How Do You Calculate Work Done on an Incline with Friction?

[Tom McCurdy]

Big Test Prep--Questions

Alright we are having a review test and I needed to go back to the stuff we did earlier this year and I am having some trouble remembering what to do...
:shy: how can i forget this stuff you ask... I don't know


7.
A 52.3 kg trunk is pushed 5.95 m at a constant speed up a 28.0 degree incline by a constant horizontal force. The coefficent of kinetic friction beeween the trunk and the incline is .19 Calculate the work done by (a) the applied force (i got b)


Answer:2160 J

 

[James R]

The work done by a force is

W = \vec{F}.\vec{s} = |\vec{F}||\vec{s}|cos \theta

where \theta is the angle between the force vector and the displacement vector.

 

[Sirus]

Just W=Fd\cos{\theta}. Can you work with the force vectors acting on the block and figure it out? Start with a FBD.

 

[Tom McCurdy]

Indded that is the forumula, however it is in calculating F, that I am having difficulty
I did just realize that it is horizontal so the cos \alpha will apply as opposed to if they were pushing directly at the 28 degree angle,

 

[Tom McCurdy]

I am being really slow with friction, its been awhile I tried setting it up like follows




\sum{x=0}P-f cos\alpha=0
\sum{y=0}N-mp-fsin\alpha=0

however I have a strong feeling this is incorrect, I have also tried 2 other ways but they are more involved and require a lot more work to show, and I know they don't work

 

[Tom McCurdy]

P, being the force
N- Normal
f= uk
there should also be spaces betwen 0 and p and 0 and N

 

[Parth Dave]

The applied force is horizontal. So you have four forces: The normal force (points directly up), friction (pointing back), gravity (pointing downwards) and your applied force, pointing horizontally. These are the equations i get:

Fy = Fn - Asin28 - mgcos28 = 0
Fx = Acos28 - Ff - mgsin28 = 0

You might want to double check that. I may have made a mistake somewhere. Solve for normal force in Fy and then plug that into Fx. Then you can solve for your applied force. When I did it, i got a slightly different answer. That may have been an algebraic mistake (hopefully). Either way, do it for yourself and see what you come up with.

 

[Parth Dave]

Yea, i made a small mistake. It does work.

 

[Tom McCurdy]

thanks for you help it worked out when I did the work

 

[Tom McCurdy]

Fy=N-Psin28-mgcos28 = 0

Fx =Pcos28 - f - mgsin28 = 0

N=Psin28-mgcos28

Pcos28-ukN-mgsin28 = 0

Pcos28-uk*(Psin28-mgcos28)-mgsin28=0

Pcos28-ukPsin28-ukmgcos28-mgsin28=0

P(cos28-uksin28)=ukmgcos28+mgsin28

P=\frac{ukmgcos28+mgsin28}{cos28-uksin28}

P=\frac{.19*52.3*9.8*cos28+52.3*9.8*sin28}{cos28*.19sin28}

P=411.474

W=P(dot)d

W=P*d*cos28

W=411.474*5.95*cos28=2161.69

thanks for the help, just wanted to show you that I actually used your advice

 

[Tom McCurdy]

The thing I need help is with comming up with the initial values the
Fy=N-Psin28-mgcos28 = 0
Fy=N-Psin28-mgcos28 = 0

 

[Parth Dave]

Well the only suggestion i can make is that you draw yourself a free-body diagram and then start labelling your forces. Resolve them into your x and y components. Then just add them up.