anonymous
  • anonymous
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student uses a force of 15 N, what is the coefficient of kinetic friction of the floor? dont tell me the answer just tell me step by step and let me do the rest till the end of the question
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1449688222607:dw| We're asked to find the coefficient of kinetic friction, which means that we must first find the frictional force. In order to do that, we have to use Newton's 2nd Law!\[\huge \sum \text{F}=ma\]This law only applies to one axis, whether it's y axis, x-axis, or any other conventional u-v axes (such as an incline). Applying this law horizontally, we get:\[\huge \sum \text{F}_x=ma=\text{F}_\text{applied}-\text{F}_\text{frictional}\]Perfect, but we're not exactly done yet. We know that we can always define the frictional force as\[\huge \text{F}_\text{frictional}=\mu \text{N}\]Well, now what's the normal force? By doing another summation in the y-direction, we can see that:\[\huge \sum \text{F}_y=ma=\text{N}-mg=0\]This particular equation is equal to 0 (ma = 0) because it is not accelerating in the y direction! Therefore, we can see that \(\text{N}=mg\) (Newton's 3rd Law) Plugging this back into our previous equation, then we have:\[\huge ma=\text{F}_\text{applied}-\mu mg\]You know everything at this point except for \(\mu\)
anonymous
  • anonymous
ill do it then post it
anonymous
  • anonymous
Sure, sounds good

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anonymous
  • anonymous
ef=12*0.2,=2.4 frictional force= applied force-f frictional force 15N-2.4 F frictional force= 12.6 N im loat here @CShrix
anonymous
  • anonymous
You have a big error here: `frictional force= applied force-f frictional force` It should be: \[\huge \sum \text{F}_x=ma=\text{F}_\text{applied}-\text{F}_\text{frictional}\] And you don't really have to even calculate the frictional force as a numerical value. We need it in terms of other variables in order to find \(\mu\) (coefficient of friction, which is kinetic in this case). I derived all the steps for you in my big post, but if you look at the last equation that I added, all you need to do is plug in values and solve for \(\mu\).
anonymous
  • anonymous
e Fx=12*0.2=f applied-Ffrictional eFx=2.4 =15-? @CShrix
anonymous
  • anonymous
I'm still not sure why you're trying to use that specific equation. \[\huge ma=\text{F}_\text{applied}-\mu mg\] Is the final equation that we have to use.
anonymous
  • anonymous
waittttttttttt i think i got it 2.4=15-u117.6 u117.6=15-2.4 u117.6=12.6 u=12.6/117.6 u=0.107
anonymous
  • anonymous
@CShrix is it correct ^
anonymous
  • anonymous
That's what I got!
anonymous
  • anonymous
yessssssssssssssssssssss
anonymous
  • anonymous
X)
anonymous
  • anonymous
im post a slightly different question in another page can u hep me @CShrix

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