Question

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.

Answer 1

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

$But, a = \frac{v-u}{t}$

$F =m (\frac{v-u}{t})$

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

$F = m(\frac{v-u}{t} )\\\\F = 52.5(\frac{0-2.25}{7.55})\\\\F = -15.65 \ N$

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