Calculate the coefficient of kinetic friction for a sliding box

[Sace Ver]

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

Homework Equations

u_k=F_k / F_N

The Attempt at a Solution

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

Thought F_k would be 350N because that's the F_t.

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

 

[SteamKing]

Ya mean, calculate the coefficient of kinetic friction?

 

[haruspex]

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

 

[Sace Ver]

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 F_g and F_N would cancel out leaving me with only F_T?

 

[haruspex]

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 F_g and F_N would cancel out leaving me with only F_T?

What about friction?

 

[Sace Ver]

What about friction?

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

 

[haruspex]

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

Why would you assume that?

 

[Sace Ver]

Why would you assume that?

Because the force of tension is 350N so the opposite side is 350N as well?

 

[haruspex]

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?

 

[Sace Ver]

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

 

[haruspex]

Fnet = mass • acceleration that's all I know

OK, and you used that to calculate F_net, 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 F_net? Do you understand what net force means?

 

[Sace Ver]

OK, and you used that to calculate F_net, 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 F_net? Do you understand what net force means?

Net force is the sum of all forces acting on an object

 

[haruspex]

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.

 

[Sace Ver]

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]

 

[haruspex]

(350N) Fx = 150N

I have no idea where you get that from.

Fx = 150N - 350N

That is not quite what I meant, but it's close. "F_x" was meant to stand for a generic horizontal force. Let's use F_f for frictional force:
$150N=ma_x=F_{x net}=\Sigma F_x=$Tension+Friction$=350N+F_f$.
So, yes, F_f has magnitude 200N.
Now deduce the coefficient of friction.

 

[Sace Ver]

I have no idea where you get that from.

That is not quite what I meant, but it's close. "F_x" was meant to stand for a generic horizontal force. Let's use F_f for frictional force:
$150N=ma_x=F_{x net}=\Sigma F_x=$Tension+Friction$=350N+F_f$.
So, yes, F_f has magnitude 200N.
Now deduce the coefficient of friction.

Do I use that entire formula and sub in all my information?

 

[haruspex]

Do I use that entire formula and sub in all my information?

Do you mean the whole of this?

$150N=ma_x=F_{x net}=\Sigma F_x=$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.

 

[Sace Ver]

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]

150N = 350N + Ff is what I meant!

Just not sure where to go after finding out that the magnitude of F_f equals 200N.

 

[insightful]

Just not sure where to go after finding out that the magnitude of F_f equals 200N.

How does your F_f relate to your F_k?

 

[haruspex]

Just not sure where to go after finding out that the magnitude of F_f equals 200N.

Use the relevant equation you quoted in your original post.

 

[Sace Ver]

Use the relevant equation you quoted in your original post.

So my F_f = 200 N would be subbed into the F_k of the relevant equation?

 

[haruspex]

So my F_f = 200 N would be subbed into the F_k of the relevant equation?

yes, it's the kinetic friction force.

 

[Sace Ver]

yes, it's the kinetic friction force.

And F_N would be 150N?

 

[haruspex]

And F_N would be 150N?

No.
What is meant by the term 'normal force'?

 

[Sace Ver]

I

No.
What is meant by the term 'normal force'?

I see I see so it would be u_k = F_f/F_N
u_k = 200N/1225N
u_k = 0.16
Thank you so much !