Answer:
Explanation:
A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .
What speed does he need at the bottom
Let the lowest point of the circle = (0,0)
Center of circle = (0, 2.6)
The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m
Highest point = (0, 5.2)
As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)
As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6
During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE
½ * m * ∆v^2 = m * g * ∆h
½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6
½ * ∆v^2 = 9.8 * 2.6
∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s
The velocity at highest point, (2.6,2.6) is 0 m/s
So the velocity at the lowest point must be 7.14 m/s
Let’s see if the skateboarder has enough KE to move upward 2.6 m.
KE at bottom = ½ * 52 * 7.14^2 = 1325.5
PE = 52 * 9.8 * h
52 * 9.8 * h = 1325.5
h = 2.6 m
The answer is OK
The speed needed by the skateboarder at the bottom is 7.8 m/s.
The given parameters;
Apply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;
[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]
Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.
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