# Dynamics: finding the coefficent of friction.

## Homework Statement

The question:
It takes a 5.0 N force to pull a 2.0kg object along the ground. What is the coefficient of Friction.

Its Pretty Blunt, and i have been stuck on it for about 10 minutes, There must be a step that i dont know or am missing because i get stuck

## Homework Equations

Sum of the Forces = m*a
Force of Friction = Coefficient of Friction*Force Normal
Fg=m*a

## The Attempt at a Solution

There must be a equation i am missing, to find out the Force of Friction
heres where i get to:
Fp= 5N
m=2kg
Fg-Fn= 0
Fg= 19.6
Fn= 19.6

I would be extremely grateful if i could get some help on this question

It takes a 5.0 N force to pull a 2.0 kg object along the ground. What is the coefficient of friction? 0.26
http://www.mouatonline.com/Teachers/BHutchinson/Physics/Phys11/Unit%203/PS36.htm" [Broken] you have the solution (No procedure). It seems that you don't take into consideration acceleration. Maybe 5 N is the force that moves the body from the rest position. Google a little bit, problems with similar text say that body is moving with constant velocity.

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Andrew Mason
Homework Helper
There must be a equation i am missing, to find out the Force of Friction
heres where i get to:
Fp= 5N
m=2kg
Fg-Fn= 0
Fg= 19.6
Fn= 19.6

$$F = \mu_kN$$

where F is the force of friction and N is the normal force. $\mu_k$ is the ratio of friction force to normal force. You are given the friction force (ie. the applied force is equal and opposite to the kinetic friction force as there is no acceleration). You can work out the normal force from the mass. Divide the former by the latter.

AM

Thanks, could you please show me the formulas and methods used to find 0.26. Its the right anwser but i still dont know how to do it.

How did you get acceleration for $$\Sigma$$forces = m*a

Andrew Mason
Homework Helper
Thanks, could you please show me the formulas and methods used to find 0.26. Its the right anwser but i still dont know how to do it.
?? I gave you the formula!! It is:

$$F = \mu_kN$$

The normal force exactly balances the gravitational force, mg. So:

$$F = \mu_kmg$$

$$\mu_k = F/mg = 5/2*9.8 = .26$$

AM