[Forces/Kinematics] How far will skateboarder go?

  • Thread starter brbrett
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  • #1
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Homework Statement


Suppose a skateboarder is coasting initially at 7.5 m/s to the right across flat, rough ground (coefficient of friction = 0.125). How far will the skateboarder go before stopping. (Kinematics and forces will be used in combination)

Homework Equations


v2=v1+at
v2=v02+2ad
v=d/t
d=v0t+(1/2)at2
F=ma

The Attempt at a Solution


I'm not certain how to approach the question. The question doesn't state what the mass is, nor any applied force apart from the initial velocity. I know that the final velocity is 0.
 

Answers and Replies

  • #2
billy_joule
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You don't need the mass, work symbolically (ie don't plug any numbers in until the end) and it'll all work out.
Does this formula ring any bells?
Ff = μN
 
  • #3
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I tried doing the problem symbolically, but I've run into the mass problem again. Here is my work:
Ff = μN
a=Ff/m
v22=v02+2ad
v22=v02+2(Ff /m)d
This time I have figured out what "a" is, and I only need to know the inside parts of it (m and Ff)
 
  • #4
billy_joule
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I tried doing the problem symbolically, but I've run into the mass problem again. Here is my work:
Ff = μN
a=Ff/m
v22=v02+2ad
v22=v02+2(Ff /m)d
This time I have figured out what "a" is, and I only need to know the inside parts of it (m and Ff)
Getting there.
What is 'N'? It's definition can be used to eliminate m.
 
  • #5
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Normal force is the reaction force to gravity.
Can I do this then?
g is gravity
Ffr= μ(gm)
a= [μ(gm)]/m
a= μg

that would cancel out the mass and leave me with 9.8
 
  • #6
billy_joule
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Normal force is the reaction force to gravity.
Can I do this then?
g is gravity
Ffr= μ(gm)
a= [μ(gm)]/m
a= μg
Your math is right but I'm not sure what you mean by this:
that would cancel out the mass and leave me with 9.8
'g' is 9.81 m/s2 but 'a' won't have that value.

You'll also have to be aware of the directions of v and a and get their signs right to get the right answer.
 
  • #7
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err, that was an error from before I altered my math. a is actually μg, so gravity times that coefficient of friction will give me the acceleration, which will be negative since we are talking about slowing an object down.
 
  • #8
billy_joule
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err, that was an error from before I altered my math. a is actually μg, so gravity times that coefficient of friction will give me the acceleration, which will be negative since we are talking about slowing an object down.
Good work. You're pretty much there. Just solve for d then plug and chug.
 
  • #9
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Thanks for the help.
I got 22.95 meters as my answer when I used one of the kinematic equations.
 
  • #10
haruspex
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Thanks for the help.
I got 22.95 meters as my answer when I used one of the kinematic equations.
I agree with that answer, but I would like to point out a serious flaw in the wording of the question.
A skateboard has wheels. Wheels roll. In rolling contact, friction is static, and static friction does no work. No matter how high the coefficient of friction, it won't slow the skateboard at all.
What slows the skateboard is rolling resistance. This consists of absorbency in the springiness of the ground and wheels, and torsional friction at the wheel axles.
 
  • #11
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Unfortunately the problem is taken from an introductory physics course, where we have yet to learn anything more about friction and how it works. My teacher often makes questions with serious flaws like this though, especially with work/energy.
 
  • #12
haruspex
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  • #13
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Thanks! I'll definitely take a look at it.
 

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