Question

a 64.0 kg skater moving initially at 3.00 m/s on rough horizontal ice comes to rest uniformly in 3.53 s due to friction from the ice. part a what force does friction exert on the skater? enter your answer as the magnitude of the force. f = n

Explanation:

Step1: Calculate acceleration

$a=\frac{v - v_0}{t}=\frac{0 - 3.00}{3.53}\approx-0.8499\ \text{m/s}^2$

Step2: Calculate friction force magnitude

$F=m|a|=64.0\times0.8499\approx54.4\ \text{N}$

Answer:

54.4