Question

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? Round your answer to three significant digits.

a. 137N
b. 2.97N
c. 1,370N
d. 15.4N

Answer 1

The minimum centripetal force necessary to keep the skateboarder on the ramp is found by using the formula for centripetal force (Fc = m * v^2 / r), which results in approximately 1370 N after rounding to three significant digits.

Step-by-step explanation:

To determine the minimum centripetal force necessary to keep the skateboarder on the ramp, we need to use the formula for centripetal force:

Fc = m * v2 / r

Where:

• m is the mass of the skateboarder (52.0 kg)
• v is the velocity (8.90 m/s)
• r is the radius of the half-pipe (3.00 meters)

Plugging in the values, we get:

Fc = 52.0 kg * (8.90 m/s)2 / 3.00 m

Fc = 52.0 kg * 79.21 m2/s2 / 3.00 m = (52.0 kg * 79.21) / 3.00

Fc = 1373.92 N

After rounding to three significant digits, the centripetal force is approximately 1370 N.

Therefore, the answer is c. 1,370N.