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A 52.0-kilogram skateboarder does a trick on a half-pipe. During the trick, he reaches a speed of 8.90 m/s. The radius of the half-pipe ramp is 3.00 meters. What is the minimum centripetal force necessary
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A 52.0-kilogram skateboarder does a trick on a half-pipe. During the trick, he reaches a speed of 8.90 m/s. The radius of the half-pipe ramp is 3.00 meters. What is the minimum centripetal force necessary to keep the skater on the ramp?

User Jesuspc
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2 Answers

3 votes

The answer to the question is 1,373

User Pragmar
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7 votes
Centripetal force = mv²/r = 52 * 8.9² / 3 = 1373 N
User Siritinga
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8.2k points
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