Kinetic Friction

  • Thread starter amcavoy
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  • #1
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I'd appreciate if someone could help me out with this:

A penguin weighing 300 N slides across an icy floor with an inital speed of 6.38 m/s and slides 32.2 m before coming to a stop. Find the frictional force between the penguin and the floor and find the coefficient of kinetic friction. Use g=10m/s2

The first thing I did was to find the mass of the penguin, which is 30 kg. Now I use the equation [itex]{v}^2={v_i}^2+2a\Delta x[/itex]. I come up with:

[tex]6.38^2+64.4a=0\implies a=-.632 m/s^2[/tex]

Now to find the force, I multiplied by the mass 30 kg and came up with the frictional force as -18.96 N. To find the coefficient of kinetic friction from here, I took -18.96 and divided it by 300 N, coming up with -.0632.

Are my steps / results correct?

Thanks,

Alex
 

Answers and Replies

  • #2
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30 kg, sounds good!

where did 64.4meters come from please? was that given in the problem?

oops now i see it there.
 
  • #3
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cyrusabdollahi said:
30 kg, sounds good!
where did 64.4meters come from please? was that given in the problem?
oops now i see it there.
[tex]2a\Delta x = 2(32.2)a = 64.4a[/tex]

Does the rest look alright?

Thanks again,

Alex
 
  • #4
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Nevermind I looked up the answer here (I didn't know I had it!) and it's correct. Thanks for the help.

Alex
 
  • #5
hotvette
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I got the same answer. You can always double check the answer by using the equation of motion for x and see if the distance traveled is consistent with what you were given.

Editorial comment: the easiest approach for me on elementary problems of motion is to draw a free body diagram, label the forces & initial conditions, pick a coordinate system, then write [itex]\Sigma F = m \ddot x[/itex] (in this case), integrate once to get velocity, again to get distance, and go from there. That way you always know that the equations you have are relevent becuase they are specific to the problem. The only equation that is needed is F=ma.
 
Last edited:
  • #6
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Sure, Just a word of caution, try to avoid using that equation, it is derived from your two equations of motion, and if you cant derive it, u shouldent use it! Try to use v=v_0 + at and y=y+v_0 t + 1/2at^2 by eliminating t and solving for a. It is more fundamental.

I.e

v=v_0 + at

0= 6.38 + at (v=0)

then -6.38/a = t

and y=y_0 + v_0t + 1/2 (at^2)

plug in,

32.2 = 0 + 6.38(-6.38/a) + 1/2 (a) (-6.38/a)^2

32.2= -40.7044/a + (40.7044/2)(1/a)

32.2 = - 40.7044/2a

solve for a:

a= -.632055 m/s^2

Thus you have used equations that make sense.
 
  • #7
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cyrusabdollahi said:
Sure, Just a word of caution, try to avoid using that equation, it is derived from your two equations of motion, and if you cant derive it, u shouldent use it! Try to use v=v_0 + at and y=y+v_0 t + 1/2at^2 by eliminating t and solving for a. It is more fundamental.
I.e
v=v_0 + at
0= 6.38 + at (v=0)
then -6.38/a = t
and y=y_0 + v_0t + 1/2 (at^2)
plug in,
32.2 = 0 + 6.38(-6.38/a) + 1/2 (a) (-6.38/a)^2
32.2= -40.7044/a + (40.7044/2)(1/a)
32.2 = - 40.7044/2a
solve for a:
a= -.632055 m/s^2
Thus you have used equations that make sense.
I know how it was derived, but I appreciate the alternative method. Thanks again to both of you.

Alex
 

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