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?
Wrong answer I tried 137N and it didn't work. The correct answer is 1,370N.
137N (:
1,370 N
1,370 N
To find the minimum centripetal force necessary to keep the skateboarder on the ramp, we can use the formula:
Centripetal force (Fc) = (mass × velocity^2) / radius
Given:
- Mass of the skateboarder (m) = 52.0 kg
- Speed of the skateboarder (v) = 8.90 m/s
- Radius of the half-pipe ramp (r) = 3.00 m
Now we can substitute these values into the formula to calculate the centripetal force:
Fc = (52.0 kg × (8.90 m/s)^2) / 3.00 m
First, we need to square the velocity:
(8.90 m/s)^2 = 79.21 m^2/s^2
Then multiply the squared velocity by the mass:
52.0 kg × 79.21 m^2/s^2 = 4118.92 N
Finally, divide this by the radius to get the centripetal force:
Fc = 4118.92 N / 3.00 m = 1372.97 N
Therefore, the minimum centripetal force necessary to keep the skateboarder on the ramp is 1372.97 Newtons.