HELP ME PLS IM STUCK i need this :): A 13.3 kg box sliding across the ground
decelerates at 2.42 m/s2. What is the
coefficient of kinetic friction?
(No unit)
(ps the answer is not 0.242)

μk = 0.247

Explanation:

To start off, let's draw a free body diagram. My teacher tells us to do that first, and I find that's it's very helpful. I have attached a free body diagram to this- please take a look at it.

Forces:

Friction- as mentioned in the problem, friction is the only horizontal force acting on the object

Normal force- whenever an object is on a surface, it has normal force.

Weight- weight always acts on objects.

Does ∑F=ma apply here?

To determine this, we need to see if the object is at equilibrium. If it is, that means the net force is zero, so this equation would not apply. If an object is at equilibrium, it is either at rest or is at constant velocity. Since we know there is an acceleration, it cannot be at equilibrium. So this equation does apply.

Find the net force (∑F)

To do this, we need to look at all the horizontal forces acting on the object. We need not look at vertical forces (weight and normal force) because the object is only moving horizontally. The only horizontal force is friction, so that is also the net force.

Rewrite the equation:

Friction = ma

Fill in what we know:

Friction = (13.3)(-2.42)

Friction = -32.186 N

Use kinetic friction formula

The formula is:

Friction = μkN

μk = coefficient

N = Normal force

Find the normal force

To do this, first see whether the object is at equilibrium VERTICALLY. Is it acceleration up/down? No. So it IS at equilibrium vertically. When an object is at equilibrium in one direction, all the forces in that direction will be equal. Since the vertical forces are weight and normal force, they are equal.

Find Weight

Weight= mg= (13.3)(9.8)= 130.34 N

Normal force= 130.34 N

Find coefficient of kinetic friction

Friction = μkN

-32.186 = μk (130.34)

*Divide both sides*

μk = 0.247

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