Work done on a system by an external force Problem

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Homework Help Overview

The problem involves calculating the work done by an external force on a block being pushed along a level floor, considering factors such as friction and the angle of the applied force. The subject area includes concepts from mechanics, specifically work-energy principles and friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of work done, the role of friction, and the influence of the angle of the applied force. Questions arise regarding the correct interpretation of the normal force and its relationship to the weight of the block. Some participants express uncertainty about the calculations and the implications for thermal energy.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem and attempting to clarify their calculations. Some guidance has been offered regarding the correct use of angles in work calculations and the relationship between work and thermal energy. There is no explicit consensus on the final answers, as participants are still working through their reasoning.

Contextual Notes

Participants mention the constraints of the homework platform, which generates new values if previous answers are incorrect. This has led to adjustments in calculations and assumptions about the normal force and applied force components.

VitaX
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Homework Statement



A worker pushed a 16 kg block 6.7 m along a level floor at constant speed with a force directed 32° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.34, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Homework Equations



W = (K2 - K1) + (Ug2 - Ug1) + fkd

The Attempt at a Solution



Part a:

W = 0 + 0 + fkd

W = Friction coefficient*n*d

In this case n=mg

W = .34*(16*9.8)*6.7 = 357.1904 J

Part b:

I'm thinking its either 0 or 357.1904 J, but I'm not sure that's why I posted this. I mean I just found the change in thermal energy in part a pretty much. So is that my answer to part b then since it is asking for the increase in thermal energy of the block-floor system? Or is there more to it?
 
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It seems my answer to part a is incorrect according to my homework site. I tried inputting 357.1904 J for both a and b since it made sense to me after reading through the book. But apparently my work for part a is off which is strange to me because the change in K is 0 because the speed is constant. The change in U is 0 because the height is left unchanged. Only thing left is just Fkd. I probably have to utilize that angle but I'm not quite sure how and what to use it in.
 
The normal force is greater than mg because the applied force has a vertcal component to it. Then work is the dot product of force and displacement. The work done by friction belongs on the left side of the equation...the heat thermal energy from this work is the negative of this work.
 
PhanthomJay said:
The normal force is greater than mg because the applied force has a vertcal component to it. Then work is the dot product of force and displacement. The work done by friction belongs on the left side of the equation...the heat thermal energy from this work is the negative of this work.
Is this picture drawn correctly?
262y8gm.png


How do I solve for F? When I write out the x and y components for the forces I get this:

Focs32 + ukn = 0
Fsin32 + n - mg = 0

I can't solve for F because I have 2 unknowns in each equation. What am I doing wrong? Or perhaps this is completely the wrong approach in finding the force that I need to find for part a to be able to find the Work.
 
Last edited:
VitaX said:
Is this picture drawn correctly?
Yes.
How do I solve for F? When I write out the x and y components for the forces I get this:

Focs32 + ukn = 0
Good.
Fsin32 + n - mg = 0
The y component of F should be negative.

I can't solve for F because I have 2 unknowns in each equation.
You have two equations and two unknowns. Just right! (Solve them together, using substitution or any other method.)
 
Ok I have the following for part a:
Fcos328 - ukn = 0
Fsin328 + n - mg = 0 ----> n = mg - Fsin328

Fcos328 - uk(mg - Fsin328) = 0 -----> F = .5185 N

.5185cos328 - .34n = 0 ------> n = 1.2933 N (Don't know if it's necessary to find for part b or not, I think yes)

W = Fd = .5185*6.7 = 3.474 J

Hows it look to be? Is my work and way I found it correct?

So is the answer to part b just the negative of that work. -3.474 J for increase in thermal energy?

I think something is wrong because Jay said the normal force is greater than mg, when in my work it is well below mg. What happened?
 
Last edited:
VitaX said:
I think something is wrong because Jay said the normal force is greater than mg, when in my work it is well below mg. What happened?
You never corrected the error that I pointed out in my last post. You have the wrong sign for the y-component of the applied force. You'll need to rewrite your equation.
 
Doc Al said:
You never corrected the error that I pointed out in my last post. You have the wrong sign for the y-component of the applied force. You'll need to rewrite your equation.

I corrected it my changing the degrees to 328. That way sin is negative.

Edit: I think I see my error and its my calculation of ukmg its supposed to be 53.312 not .53312
When I fixed that my Force is 51.8515 N and Normal force is 129.3312 N. Either way the weight force is still greater then the normal force, but that should be shouldn't it? Since it's keeping it on the ground.
 
Last edited:
VitaX said:
I corrected it my changing the degrees to 328. That way sin is negative.
Ah... OK.
 
  • #10
Doc Al said:
Ah... OK.

How do you find part b? Is it the same value calculated for work in part a just negative that or is there more to it then that?
 
Last edited:
  • #11
VitaX said:
Edit: I think I see my error and its my calculation of ukmg its supposed to be 53.312 not .53312
When I fixed that my Force is 51.8515 N and Normal force is 129.3312 N. Either way the weight force is still greater then the normal force, but that should be shouldn't it? Since it's keeping it on the ground.
Please redo your calculations. Your value for F is incorrect. And the fact that your normal force is less than the weight indicates a problem.

Just start with your two equations and redo your work carefully.
 
  • #12
I did the question with new values (Since on Wiley if you input an answer and it's wrong it gives you new values to re do the question sadly) my normal force is greater than the weight force now so it seems my work is correct (W = Fd). For part b I'm thinking it is either the negative of the work calculated in part a or its uk*n*d. Not sure which one is correct though so if someone could give me a heads up that'd be great.
 
  • #13
VitaX said:
I did the question with new values (Since on Wiley if you input an answer and it's wrong it gives you new values to re do the question sadly) my normal force is greater than the weight force now so it seems my work is correct (W = Fd).
W = Fd is not correct; don't forget the angle between F and d.
 
  • #14
Doc Al said:
W = Fd is not correct; don't forget the angle between F and d.

Ok W = Fdcos(theta) Should I use a theta value of 360 - given angle. Don't think it matters either way but I guess I'll do 360 - given angle. Ok so that finds Work. Is part b increase in thermal energy = fkd = uk*n*d ?
 
  • #15
VitaX said:
Ok W = Fdcos(theta) Should I use a theta value of 360 - given angle. Don't think it matters either way but I guess I'll do 360 - given angle.
The angle between the force and the displacement is given as 32 degrees.
Ok so that finds Work. Is part b increase in thermal energy = fkd = uk*n*d ?
Yes.
 
  • #16
Doc Al said:
The angle between the force and the displacement is given as 32 degrees.

Yes.

K thanks for the help, entered my answers in and both are correct.
 

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