Answer:
The magnitude of the average force of friction acting on the ice skater while she slows to a stop is 15.65 N
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
[tex]But, a = \frac{v-u}{t}[/tex]
[tex]F =m (\frac{v-u}{t})[/tex]
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
[tex]F = m(\frac{v-u}{t} )\\\\F = 52.5(\frac{0-2.25}{7.55})\\\\F = -15.65 \ N[/tex]
Thus, the magnitude of the average force of friction acting on the ice skater while she slows to a stop is 15.65 N
