A skateboarder is skating through a half-pipe with a radius of 4 m. Initially sh

Question

A skateboarder is skating through a half-pipe with a radius of 4 m. Initially she is in a crouched position with her center of mass 0.5 m above the surface of the half-pipe. She starts from rest with her center of mass a height of 3 m above the base of the half-pipe and experiences no loss of energy from friction as she moves. At the bottom of the half-pipe she stands up and lifts her arms into the air, thereby raising her center of mass to 1.0 m. She then continues up the other side of the half-pipe. She had to perform work to lift her arms while skating. How much work (in J) did she do if her mass is 78 kg? You may model the skateboarder as a point particle.

Explanation / Answer

given

R = 4 m

y1 = 0.5 m

y2 = 1 m

h = 3 m

m = 78 kg

using equation

PE = m g Hmax

and KE = 1/2 m vo2

KE = PE

1/2 m vo2 = m g Hmax

Hmax = h ( R - y1 / R - y2 )2

= 3 ( 4 - 0.5 / 4 - 1 )2

Hmax = 4.08333 m

the work she do is W = m g Hmax

W = 78 X 9.8 X 4.08333

W = 3121.297 J

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