Cyclist competes in a one-lap race around a flat, circular course of radius 140 m . starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60 s . the mass of the bicycle (including the rider) is 76 kg . what is the magnitude of the net force fnet acting on the bicycle as it crosses the finish line?

Refer to the diagram shown below.

The angular distance traveled in one lap is θ = 2π radians.

Let  α =  the angular acceleration, rad/s².
Because 1 lap was completed in t = 60 s, therefore the angular acceleration is given by
θ = (1/2)*α*t²
That is,
2π rad = 0.5*(α rad/s²)*(60 s)² 
2π = 1800 α
α = 3.49 x 10⁻³ rad/s²

The angular velocity at the end of the lap is
ω = αt
    = (3.49 x 10⁻³ rad/s²)*(60 s)
    = 0.2094 rad/s

The tangential velocity is
v = rω = (140 m)*(0.2094) = 29.32 m/s

The centripetal force acting on the cyclist at the finish line is
F = m*r*ω²
   = (76 kg)*(140 m)*(0.2094 rad/s)²
   = 466.5 N

466.5 N

If there was no gravity then their would be no downword force on the football to have it return the the grownd, and if there was no friction (mainly air friction) then there would be no opposing force the stop the ball from moving through the air allowing the ball to travle ∞.

As newton's first law states a object in motion stays in motion