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An ice skater with a mass of 52.5 kg is gliding across a smooth lake with a speed of 2.25 m/s when she hits a spot of ice covered with snow. After a time of 7.55 s she has slowed to a stop. Using your knowledge of momentum, determine the magnitude of the average force of friction acting on the ice skater while she slows to a stop.

User BasTaller
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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)


F =m ((v-u)/(t))

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

Thus, the magnitude of the average force of friction acting on the ice skater while she slows to a stop is 15.65 N

User Henry Yik
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