Physics help with friction! the problem is "You push a 25kg wooden box across a wooden floor at a constant velocity of 1.0 m/s. If the coefficient of friction is .20, how much force do you exert on the box?"

- anonymous

- schrodinger

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- anonymous

how do you set up the question?

- anonymous

I know that m= 25kg

- anonymous

F=\[nuN\]

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## More answers

- anonymous

First, you draw the free body diagram.
And note that the question tells you:
there is no acceleration(constant velocity)
Then you apply Newton's second law to it.

- anonymous

ok
so newtons second law you would use f=ma?

- anonymous

yes. \(\Sigma F = m\Sigma a\)

- anonymous

oh ok a

- anonymous

so we know that m=25 kg and what is the a?

- anonymous

|dw:1352779234119:dw|

- anonymous

first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25

- anonymous

well, since it is constant velocity, a=0.

- anonymous

for your a you should get 1.96 m/s^2

- anonymous

oh ok

- anonymous

so how do you get 1.96 m/s^2

- anonymous

Hmm... @Sea15era why is there acceleration from the total force?

- anonymous

|dw:1352779486046:dw|

- anonymous

but the box is moving with constant velocity, i.e. no acceleration, right?

- anonymous

thats just a velocity acceleration is in m/s^2 not m/s i believe.

- anonymous

F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.

- anonymous

oh ok

- anonymous

I hope this helps you , good luck and if by chance it's wrong I'm sorry!

- anonymous

Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.

- anonymous

so how do you find the force that is being exerted?

- anonymous

\(\Sigma F=ma\)
Start with this picture. |dw:1352781084312:dw|
As shown, the forces are F_x and F_f, of different directions.
Applying the second law,
\(F_x - F_f =ma=0\) as it is constant velocity.

- anonymous

Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.|dw:1352781249983:dw|
Do you follow?

- anonymous

kinda

- anonymous

im still a bit confused about the Fn= Fg

- anonymous

Now, as \(F_g = m g\), where m=25kg and g=9.8, so you get \(F_g\)
Now after you get \(F_g\) you can get \(F_N\)
After you get \(F_N\), You get \(F_f\)
Then since \( F_x-F_f=0\), \(F_x=F_f\).

- anonymous

On the y-axis, v=0,(it doesn't move vertically), so a=0 too.
Since \(\Sigma F_y=F_N-F_g= 0\),it follows that \(F_N=F_g\)

- anonymous

oh ok i got you so far

- anonymous

Step by step, have you got the value of \(F_N\)?

- anonymous

im still confused on how to get Fn

- anonymous

i got Fg= 245

- anonymous

yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.

- anonymous

oh ok

- anonymous

yeah, so next you will get the value of friction force...

- anonymous

oh ok so how do you find the friction force

- anonymous

\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.

- anonymous

oh ok

- anonymous

(.20)(25) ?

- anonymous

nvm i mean (245)(.2)

- anonymous

yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.

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