HELP w/ coefficient of kinetic friction problem

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SUMMARY

The coefficient of kinetic friction between the ice and the puck is calculated using the formula u = a / g, where 'a' is the acceleration of the puck (0.605 m/s²) and 'g' is the acceleration due to gravity (9.80 m/s²). This results in a coefficient of friction (u) of 0.0617. The discussion emphasizes that the mass of the puck is not necessary for this calculation, as the forces can be equated to eliminate mass from the equations. The key takeaway is that understanding the relationship between acceleration and friction allows for the determination of the coefficient without needing the mass.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly F = ma.
  • Knowledge of basic physics concepts such as friction and normal force.
  • Familiarity with the acceleration due to gravity (g = 9.80 m/s²).
  • Ability to manipulate algebraic equations to isolate variables.
NEXT STEPS
  • Study the derivation of the frictional force equation F_friction = μF_n.
  • Learn how to draw and interpret free body diagrams in physics.
  • Explore real-world applications of kinetic friction in sports and engineering.
  • Investigate the effects of different surfaces on the coefficient of friction.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of friction in motion. Additionally, athletes and coaches interested in the dynamics of sports equipment on ice surfaces would benefit from this discussion.

WCU
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ok, i have a problem and w/o a given mass i am lost on how to solve it.

"What is the coefficient of kinetic friction between the ice and the puck in the problem above?"

problem above states, "A Carolina Hurricanes player hits a puck with his hockey stick during a practice, giving the puck an initial speed of 5.50m/s. The puck decelerates uniformly and comes to rest after traveling 25.0m. What is the rate of acceleration for the puck?"
-answer is 0.605m/s^2

Please Help Me!
 
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Why don't you draw yourself a free body diagram of the puck? Try to figure out what forces are working in the horizontal direction. And remember, F = ma.
 
yes, i am aware of that, but the MASS is UNKNOWN

if anyone knows an alternate formula (one that doesn't require mass) i would be most grateful

thanks for your help...
 
Last edited:
you don't need the mass

watch

now that you know the acceleration due to the friction (since the friction force stops it)

F = ma = \mu F_{n} = \mu mg

where Fn is normal force of the puck and mu is the co efficient of friction.
 
thanks for the help, the formula makes sense, sorta, but i was never taught how to get the normal force without the mass, so I'm still lost, but not as lost

i do know the answer is 0.0617 ... i just don't know how to get there ...
 
WCU said:
thanks for the help, the formula makes sense, sorta, but i was never taught how to get the normal force without the mass, so I'm still lost, but not as lost

i do know the answer is 0.0617 ... i just don't know how to get there ...

normal force without mass?? come again??

in that equality i have simply used Newtons first law that F = ma and frictional force is F = mu Fn = mu mg and equated the two together and eliminate the mass factor
 
I thought m = mass... i need the coefficient of friction (u), and i don't have F, Fn, or m ... what units do i use? i feel like the answer is right in front of me and I'm just too blind to see it
 
ok realize this FOR ANY FORCE F = ma

secondly Force of friction = u Fn = u mg (because Force of friction is mg when there is no angle involved)

now use the first and second lines and you can equate ma = u mg

a = u g

u = a / g
 
so (0.605m/s^2)/(9.80m/s^2)=u=0.0617, which happens ot be the answer i was looking for.

thankyou so much
 

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