A skateboarder with his board can be modeled as a particle of mass 71.0 kg, loca
ID: 2176798 • Letter: A
Question
A skateboarder with his board can be modeled as a particle of mass 71.0 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 (point ). The half-pipe is one half of a cylinder of radius 7.00 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 6.50 m. Find his speed at the bottom of the half-pipe (point ). m/s Immediately after passing point , he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.900 m above the concrete (point ). Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 6.10 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 4.85 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 ?Your response is within 10 % of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. J How high above point does he rise?Explanation / Answer
a) v = sqrt (2*g*r) => sqrt (2*9.81*6.5) = 11.292 m/s b) KE = 0.5*MV^2 KE = 0.5 * 71 kg * (4.85)^2 KE = 835.04 J c) mgh = .5mv^2 => h = (v^2) / 2g so (4.85^2) / 2*9.81 = 1.1989 m