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3 years ago

6

An ice hockey player is skating on an ice rink. The rink has a coefficient of kinetic friction of roughly 0.1. If the normal for

ce on the hockey player is 800. N, what is the frictional force acting on the hockey player?

1 answer:

AlekseyPX3 years ago

3 0

Answer:yes

Explanation:he divided by the numnebr of hockey pucks

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Answer:

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Given data:

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From figure

balancing force

$mgsin\theta - f = ma$ .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

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from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

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$F =\mu N$

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$= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for coefficent of friction we get

$\mu = 0.31$

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