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How far does the car skid before coming to a rest

  • Thread starter alex7298
  • Start date
  • #1
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Homework Statement


Traveling at a speed of 58.0 km/h, the driver of a car locks the wheels by slamming on the brakes. The coefficient of kinetic friction is 0.720. How far does the car skid before coming to a rest


Homework Equations


f=ma?
uk=Fk/Fn
??


The Attempt at a Solution



i dont know where to start because no mass is given
 

Answers and Replies

  • #2
Let the mass equal an arbitrary value m. Start working with what you know and see if anything happens to the mass. You might not need it after all.
 
  • #3
27
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i think i should be using an equation that would make the masses cancel, but i dont know what equation that is
 
  • #4
What do you know about the frictional force?

You probably (or you probably should) know that it is equal to uN where u = coefficient of friction and N = mg.

Being that this is the only force acting in the horizontal direction, what can you say about the net force and, thus, the acceleration of the car?

Take it one step further and you've got the solution.
 
  • #5
27
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lol im still drawing a blank because of that mass
Fk=(.720)(m)(9.8m/s^2)
you cant put zero for mass and i dont see how can go away or how to solve for it.....??
 
  • #6
Kurdt
Staff Emeritus
Science Advisor
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If you know friction is the only force acting and that is [itex] F = \mu_k mg [/itex] and that force is also [itex] F= ma[/itex] I can easily see how the mass will cancel and you can solve for acceleration.
 
  • #7
27
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ok
so ukmg=ma
uk(g)=a
a=(.720)(9.8m/s^2), a=7.056m/s^2
the question is find the distance it takes to stop...i know final velocity, acceleration, and am looking for distance which means i need initial velocity....?
 
  • #8
27
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o lol
nvm
forgot to read back through to find initial velocity
 
  • #9
1,341
3
ok
so ukmg=ma
uk(g)=a
a=(.720)(9.8m/s^2), a=7.056m/s^2
the question is find the distance it takes to stop...i know final velocity, acceleration, and am looking for distance which means i need initial velocity....?
Which is given, can you solve it now?
 
  • #10
27
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yep
thanks....there may be more questions coming soon.
have a huge test tomorrow and i dont really understand this stuff
 
  • #11
1,752
1
well this is what i did

[tex]v^{2}=v_{0}^{2}+2a_{x}(x-x_{0})[/tex]

solve for x

x-initial & v-final = 0

[tex]\Sigma{F_{x}}=ma_{x}[/tex]
[tex]-\mu_{k}N=ma_{x}[/tex]

[tex]\Sigma{F_{y}}=0[/tex]
[tex]N=mg[/tex]

plug in normal in the X solve for acceleration, and plug that into the first equation, and don't forget to convert your initial velocity.
 
  • #12
27
0
what about
"a 250lb block rests on a horizontal table. The coefficient of kinetic friction between the block and table is .30. What force is required to pull the block at a constant speed if (a) a horizontal rope is attached, or if (b) the rope makes an angle of 25 above the horizontal.

for part (a) used (.30)(250)=75N whichh is the frictional force..is that the final answer?

For part (b) i used Uk=(Fk)/(Fn) where Uk=.30, and Fn=Wsin(theta).....which now that i look at it doesnt seem right
 
  • #13
1,752
1
you must convert your weight ... lb to kg

N = (kg*m)/s^2
 
  • #14
27
0
i did, its 25.5kg
 
  • #16
27
0
how about 113.40kg.
but for the inlined rope the x-comp. is Wsin(25) and the y-comp. is Wcos(25) so x comp=105.65 and y-comp=229.58 and pythag gets you back to 250.0008278.....? im lost
 
  • #17
1,752
1
you don't need the Pythagorean theorem. solve it the same way you solve for F in part a.

look at your forces in the X and Y, and then solve 1 equation and plug into the other.
 
  • #18
27
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Uk=(Fk)/(Fn)
Fn would be Wcos(25) so (.30)(250cos(25))=Fk=67.97 which is less than 75 (from part a) and my teacher said it would be SLIGHTLY less, so maybe...? assuming my part a is right
 
  • #19
27
0
anyone?
 

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