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?

Question

Nathalie Warner

Physics

18 January, 06:45

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?

+4

Answers (1)

Know the Answer?

Jamison Owens · Answer 1

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.