Big Test Prep-Questions

  • #1
1,010
1
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
 

Answers and Replies

  • #2
James R
Science Advisor
Homework Helper
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The work done by a force is

[tex]W = \vec{F}.\vec{s} = |\vec{F}||\vec{s}|cos \theta[/tex]

where [itex]\theta[/itex] is the angle between the force vector and the displacement vector.
 
  • #3
574
2
Just [tex]W=Fd\cos{\theta}[/tex]. Can you work with the force vectors acting on the block and figure it out? Start with a FBD.
 
  • #4
1,010
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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 [tex] cos \alpha [/tex] will apply as opposed to if they were pushing directly at the 28 degree angle,
 
  • #5
1,010
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I am being really slow with friction, its been awhile I tried setting it up like follows




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

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
 
  • #6
1,010
1
P, being the force
N- Normal
f= uk
there should also be spaces betwen 0 and p and 0 and N
 
  • #7
298
0
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.
 
  • #8
298
0
Yea, i made a small mistake. It does work.
 
  • #9
1,010
1
thanks for you help it worked out when I did the work
 
  • #10
1,010
1
[tex]Fy=N-Psin28-mgcos28 = 0[/tex]

[tex]Fx =Pcos28 - f - mgsin28 = 0[/tex]

[tex] N=Psin28-mgcos28 [/tex]

[tex] Pcos28-ukN-mgsin28 = 0 [/tex]

[tex] Pcos28-uk*(Psin28-mgcos28)-mgsin28=0 [/tex]

[tex] Pcos28-ukPsin28-ukmgcos28-mgsin28=0 [/tex]

[tex] P(cos28-uksin28)=ukmgcos28+mgsin28 [/tex]

[tex] P=\frac{ukmgcos28+mgsin28}{cos28-uksin28} [/tex]

[tex]P=\frac{.19*52.3*9.8*cos28+52.3*9.8*sin28}{cos28*.19sin28} [/tex]

[tex]P=411.474[/tex]

[tex] W=P(dot)d [/tex]

[tex] W=P*d*cos28 [/tex]

[tex] W=411.474*5.95*cos28=2161.69 [/tex]

thanks for the help, just wanted to show you that I actually used your advice
 
  • #11
1,010
1
The thing I need help is with comming up with the initial values the
[tex]Fy=N-Psin28-mgcos28 = 0[/tex]
[tex]Fy=N-Psin28-mgcos28 = 0[/tex]
 
  • #12
298
0
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.
 

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