# If a #3# #kg# object moving at #30# #ms^-1# slows to a halt after moving #9# #m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

##### 2 Answers

#### Answer:

There are a couple of possible approaches to beginning this question: I have briefly outlined both below. The later stages of the solution are the same. The frictional coefficient calculated by either method is

#### Explanation:

**Method 1:** starting by finding the force using the acceleration

Rearranging:

The minus sign just means the acceleration is in the opposite direction to the initial velocity, which makes sense because the object slows down. Understanding the direction, we don't need to use the minus sign in the rest of our calculations.

Knowing the mass, we can use Newton's Second Law to find the force acting, which is the frictional force:

From this point onward the approach is the same in both methods:

The frictional force is related to the normal force by the expression:

The normal force is just the weight force of the object,

Frictional coefficients relate forces in

**Method 2:** starting by finding the force from the work done

The work done to stop a moving object is equal to the change in kinetic energy of the object. This one has

The work done is given by

Now we know the frictional force - and we're very happy that it's the same as the force we calculated using a different method! We can go through the rest of the steps in the same way and calculate the coefficient of friction.

(Side note: