(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
If the puck strikes the goalie's pads and stops in a distance of 4.14 cm, what average force is exerted on the pads? Answer in units of N.
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
1 answer
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