Finding static and kinetic friction coefficients for a box on concrete

[Jazzybelle4]

Homework Statement

A force of 48.0 N is required to start a 5.0 kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor?
(b) if the 48.0N force continues the box accelerates at 0.70m/sec2. what is the coefficient of kinetic friction


I have no idea how to set up up problem (a), all i know is the Ffr=us(mg)

 

[Kurdt]

You assume the force that moves the box is only just enough to overcome static friction and thus you can set the max force of static friction equal to the force applied to the box.

 

[Jazzybelle4]

so then the equation set up would look like:
Ffr max= us(mg)
48N=us(mg)

 

[Kurdt]

Yes. You can then find the coefficient.

 

[Jazzybelle4]

I found the coefficient it is .98 now i do not know what to do at all for the second part of the question (b)

 

[Kurdt]

What force would be required to accelerate the box at that rate without any friction?

 

[Jazzybelle4]

any force about 48.0? there will always be friction. the coefficient needs to be smaller at .98

 

[Kurdt]

What about Newton's second law. You can work out the force required to make a mass the same as the box accelerate at the value you're given. How does that compare to the force you're having to use?

 

[Jazzybelle4]

second law is
F=ma
48=m(.70)
I am so lost... my teacher doesn't explain anything

 

[Kurdt]

You're finding the force required to accelerate a 5 kg mass at 0.7 ms[su[]-2[/sup]. You don't need the 48 in there. Hang in there all will become clear.