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?

1 Answer
Dec 12, 2017

The force of friction is #F=46.7N#

Explanation:

The mass of the skater is #m=68.5kg#

The initial speed is #u=2.4ms^-1#

The final speed is #v=0ms^-1#

The time is #t=3.52s#

Applying the equation of motion

#v=u+at#

The acceleration is

#a=(v-u)/t=(0-2.4)/3.52=-0.68ms^-2#

According to Newton's Second Law

#F=ma#

The force of friction is

#F=68.5*0.68=46.7N#