Finding coefficient of kinetic friction from thermal energy?

[Eclair_de_XII]

Homework Statement

"A rope is used to pull a $\frac{357}{100}kg$ block at constant speed $\frac{203}{50}m$ across a horizontal floor. The force on the block from the rope is $\frac{768}{100}N$ and directed $15°$ above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and the floor?"

Homework Equations

$W=ΔE_mec+ΔE_th=(ΔK+ΔU)+ΔE_th$
$ΔU=0$
$K=\frac{1}{2}mv^2$
$f_k=(mg)(sin\theta)$

The Attempt at a Solution

(a) $W=ΔK=F⋅Δx=(F)(cos\theta)(Δx)=(\frac{768}{100}N)(cos15°)(\frac{203}{50}m)=30.12J$.

(b) Then I use the relation $W_{block-floor system}=-ΔE_{mec}+ΔE_{th}=0$ to equate those two terms to each other, since the block releases mechanical energy, and gains thermal energy. Someone please correct me because I'm very sure that I am incorrect with this reasoning.

$ΔE_{th}=ΔE_{mec}=30.12J$

(c) Now I move onto finding the coefficient of static friction using the relation:

$ΔE_{th}=(f_k)(Δx)=(mg)(sin\theta)(μ_k)(Δx)$
$μ_k=\frac{ΔE_{th}}{(mg)(sin\theta)(Δx)}=\frac{(F)(cos\theta)(Δx)}{(mg)(sin\theta)(Δx)}=\frac{(F)(cot\theta)}{mg}=0.82$

I don't understand what it is that I'm doing wrong.

 

[haruspex]

(mg)(sinθ)

Please explain these factors.

 

[Eclair_de_XII]

That's the normal force.

 

[haruspex]

That's the normal force.

How do you deduce that?

 

[Eclair_de_XII]

It acts perpendicular to the motion. So I suppose that it is just $F_N=mg$.
Edit: According to Wikipedia, it is "perpendicular to the surface of contact".
I'm still not getting the right numbers:

$μ_k=\frac{(F)(cos\theta)}{mg}=0.212$

Sorry for the multiple edits.

 

[haruspex]

just $F_N=mg$.

Better, but still wrong. What forces acting on the block have at least some vertical component?

 

[Eclair_de_XII]

I'm guessing it's the rope tugging on the block. So is normal force: $F_N=(mg)-(sin\theta)(F)$?

$μ_k=\frac{(F)(cos\theta)}{(mg)-(sin\theta)(F)}=0.225$

Awesome. Thanks!

 

[haruspex]

I'm guessing it's the rope tugging on the block. So is normal force: $F_N=(mg)-(sin\theta)(F)$?

$μ_k=\frac{(F)(cos\theta)}{(mg)-(sin\theta)(F)}=0.225$

Awesome. Thanks!

Ok!

 

[Eclair_de_XII]

Oh, one more thing. I need to verify if this is correct or not:

Then I use the relation $W_{block−floorsystem}=−ΔE_{mec}+ΔE_{th}=0$ to equate those two terms to each other, since the block releases mechanical energy, and gains thermal energy. Someone please correct me because I'm very sure that I am incorrect with this reasoning.

 

[kuruman]

(b) Then I use the relation $W_{block−floorsystem}=−ΔE_{mec}+ΔE_{th}= 0$ to equate those two terms to each other, since the block releases mechanical energy, and gains thermal energy. Someone please correct me because I'm very sure that I am incorrect with this reasoning.

It looks like you are missing the point of the energy conservation equation. The block is moving at constant velocity on a horizontal surface. Just on that piece of information, what is its change in mechanical energy?

 

[Eclair_de_XII]

The block is moving at constant velocity on a horizontal surface. Just on that piece of information, what is its change in mechanical energy?

Let's see... if there is no change in velocity, there is no change in kinetic energy. So assuming no change in potential energy, the mechanical energy does not change?

 

[Eclair_de_XII]

So assuming no change in potential energy, the mechanical energy does not change?

Hold on; this is a flat surface. There is no (gravitational) potential energy.

 

[kuruman]

Let's see... if there is no change in velocity, there is no change in kinetic energy. So assuming no change in potential energy, the mechanical energy does not change?

Correct. The mechanical energy does not change, i.e. $\Delta E_{mec}=0$. If the relation $W_{block−floorsystem}=−ΔE_{mec}+ΔE_{th}=0$ were correct, it would imply that $\Delta E_{th}=0$ which is not happening. Look again at the very first equation you posted under "Relevant Equations". It becomes $W=ΔE_{th}$. That's the equation to start from. It says that the work done on the block is entirely converted into thermal energy when the mechanical energy is not changing.

 

[Eclair_de_XII]

I see. Thank you for taking the time to explain this to me. It's hard to learn this without collaboration.