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Answer :
Final answer:
The speed of the roller coaster car at the top of a 16 m high loop, starting from rest on a 40.5 m high track and with no resistive forces, is found to be approximately 20.6 m/s using the conservation of mechanical energy.
Explanation:
The student's question involves calculating the speed of a roller coaster car at a certain point on its track, which is a typical problem addressed in physics, particularly in the area of energy conservation and mechanics. To determine the speed at the top of a 16 m high loop, we can use the principle of conservation of mechanical energy since resistive forces such as friction and air resistance are negligible. The total mechanical energy at the starting point (potential energy due to height) will be equal to the sum of the potential energy and kinetic energy at the top of the loop.
The initial potential energy (PEinitial) is given by m * g * h, where m is the mass of the roller coaster car (4,397 kg), g is the acceleration due to gravity (9.81 m/s2), and h is the initial height (40.5 m). The final potential energy (PEfinal) at the top of the loop is given by m * g * hloop, with hloop being 16 m. As the roller coaster car starts from rest, its initial kinetic energy (KEinitial) is 0. The kinetic energy at the top of the loop (KEfinal) can be expressed as 0.5 * m * v2, with v being the velocity we want to find. Applying the conservation of energy:
PEinitial = PEfinal + KEfinal
m * g * h = m * g * hloop + 0.5 * m * v2
Solving for v gives us v = sqrt(2 * g * (h - hloop)). Substituting the given values results in [tex]v = \sqrt(2 * 9.81 m/s2 * (40.5 m - 16 m))[/tex], which calculates to approximately 20.6 m/s, corresponding to answer choice (a).
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