Answer:
The magnitude of the average force of friction acting on the ice skater while she slows to a stop is 15.65 N
Step-by-step explanation:
Given;
mass of the ice skater, m = 52.5 kg
speed of the ice skater, u = 2.25 m/s
time for her gliding, t = 7.55 s
To determine the magnitude of the average force of friction acting on the ice skater while she slows to a stop, we apply Newton's second law of motion;
F = ma
![But, a = (v-u)/(t)](https://img.qammunity.org/2021/formulas/physics/college/bczxow62quo67jviqvowf5fixyhk1aiqqz.png)
![F =m ((v-u)/(t))](https://img.qammunity.org/2021/formulas/physics/college/f9bmq3xfelumouxd6zvbfhha6vros82nwr.png)
where;
F is average force of friction acting on the ice skater
v is the final speed speed of the ice skater = 0
u is the initial speed of the ice skater
t is time
![F = m((v-u)/(t) )\\\\F = 52.5((0-2.25)/(7.55))\\\\F = -15.65 \ N](https://img.qammunity.org/2021/formulas/physics/college/szqanf22quyldbrkh9ebduyc13x4d7te8v.png)
Thus, the magnitude of the average force of friction acting on the ice skater while she slows to a stop is 15.65 N