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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?"
 one year ago
 one year ago
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?"
 one year ago
 one year ago

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abannavongBest ResponseYou've already chosen the best response.0
how do you set up the question?
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
I know that m= 25kg
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
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.
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
ok so newtons second law you would use f=ma?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
yes. \(\Sigma F = m\Sigma a\)
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
so we know that m=25 kg and what is the a?
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
dw:1352779234119:dw
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
well, since it is constant velocity, a=0.
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
for your a you should get 1.96 m/s^2
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
so how do you get 1.96 m/s^2
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
Hmm... @Sea15era why is there acceleration from the total force?
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
dw:1352779486046:dw
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
but the box is moving with constant velocity, i.e. no acceleration, right?
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
thats just a velocity acceleration is in m/s^2 not m/s i believe.
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.
 one year ago

Sea15eraBest ResponseYou've already chosen the best response.0
I hope this helps you , good luck and if by chance it's wrong I'm sorry!
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
so how do you find the force that is being exerted?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
\(\Sigma F=ma\) 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.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.dw:1352781249983:dw Do you follow?
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
im still a bit confused about the Fn= Fg
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
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 \( F_xF_f=0\), \(F_x=F_f\).
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
On the yaxis, v=0,(it doesn't move vertically), so a=0 too. Since \(\Sigma F_y=F_NF_g= 0\),it follows that \(F_N=F_g\)
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
oh ok i got you so far
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
Step by step, have you got the value of \(F_N\)?
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
im still confused on how to get Fn
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
yeah, so next you will get the value of friction force...
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
oh ok so how do you find the friction force
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
nvm i mean (245)(.2)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.
 one year ago
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