Answer to Question #279327 in Mechanics | Relativity for btch

Question

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?

Answer 1

Solution:

To determine the friction force exerted in the skater, we use Newton's Second law of motion which relates force, acceleration, and 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 by the total time. Thus,

$a=\frac{2.40(\frac{m}{s})}{3.52 s}=0.682(\frac{m}{s^2})$

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

F = m (a) = 68.5 kg ( 0.682 $(\frac{m}{s^2})$ ) = 46.7 N 

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