Sunday

April 20, 2014

April 20, 2014

Posted by **Alexandra** on Sunday, November 25, 2012 at 4:02pm.

This is part three of three parts.

1st Part: A player strikes a hockey puck giving it a velocity of 39.486 m/s. The puck slides across the ice for 0.231s after which time its velocity is 38.666 m/s. The acceleration of gravity is 9.8 m/s^2. If the mass of the puck is 209 g, what is the average drag force on it by the ice? The answer is: -0.74190 N.

2nd Part: What is the coefficient of kinetic friction for the puck sliding on ice? The answer is 0.3622228112.

I would REALLY appreciate any help! Thanks<3

- Physics -
**Elena**, Sunday, November 25, 2012 at 4:57pm(a)

v=v₀-at

a=( v-v₀)/t=(38.666-39.486)/0.231 =-3.55 m/s²

F(dr)=ma=-0.209•(-3.55)=-0.7419 N

(b)

F(dr)=μN=μmg

μ=F(dr)/mg=0.7419/0.209•9.8=0.362

(c)

Work-energy theorem

ΔKE=W(dr)

0-mv²/2=F•s.

F= - mv²/2s= - 0.209• 39.486²/2•0.041=-3935N

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