Question

A go-cart travels around a flat, horizontal,

circular track with a radius of 25 meters. The
mass of the go-cart with the rider is
200. kilograms. The magnitude of the maximum
centripetal force exerted by the track on the
go-cart is 1200. newtons.
What is the maximum speed the 200.-kilogram
go-cart can travel without sliding off the track?
(1) 8.0 m/s (3) 150 m/s
(2) 12 m/s (4) 170 m/s

Answer 1

Centripetal force = mv^2/r

1200/200=v^2/25

1200*25/200=v^2
v^2=150
v=sqrt(150)=12.2

Therfore the maximum speed it can go is 12m/s.

Answer 2

♥ Centripetal force formula ⇒ mv^2/r
♥ Let's solve:
⇒1200/200=v^2/25
⇒1200*25/200=v^2
⇒v^2=150
⇒v=SQRT(150)
⇒=12.2
Keep in mind anything right after the carrot sign is considered a "exponent" or "to the power of."
♥ Maximum speed it can go is 12m/s.