A 55kg skateboarder wants to just make it to the upper edge of a quarter pipe, a pipe that is one-quarter of a circle with a radius of 3.10m. What speed does he need at the bottom?

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 small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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