A68.5-kg skater moving initially at 2.40 m> s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. what force does friction exert on the skater?

To determine the friction force exerted in the skater, we use Newton's Second law of motion which relates force, acceleration and the mass. It is expressed as follows: 

F = m (a) 

where
m = mass of the body = 68.5 kg 

The acceleration, a,  of the moving body  is calculated by dividing the initial velocity with the total time. Thus,

a = 2.40 m/s / 3.52 s = 0.682 m/s^2 

Substituting the values to the equation of force, we will have

F = m (a) = 68.5 kg ( 0.682 m/s^2 ) = 46.7 N 

Thus, the friction force that is being exerted by the ice  to the skater ould be 46.7 N.