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Work Problem - Distance of particle moving up an incline

  1. Jul 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi! I'm having issues with this practice problem. Any advice on what I'm doing wrong would help!!!

    A 2.5 kg particle is projected with an initial speed of 3.3m/s along a surface for which the coefficient of friction is 0.7. Find the distance it travels given that the particle moves up a 26 degree incline.

    2. Relevant equations
    W=ΔKE
    W=F*d
    Ff=μFn
    W=ΔKE+ΔPE
    KE=1/2mv2
    PE=mgh

    3. The attempt at a solution
    mg=2.5kg*-9.8m/s2=-24.5N

    -24.5Nsin26=-10.74 N = mgx
    -24.5Ncos26=-22.02N = mgy

    ΣFy=0=Fn-mgy
    Fn= 22.02 N

    Ff = Fnμ
    Ff = 22.02N*0.7
    Ff=15.14 N

    W= KEf-KEf+PEf-PEi
    W=.5mv2-.5mv2+mgh-mgi
    W=0.5*0.3kg*(3.3m/s)2-0+2.5kg*9.8*h-0
    W=F*d

    Here I assumed KEf was 0 because the final velocity is 0, and I assumed PEi was 0 because the inital height is 0.

    The forces applied that aren't in equilibrium are mgx, and Ff.

    W= (-15.14N-10.74N)*d

    -26.15N*d=0.5*0.3kg*(3.3m/s)2-0+2.5kg*9.8*h

    So instead of having 2 unknown variables, I made h equal to...

    h= d*sin26

    So.. I inserted that in the previous equation:

    -26.15N*d=0.5*0.3kg*(3.3m/s)2-0+2.5kg*9.8*d*sin26

    Factored out d...

    d(-26.15+24.5sin26)=-13.6125

    d=-13.6125/(-26.15+24.5sin16)
    d=0.88m

    And this answer is incorrect. I am unsure where I went wrong... Any thoughts? Thank you!!!

    For an image of what my free body diagram looks like, I posted it below!!
     
    Last edited: Jul 9, 2017
  2. jcsd
  3. Jul 9, 2017 #2
    I'm not really sure what you are doing here. You have effectively already calculated the normal force here
    -24.5Ncos25=-22.02N = mgy
    although the angle is wrong. It is also unclear why you are setting this equal to mgy.
     
  4. Jul 9, 2017 #3
    Hi! Thanks for taking the time to answer.
    The 25 degrees was a typo, I did use 26 degrees to calculate mgy.
    I calculated the horizontal gravitational force, mgx, as I believe (its force applied to the object + the friction force) multiplied by the distance, are equal to work.
    A.k.a. the applied forces to the object which are not in equilibrium (and the x axis is not in equilibrium)

    Is that an incorrect assumption? If so, any pointers?
     
  5. Jul 9, 2017 #4
    I was calculating the gravitational force, the mgy = gravitational force along the y (perpendicular to the incline)
    I then made the value positive later and said that it was equal to the normal (Fn). Because ΣFy=0
     
  6. Jul 9, 2017 #5

    SammyS

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    Please explain how you have defined your coordinate system.

    Using a free-body-diagram would be helpful.
     
  7. Jul 9, 2017 #6
  8. Jul 9, 2017 #7
    IMG_4919.JPG
     
  9. Jul 9, 2017 #8

    SammyS

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    You are using two different expressions for work.

    The net work is given by Wnet = Fnet⋅d ,
    and is related to the change in kinetic energy by: Wnet = ΔKE .

    The work done by non-conservative forces is related to the change in PE and KE by: Wnon-con = ΔKE+ΔPE .

    You were using these interchangeably, and therefore accounted gravity's effect twice.

    The above is true for the work done by non-conservative forces. The PE accounts for effects of gravity.

    The following gives the net Force and thus the net Work.
    That last equation has effects of gravity taken into account twice.
     
  10. Jul 9, 2017 #9
    Ohhh! So.. Work will be

    Wnet=Fnet*d

    and Fnet will only be equal to that of the frictional force?
     
  11. Jul 9, 2017 #10
    Using the previous equation, I have tried this instead:

    Fnet=-15.41 N

    Fnet*d=½(0.3kg)(3.3m/s)2+(2.5kg)(9.8m/s2)(dsin26°)

    d(-15.41N+(2.5kg)(9.8m/s2))=½(0.3kg)(3.3m/s)2

    d=½(0.3kg)(3.3m/s)2/((-15.41N+(2.5kg)(9.8m/s2))

    d=2.91m

    This is incorrect as well. :( Any further pointers? Thank you for your time, effort, and knowledge!!
     
  12. Jul 9, 2017 #11
    No, the gravitational force is still there.

    Both friction and gravity do work here. You know both of these forces and the initial kinetic energy. So how much work is required by these forces to bring the kinetic energy to zero?
     
  13. Jul 9, 2017 #12

    SammyS

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    The net Force: That's the sum of all the forces not in equilibrium. You previously had that correct. Use this for net Work, Wnet.
    Work - Energy Theorem: Wnet = ΔKE.​

    Otherwise, use the WN.C. work done only by non-conservative forces.
    The only non-conservative force here is friction.
    WN.C. = ΔKE+ΔPE​

    It appears that you keep mixing these two this together ..
     
  14. Jul 9, 2017 #13
    Oh. So I should do this..?

    Wnet=F*d

    F being friction and mgx

    And then:

    F*d=0-1/2mvi^2

    Since KEf is 0??

    (Thanks again)
     
  15. Jul 9, 2017 #14
    Hi, thanks for popping in to help.

    So does this make more sense?

    Wnet=F*d

    F being friction and mgx

    And then:

    F*d=0-1/2mvi^2

    Since KEf is 0??

    (Thanks)
     
  16. Jul 9, 2017 #15

    SammyS

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    Other than mislabeling and some sign errors this should work.

    What you call Fnet here is just the friction force: That's the Non-Conservative force, not the net force. The work done by this force is equal to the change in KE+PE .

    You messed up the change in KE. Final minus Initial is negative, Right?

    Finally, you have an Algebra error: added PE rather than subtracting .
     
  17. Jul 9, 2017 #16
    Yes
     
  18. Jul 9, 2017 #17
    Thank you!! It is getting late where I am, but I will try this in the morning.

    Thank you x 10000.
     
  19. Jul 10, 2017 #18
    This solved my problem!! THANK YOU
     
  20. Jul 10, 2017 #19

    SammyS

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    So, what are those last few steps as well as you final answer ?

    (We like feedback.)
     
  21. Jul 10, 2017 #20
    This is what I did, based on my conversation with Nfuller

    Wnet=f*d

    f=Ff+mgx

    Ff+mgx=-1/2mvi^2

    (-15.14N-10.74N)*d=-(1/2)(2.5kg)(3.3m/s)^2

    -25.88N*d=-13.61

    d=0.52m
     
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