Final answer:
By applying conservation of energy, the speed of the roller coaster at the top of a 10.8 m high loop is found to be approximately 22.44 m/s, assuming negligible friction.
Step-by-step explanation:
To solve for the speed of the roller coaster at the top of a 10.8 m high loop starting from rest at an initial height of 36.5 m, we can use the principle of conservation of energy. According to this principle, the total mechanical energy of the system (kinetic plus potential energy) should remain constant if there are no non-conservative forces (like friction) doing work on the system.
We set the potential energy at the top of the starting height (36.5 m) equal to the sum of kinetic energy and potential energy at the top of the loop (10.8 m high). The potential energy (PE) at any height h is given by PE = m*g*h, where m is the mass of the roller coaster car, g is the acceleration due to gravity (9.81 m/s2), and h is the height. The kinetic energy (KE) is given by KE = 0.5*m*v2, where v is the velocity of the roller coaster car.
At the start, the roller coaster car has no kinetic energy because it is at rest (v = 0 m/s), so all of its mechanical energy is potential energy at the height of 36.5 m: PEinitial = m*g*36.5. At the top of the 10.8 m loop, the car will have both potential and kinetic energy: PEfinal = m*g*10.8 and KEfinal = 0.5*m*v2.
Setting these equal (PEinitial = PEfinal + KEfinal), we can solve for v:
m*g*36.5 = m*g*10.8 + 0.5*m*v2
g*36.5 = g*10.8 + 0.5*v2
v2 = 2*g*(36.5 - 10.8)
v2 = 2*9.81*(25.7)
v2 = 503.562
v = √503.562
v ≈ 22.44 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 22.44 m/s.