A 68.5-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?

Solution 7E First find the acceleration assuming it is uniform. Given data: Initial velocity of skatter v i 2.40 m/s. Final velocity of skatter v =f0 m/s. Time of ice comes to rest t = 3.52 s. Mass of skatter m = 68.5 kg. a = (vf v 0/t a = (0 2.40 m/s )/3.52 s a = 0.682 m/s 2 According to newton's second law F = ma 2 F = ( 0.682 m/s ) 68.* kg F = 46.70 N