Ago-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

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.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j