb
mgh1 = mgh2 + 0.5mv^2
Canceling the mass on both sides:
gh1 = gh2 + 0.5v^2
Rearranging the equation and solving for v^2:
0.5v^2 = gh1 - gh2
v^2 = 2(g(h1 - h2))
Substituting the given values:
v^2 = 2(9.8(23.3 - 8.4))
v^2 = 2(9.8(14.9))
v^2 = 2(146.02)
v^2 = 292.04
Taking the square root of both sides to find v:
v ≈ 17.08 m/s
Therefore, the speed of the cart at point B, ignoring the effect of friction, is approximately 17.08 m/s.
To find the speed of the cart at point B, we can use the principles of conservation of energy.
The potential energy (PE) of the cart at point A can be calculated as:
PE1 = m * g * h1
Similarly, the potential energy of the cart at point B can be calculated as:
PE2 = m * g * h2
Since the cart starts from rest, it has no initial kinetic energy (KE1 = 0). Therefore, the total mechanical energy at point A is equal to the potential energy at point A:
E1 = PE1
At point B, the mechanical energy is the sum of the kinetic energy (KE2) and the potential energy (PE2):
E2 = KE2 + PE2
According to the law of conservation of energy, the total mechanical energy remains constant. Therefore, E1 = E2:
E1 = E2
PE1 = KE2 + PE2
Substituting the expressions for potential energy:
m * g * h1 = KE2 + m * g * h2
Now, we can solve for the kinetic energy at point B, which is equal to the mechanical energy at point A:
KE2 = m * g * h1 - m * g * h2
To find the speed of the cart at point B, we can use the equation for kinetic energy:
KE2 = (1/2) * m * v^2
Rearranging the equation to solve for the velocity (v):
v^2 = (2 * KE2) / m
v = √[(2 * KE2) / m]
Now, plugging in the numerical values given in the problem, we can calculate the speed of the cart at point B:
m = 293 kg
h1 = 23.3 m
h2 = 8.4 m
g = 9.8 m/s^2
KE2 = (293 kg * 9.8 m/s^2 * 23.3 m) - (293 kg * 9.8 m/s^2 * 8.4 m)
KE2 = 64724.58 J
v = √[(2 * 64724.58 J) / 293 kg]
v ≈ 16.13 m/s
Therefore, the speed of the cart at point B, ignoring the effect of friction, is approximately 16.13 m/s.
g (change in h) = (1/2)v^2
or
v = sqrt (2 g*change in h)
(note - m is irrelevant. That is lucky or a full car would be faster than a light one)