Finding friction force and coefficient when pushing a box at constant velocity

[tobyguy]

A man is pushing a box at a constant velocity. The weight of the box is 30 N and the force is 15 N. How do you arrive at the force of friction and the coefficient of sliding friction.

 

[Sirus]

How much net work is being done on the box? What does that tell you about the frictional force, and consequently the coefficient of kinetic friction?

 

[tobyguy]

Totally lost here but thanks for trying to help.

 

[gnome]

"constant velocity"

What does that tell you about the forces acting on the box?

 

[tobyguy]

I really don't know. Sorry. Thats why I am asking here. Thanks though for trying to help. I appreciate it.

 

[schattenjaeger]

Constant velocity means there is no acceleration, hence a=0, so the force of friction is cancelling out that applied force...or something like that, I think

 

[gnome]

Read the chapter again. Sorry if that sounds harsh, but it's really what you need to do. Otherwise you'll be just as lost with every question.

 

[tobyguy]

See that's the problem there is only a very small section on this and nothing that helped but again thanks for your suggestion.

 

[Sirus]

Consider the following formulas.

$$F_{net}=ma$$

$$W=Fd\cos{\theta}$$

W is work done by a force F over a distance d, in this case. Now consider the other posts in this thread. Does it make any sense?

 

[tobyguy]

I don't know. I have been researching this on the Net and as close as I have come is possibly the coefficient would be 0.5 and force would be 15 =30 x .5 so force would be 15 but I am not sure. Anyway thanks for the help. Bedtime for me. :)

 

[Sirus]

The main concept to understand here is that, because of the first equation in my last post, no net force acts on the box since its velocity is not changing. No work is being done on the box since the kinetic energy is constant, so the frictional forces must be equal in magnitude and opposite in direction to the applied force. Your answers above are correct, therefore.

 

[gnome]

tobyguy: you're right, but you obviously have to work on it some more to convince yourself of that. Is there anything in particular that's puzzling you?

Sirus: his question and today's date both suggest that he is one or two days into a first course in physics, so any talk about work and energy is probably a few weeks premature.

 

[tobyguy]

Correct gnome :) Thanks for your help all !