Ask Question
18 March, 08:53

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
Answers (1)
  1. 18 March, 10:05
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers