Work done on a system by an external force Problem

[VitaX]

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 coefficent*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?

 

[VitaX]

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.

 

[PhanthomJay]

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.

 

[VitaX]

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?

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.

 

[Doc Al]

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.)

 

[VitaX]

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?

 

[Doc Al]

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.

 

[VitaX]

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.

 

[Doc Al]

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

Ah... OK.

 

[VitaX]

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?

 

[Doc Al]

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.

 

[VitaX]

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.

 

[Doc Al]

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.

 

[VitaX]

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 ?

 

[Doc Al]

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.

 

[VitaX]

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.