A skateboarder with his board can be modeled as a particle of mass 74.0 kg, loca

Question

A skateboarder with his board can be modeled as a particle of mass 74.0 kg, located at his center of mass (which we will study in a later chapter). As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe. The half-pipe is one half of a cylinder of radius 6.30 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 5.80 m. Find his speed at the bottom of the half-pipe. m/s Immediately after passing point ®, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.880 m above the concrete. Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 5.42 m. His body is horizontal when he passes point, the far lip of the half-pipe. As he passes through point, the speed of the skateboarder is 4.73 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy in the skateboarder-Earth system when he stood up at point ? How high above point does he rise? Caution: Do not try this stunt yourself without the required skill and protective equipment.

Explanation / Answer

a) Using energy of conversation,

initial total energy (PE+ KE) = final energy

mgh + 0 = 0 + mv^2 /2

v = sqrt(2gh) = sqrt(2 x 9.8 x 6.30) = 11.11 m/s

b) change in PE from C to D = - mg(5.42)

     = - 74 x 9.8 x 5.42 = - 3930.58 J

initial KE = 74 x 11.11^2 /2 = 4567 J

final KE = 74 x 4.73^3 / 2 = 1214.82 J

using energy conservation,

energy needed = 4567 - 1214.82 - 3930.58 = 578.4 J

c) Using vf^2 - vi^2 = 2 a d

at maximum point, he stops momentarly. so vf = 0

0^2 - 4.73^2 = 2 (-9.8) d

d = 1.14 m

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