Will a 36 N force overcome static friction on a 45 N block?

[pinky2468]

Ok, here is the question, which I thought was easy at first but...
A block whose weight is 45.0N rests on a horizontal table. A horizontal force of 36.0N is applied to the block. The coefficients of the static and kinetic friction are 0.650 abd 0.420 respectively. Will the block move under the influence of the force and if so what wiil be the block's acceleration?
(u=the coefficient)
I used the fs=us(Fn)= .650(45.0N)= 29.3N
fk=uk(Fn)= .420(45.0)N= 18.9N

I know that this will make the box move, but I am a little confused on where to go from here. Do I use F=MA?
Thanks!

 

[Gokul43201]

Draw a free-body force diagram. Show all the forces acting on the block. Any resultant force in a given direction will cause an acceleration given by F = ma, in that direction.

When the block is moving, which coefficient should you use, static or kinetic ?

 

[pinky2468]

I should use the coefficient of kinetic friction. Also since W=mg, do I divide the weight by 9.80 to get the mass?

 

[Pyrrhus]

I should use the coefficient of kinetic friction. Also since W=mg, do I divide the weight by 9.80 to get the mass?

Yes, you have to see if the force applied will overcome static friction, if it does it will move, and friction is less strong

 

[pinky2468]

I am not getting the right answer when I use M=W/g to get 4.59kg and then a=F/m
should I use 18.9/4.59? It does not give me the right answer of a=3.72m/s(sqr)

 

[Gokul43201]

When you say F = ma, F represents the resultant force in some direction. In this case, we are interested in the horizontal direction.

What are all the forces in the horizontal direction ? What is their resultant ?

You MUST draw a force diagram, else you'll get nowhere.

PS : m = 4.59 kg is correct.