1) A 2,494-kilogram roller coaster car is approaching a hill. At its top, the hi
ID: 1596023 • Letter: 1
Question
1) A 2,494-kilogram roller coaster car is approaching a hill. At its top, the hill can be described by a circle with radius 9 m. Draw a proper freebody diagram for the car at the top of the hill (point B). Employ Newton's Law to find the speed that the car must have at point B such that the normal force acting on the car is 73% of the weight. Now employ conservation of energy to find the speed required at point A, such that the car passes point B at the required speed. Draw an energy bar graph. Below answer with the required speed at point A with units of m/s. Make sure to practise how to explain your work and also how to evaluate your result.
2) A 2,171-kilogram roller coaster car is approaching a valley. At its bottom, the valley can be described by a circle with radius 6 m. Find the speed required at point A, such that the normal force acting on the car in point B is 234% more than the weight. Below answer with the required speed at point A with units of m/s. Make sure to practise how to explain your work and also how to evaluate your result. Draw proper diagrams.
3) A 1,404-kilogram roller coaster car starts at point A almost at rest. It first runs downhill and then is approaching a hill. At its top, the hill can be described by a circle with radius 8 m. Find the height required at point A, such that the normal force acting on the car in point B is 38% of the weight. Make sure to practise how to explain your work and also how to evaluate your result. Draw proper diagrams to support your work. Below answer with the height.
4) A 2,458-kilogram roller coaster car starts at point A almost at rest. It runs downhill into a valley. At its bottom, the valey can be described by a circle with radius 11 m. Find the height required at point A, such that the normal force acting on the car in point B is 324% of the weight. Make sure to practise how to explain your work and also how to evaluate your result. Draw proper diagrams to support your work. Below answer with the height.
Explanation / Answer
1)
vb = speed at B
r = radius of circle = 9 m
m = mass = 2494 kg
force equation at B is given as
mg - Fn = m vb2/r
given that Fn = 0.73 mg
mg - 0.73 mg = m vb2/r
(0.27) (9.8) = vb2/9
vb = 4.88 m/s
using conservation of energy
(0.5) m va2 = mg(2r) + (0.5) m vb2
(0.5) va2 = g(2r) + (0.5) vb2
(0.5) va2 = (9.8)(2 x 9) + (0.5) (4.88)2
va = 19.4 m/s
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