Consider a roller coaster that moves along the track shown in the figure below. Assume all friction is negligible and ignore the kinetic energy of the roller coaster's wheels. Assume the roller coaster starts with a speed of
19 m/s at point A. Find the speed of the roller coaster when it reaches locations B and C.(Assume hA = 40 m and hB = 20 m.)
VB=____m/s
VC=____m/s
To find the speed of the roller coaster at locations B and C, we can use the principle of conservation of energy. At point A, the roller coaster has gravitational potential energy and kinetic energy.
The total mechanical energy at point A is given by:
E_total = E_potential + E_kinetic
E_potential = mghA, where m is the mass of the roller coaster, g is the acceleration due to gravity (approximately 9.8 m/s^2), and hA is the height at point A.
E_kinetic = 1/2 * m * v^2, where v is the initial speed of the roller coaster at point A (19 m/s).
Since we are assuming negligible friction and kinetic energy of the wheels, the total mechanical energy is conserved along the track.
At point B, the roller coaster has gravitational potential energy and kinetic energy.
E_total = E_potential + E_kinetic
E_potential = mghB, where hB is the height at point B.
E_kinetic = 1/2 * m * vB^2, where vB is the speed of the roller coaster at point B.
Since the total mechanical energy is conserved, we can equate the expressions for the total mechanical energy at point A and point B:
mghA + 1/2 * m * v^2 = mghB + 1/2 * m * vB^2
Canceling out the mass, we get:
ghA + 1/2 * v^2 = ghB + 1/2 * vB^2
Simplifying the equation, we get:
vB^2 = v^2 + 2g(hA - hB)
Plugging in the given values:
v = 19 m/s
hA = 40 m
hB = 20 m
g = 9.8 m/s^2
vB^2 = (19 m/s)^2 + 2 * 9.8 m/s^2 * (40 m - 20 m)
vB^2 = 361 m^2/s^2 + 2 * 9.8 m/s^2 * 20 m
vB^2 = 361 m^2/s^2 + 392 m^2/s^2
vB^2 = 753 m^2/s^2
Taking the square root of both sides, we get:
vB = sqrt(753) m/s
vB ≈ 27.45 m/s
Therefore, the speed of the roller coaster at point B is approximately 27.45 m/s.
To find the speed at point C, we can use the same equation:
vC^2 = v^2 + 2g(hA - hC)
Plugging in the given values:
v = 19 m/s
hA = 40 m
hC = 0 m (assuming C is at ground level)
g = 9.8 m/s^2
vC^2 = (19 m/s)^2 + 2 * 9.8 m/s^2 * (40 m - 0 m)
vC^2 = 361 m^2/s^2 + 2 * 9.8 m/s^2 * 40 m
vC^2 = 361 m^2/s^2 + 784 m^2/s^2
vC^2 = 1145 m^2/s^2
Taking the square root of both sides, we get:
vC = sqrt(1145) m/s
vC ≈ 33.85 m/s
Therefore, the speed of the roller coaster at point C is approximately 33.85 m/s.