Find the coefficient of kinetic friction between the block

[nelone]

A horizontal force of 45N is required to accelerate a 5kg block by 4m/s2 on a horizontal surface. I need to find the coefficient of kinetic friction between the block and the surface by assuming g= 10m/s2. How would I go about doing this?

 

[Parth Dave]

I always start by writing what I am given first:
a = 4 m/s^2
m = 5 kg
F = 45 N

There are only two forces acting on the object in the x-direction: 45N force and kinetic friction.

You also know that Fnet = ma = 45 - Fk (kinetic friction)
you know what ma is, so you can solve for Fk.
What is your formula for Fk? Use that to find out what the coefficient of Fk is.

 

[maverick280857]

You may skip drawing freebody diagrams for such simple problems, but sometimes a wrong sign can cause errors (propagation) and obviously lead to an incorrect answer. So I suggest that you draw a FBD (even if it be a rough one). That way you can always account for the forces and their directions relative to your chosen frame of reference. You are less likely to make mistakes in complicated problems once you develop a consistent habit to draw them :-)

(Supposing I changed your problem so that the force is not horizontal but inclined with the horizontal at an angle. Now you will have to account for the fact that F has a vertical component that adds to the normal reaction force in countering the weight. It is obviously easier to do so using the diagram.)

Hope that helps

 

[Banana]

Always have FuN with friction. (F = u N). F will always be your horizontal force (or horizontal component if on a ramp), u is the coefficient of friction, and N is the normal force. This is an easy way to remember the formula and will help you remember not to try to use mass. Remember the coefficient of friction is without a unit, so you must have to divide Newtons/Newtons to find it. In this particular problem, you'll need to find the net horizontal force before you have FuN.

Maverick is right; it is always a good idea to draw a diagram.