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Hockey Puck- Physics Friction Problem

  • Thread starter IamMoi
  • Start date
  • #1
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Homework Statement


A 0.170 kg hockey puck is initially moving at 21.2 m/s [W] along the ice. The coefficient
of kinetic friction for the puck and the ice is 0.005.

(a) What is the speed of the puck after traveling 58.5 m? ans: 21.1 m/s
(b) After being played on for a while, the ice becomes rougher and the coefficient of
kinetic friction increases to 0.047. How far will the puck travel if its initial and final
speeds are the same as before?


Homework Equations


Fnet=ma
Fg=mg
Fk=μk.Fn
Vf[itex]^{2}[/itex]=Vi[itex]^{2}[/itex]+2ad

The Attempt at a Solution


a.) Fnet=Fk
ma=μk.Fn
ma=μk.mg
a=[itex]\frac{(0.05)(0.170kg)(9.8m/s^{2})}{0.170kg}[/itex]
=0.049m/s[itex]^{2}[/itex]

Vf[itex]^{2}[/itex]=Vi[itex]^{2}[/itex]+2ad
=[itex]\sqrt{(21.2)^{2}-(0.049)(58.5)}[/itex]
=21.13m/s

b)Fnet=Fk
ma=μk.Fn
ma=μk.mg
a=[itex]\frac{(0.047)(0.170kg)(9.8m/s^{2})}{0.170kg}[/itex]
=0.04606m/s[itex]^{2}[/itex]

.. then i don't know what's next..
 

Answers and Replies

  • #2
33,792
9,506
At (a), the coefficient of friction is given as 0.005, but your calculation uses 0.05.

For (b), you can use the same formula as in (a). This time, the velocities are known and the distance is unknown.
 
  • #3
13
0
At (a), the coefficient of friction is given as 0.005, but your calculation uses 0.05.

For (b), you can use the same formula as in (a). This time, the velocities are known and the distance is unknown.
oh sorry,your right, i typed it wrong..
that's my problem, i always have two unknown variables whenever i used different formula..
 
  • #4
33,792
9,506
In b, the distance is the only unknown variable in the formula.
 
  • #5
13
0
yes, i tried to do it like in part a) but i did not get the right answer.. the right answer is 6.24 m
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,347
4,883
pls post your working and the answer you got.
 
  • #7
PeterO
Homework Helper
2,425
46
yes, i tried to do it like in part a) but i did not get the right answer.. the right answer is 6.24 m
For part (b), the initial and final speeds are the same as for part (a) - so the average speed is the same as part (a)

Since the coefficient of friction is nearly 10 times as large, the puck will take only (approx) one tenth the time to slow, so it has the same average speed for 1/10th the time, so should cover (approx) 1/10th the distance.

58.5 / 10 = 5.85m

Of course the new μ is not exactly 10 times the original, so that answer is only approximate.
 

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