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anonymous
 3 years 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?"
anonymous
 3 years 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?"

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do you set up the question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First, 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok so newtons second law you would use f=ma?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. \(\Sigma F = m\Sigma a\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we know that m=25 kg and what is the a?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352779234119:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, since it is constant velocity, a=0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for your a you should get 1.96 m/s^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so how do you get 1.96 m/s^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... @Sea15era why is there acceleration from the total force?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352779486046:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the box is moving with constant velocity, i.e. no acceleration, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats just a velocity acceleration is in m/s^2 not m/s i believe.

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so how do you find the force that is being exerted?

anonymous
 3 years ago
Best 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im still a bit confused about the Fn= Fg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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 \( F_xF_f=0\), \(F_x=F_f\).

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok i got you so far

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Step by step, have you got the value of \(F_N\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im still confused on how to get Fn

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, so next you will get the value of friction force...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok so how do you find the friction force

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup. 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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