# Calculate the coefficient of kinetic friction for a sliding box

• Sace Ver
In summary: It leads to result that ##150N=350N+F_f##, which I assume is what you meant by your equation in post #14.So, to find F_f, you would need to use the coefficient of friction.
Sace Ver
Mentor's note: Thread title changed to be decriptive of the problem

## Homework Statement

An electric motor is used to pull a 125 kg box across a floor using a long cable. The tension in the cable is 350N and the box accelerates at 1.2m/s2 (forward) for 5.0s. The cable breaks and the box slows down and stops. Calculating the coefficient of kinetic force?

Given:
m=125 kg
a= 1.2 m/s2
t= 5.0s
Ft=350 N

uk=Fk / FN

## The Attempt at a Solution

Fnet = ma
Fnet = (125kg)(1.2m/s2)
Fnet = 150

Thought Fk would be 350N because that's the Ft.

And then Uk would equal 350N divided by 150 but I got 2.3 when the answer is 0.16 not sure how to get answer?
[/B]

Ya mean, calculate the coefficient of kinetic friction?

You correctly got Fnet=150N, what forces is that the net of?

I thought that since the object is 125 kg it would have a force of gravity of 1225N because 125kg•9.8m/s2
so that'd mean that the normal force would also be 1225N and I know that the force of tension would be 350N. I think that Fg and FN would cancel out leaving me with only FT?

Sace Ver said:
I thought that since the object is 125 kg it would have a force of gravity of 1225N because 125kg•9.8m/s2
so that'd mean that the normal force would also be 1225N and I know that the force of tension would be 350N. I think that Fg and FN would cancel out leaving me with only FT?

haruspex said:
friction would be 350N as well I'm assuming?

Sace Ver said:
friction would be 350N as well I'm assuming?
Why would you assume that?

haruspex said:
Why would you assume that?
Because the force of tension is 350N so the opposite side is 350N as well?

Sace Ver said:
Because the force of tension is 350N so the opposite side is 350N as well?
That would make the net horizontal force zero, but you have already calculated that the net horizontal force is not zero.

Do you understand what the equation ##\Sigma F_x = ma_x## means?

haruspex said:
That would make the net horizontal force zero, but you have already calculated that the net horizontal force is not zero.

Do you understand what the equation ##\Sigma F_x = ma_x## means?

Fnet = mass • acceleration that's all I know

Sace Ver said:
Fnet = mass • acceleration that's all I know
OK, and you used that to calculate Fnet, which is to say the sum of the horizontal forces. The value was not zero. So why did you suggest in post #8 that the horizontal forces should balance, which would give a zero Fnet? Do you understand what net force means?

haruspex said:
OK, and you used that to calculate Fnet, which is to say the sum of the horizontal forces. The value was not zero. So why did you suggest in post #8 that the horizontal forces should balance, which would give a zero Fnet? Do you understand what net force means?
Net force is the sum of all forces acting on an object

Sace Ver said:
Net force is the sum of all forces acting on an object
Right.
As you said, the vertical forces balance, so it just leaves the horizontal forces: tension and friction.
Write out the ##\Sigma F_x=F_{net}## equation using the known horizontal forces and net force.

haruspex said:
Right.
As you said, the vertical forces balance, so it just leaves the horizontal forces: tension and friction.
Write out the ##\Sigma F_x=F_{net}## equation using the known horizontal forces and net force.

(350N) Fx = 150N
Fx = 150N - 350N
Fx = 200N [backwards]

Sace Ver said:
(350N) Fx = 150N
I have no idea where you get that from.
Sace Ver said:
Fx = 150N - 350N
That is not quite what I meant, but it's close. "Fx" was meant to stand for a generic horizontal force. Let's use Ff for frictional force:
##150N=ma_x=F_{x net}=\Sigma F_x=##Tension+Friction##=350N+F_f##.
So, yes, Ff has magnitude 200N.
Now deduce the coefficient of friction.

haruspex said:
I have no idea where you get that from.

That is not quite what I meant, but it's close. "Fx" was meant to stand for a generic horizontal force. Let's use Ff for frictional force:
##150N=ma_x=F_{x net}=\Sigma F_x=##Tension+Friction##=350N+F_f##.
So, yes, Ff has magnitude 200N.
Now deduce the coefficient of friction.
Do I use that entire formula and sub in all my information?

Sace Ver said:
Do I use that entire formula and sub in all my information?
Do you mean the whole of this?
haruspex said:
##150N=ma_x=F_{x net}=\Sigma F_x=##Tension+Friction##=350N+F_f##.
No, that expresses the chain of logic. Do you understand each of the steps in there?
It leads to result that ##150N=350N+F_f##, which I assume is what you meant by your equation in post #14.
It remains to express ##F_f## in some way involving the coefficient of friction.

haruspex said:
Do you mean the whole of this?

No, that expresses the chain of logic. Do you understand each of the steps in there?
It leads to result that ##150N=350N+F_f##, which I assume is what you meant by your equation in post #14.
It remains to express ##F_f## in some way involving the coefficient of friction.
150N = 350N + Ff is what I meant!

Sace Ver said:
150N = 350N + Ff is what I meant!
Just not sure where to go after finding out that the magnitude of Ff equals 200N.

Sace Ver said:
Just not sure where to go after finding out that the magnitude of Ff equals 200N.

Sace Ver said:
Just not sure where to go after finding out that the magnitude of Ff equals 200N.
Use the relevant equation you quoted in your original post.

haruspex said:
Use the relevant equation you quoted in your original post.
So my Ff = 200 N would be subbed into the Fk of the relevant equation?

Sace Ver said:
So my Ff = 200 N would be subbed into the Fk of the relevant equation?
yes, it's the kinetic friction force.

haruspex said:
yes, it's the kinetic friction force.
And FN would be 150N?

Sace Ver said:
And FN would be 150N?
No.
What is meant by the term 'normal force'?

I
haruspex said:
No.
What is meant by the term 'normal force'?
I see I see so it would be uk = Ff/FN
uk = 200N/1225N
uk = 0.16
Thank you so much !

## 1. What is the coefficient of kinetic friction?

The coefficient of kinetic friction is a measure of the resistance between two surfaces in contact when one is in motion. It is a dimensionless value that ranges from 0 to 1, with 0 being no friction and 1 being very high friction.

## 2. How is the coefficient of kinetic friction calculated?

The coefficient of kinetic friction can be calculated by dividing the force required to keep an object in motion by the normal force, or the force exerted by one surface on another perpendicular to the surface.

## 3. What factors affect the coefficient of kinetic friction?

The coefficient of kinetic friction is affected by the nature of the two surfaces in contact, their roughness, and the force pushing them together. It also depends on the speed at which the object is moving and the temperature of the surfaces.

## 4. How is the coefficient of kinetic friction different from the coefficient of static friction?

The coefficient of kinetic friction is the measure of resistance between two surfaces when one is already in motion, while the coefficient of static friction is the measure of resistance between two surfaces when one is not yet in motion. The coefficient of kinetic friction is typically lower than the coefficient of static friction.

## 5. Why is calculating the coefficient of kinetic friction important?

Calculating the coefficient of kinetic friction is important in many real-world applications, such as designing machinery and vehicles, understanding the movement of objects on different surfaces, and predicting the behavior of materials in various conditions. It also helps in minimizing friction and improving efficiency in various processes.

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