Finding coefficient of kinetic friction from thermal energy?

In summary, a rope is used to pull a 357/100 kg block across a horizontal floor at a constant speed of 203/50 m. The force on the block from the rope is 768/100 N and directed 15° above the horizontal. The work done by the rope's force is 30.12 J and the increase in thermal energy of the block-floor system is also 30.12 J. The coefficient of kinetic friction between the block and the floor is 0.225. The relation W=ΔE_th is used to equate the work done by the rope's force to the increase in thermal energy, as the block does not experience a change in mechanical energy.
  • #1
Eclair_de_XII
1,083
91

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


Wwwah4q.png

(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.
 
Physics news on Phys.org
  • #2
Eclair_de_XII said:
(mg)(sinθ)
Please explain these factors.
 
  • #3
That's the normal force.
 
  • #4
Eclair_de_XII said:
That's the normal force.
How do you deduce that?
 
  • #5
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.
 
Last edited:
  • #6
Eclair_de_XII said:
just ##F_N=mg##.
Better, but still wrong. What forces acting on the block have at least some vertical component?
 
  • #7
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!
 
  • #8
Eclair_de_XII said:
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!
 
  • #9
Oh, one more thing. I need to verify if this is correct or not:

Eclair_de_XII said:
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.
 
  • #10
Eclair_de_XII said:
(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?
 
Last edited:
  • #11
kuruman said:
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?
 
  • #12
Eclair_de_XII said:
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.
 
  • #13
Eclair_de_XII said:
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.
 
  • #14
I see. Thank you for taking the time to explain this to me. It's hard to learn this without collaboration.
 

Related to Finding coefficient of kinetic friction from thermal energy?

1. How is the coefficient of kinetic friction related to thermal energy?

The coefficient of kinetic friction is a measure of the amount of resistance between two surfaces in contact when one of the surfaces is in motion. This resistance is caused by the conversion of kinetic energy into thermal energy, which is why the coefficient of kinetic friction is directly related to thermal energy.

2. What is the formula for finding the coefficient of kinetic friction?

The formula for finding the coefficient of kinetic friction is μk = Fk/N, where μk is the coefficient of kinetic friction, Fk is the force of kinetic friction, and N is the normal force between the two surfaces in contact.

3. How can the coefficient of kinetic friction be measured?

The coefficient of kinetic friction can be measured by conducting experiments in which the force of kinetic friction and the normal force are both known. By plugging these values into the formula μk = Fk/N, the coefficient of kinetic friction can be calculated.

4. What factors can affect the coefficient of kinetic friction?

The coefficient of kinetic friction can be affected by factors such as the types of materials in contact, the roughness or smoothness of the surfaces, the temperature, and the presence of any lubricants.

5. Why is it important to know the coefficient of kinetic friction in certain situations?

Knowing the coefficient of kinetic friction can help in predicting the behavior of objects in motion, such as the amount of force needed to move an object or the distance it will travel. It is also important in engineering and design to ensure the efficiency and safety of various systems and structures.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
723
  • Introductory Physics Homework Help
2
Replies
55
Views
3K
  • Introductory Physics Homework Help
Replies
28
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
2
Replies
48
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
2K
Back
Top