Question

A 55 kg ice skater is gliding along at 3.5 m/s. Five seconds later her speed has dropped to 2.8m/s. What is the magnitude of the kinetic friction acting on her skates?

Answer 1

Answer:

7.7 N

Explanation:

∑F = ma

F = m Δv/Δt

F = (55 kg) (2.8 m/s − 3.5 m/s) / 5 s

F = -7.7 N

The magnitude of the force of kinetic friction is 7.7 Newtons.

Answer 2

The magnitude of the kinetic friction is  -7.7 N.

The given parameters;

mass of the ice, m = 55 kg

initial velocity of the ice, u = 3.5 m/s

final velocity of the ice, v = 2.8 m/s

time of motion, t = 5 s

The magnitude of the kinetic  friction is calculated as;

$F = ma = \frac{m(v-u)}{t} \\\\F = \frac{55(2.8 -3.5)}{5} \\\\F = -7.7 \ N$

Thus, the magnitude of the kinetic friction is  -7.7 N.

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