Question

a hockey skater turns in a circular path with a radius of 31.0 m while going 10.5 m/s. If he experiences a centripetal force of 565 N, what is the skater’s mass?

Answer 1

From the second Newton's law, the mass can be expressed as follows:

$$m = \dfrac{F}{a}$$

where $F = 565N$ is the net force acting on the skater, $a$ is they acceleration. In turn, the centripetal acceleration is given as follows:

$$a = \dfrac{v^2}{R}$$

where $v = 10.5m/s$ is the speed, and $R = 31m$ is the radius. Thus, obtain:

$$m = \dfrac{FR}{v^2}\\ m = \dfrac{565\cdot 31}{10.5^2}\approx 158.9kg$$

Answer. 158.9 kg.