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Block pushed up an incline, velocity using work

  1. Feb 8, 2014 #1
    this question has 4 parts, im having trouble on the last part but i will post all parts to give context.

    Part A

    How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure(Figure 1) when the force pushes the block 5.5m up along the 32∘ incline? Assuming a coefficient of friction μk = 0.14. [Hint: Work-energy involves net work done.]

    for WFp i got 700 J

    Part B

    How much work is done by the gravitational force on the block during this displacement?

    for Wgr i got -510J

    Part C

    How much work is done by the normal force?

    for WN i got 0 J

    Now the part that im having trouble

    Part D

    What is the speed of the block (assume that it is zero initially) after this displacement?

    To do this part I did

    initially I used the equation Wnet = 0.5mv2, with Wnet = WFp + Wgr + WN so Wnet = 190

    and i solved for v by v = sqrt(2W/m) and got 4.6 m/s

    that answer is incorrect. After consulting with my professor she said to take friction into account but now im not sure how to do so.

    I know that the normal force is mgcosθ + Fpsinθ and then the work of friction is equal to μk*FN*d*cosθ

    I tried that and ended up with the velocity = 6.1 m/s

    is there something im not seeing?
     
  2. jcsd
  3. Feb 8, 2014 #2
    Yes, you have to add in friction -- you can do this because friction is a force, so it also does work on the block. First find this work (remember that friction works anti-parallel to displacement, so along the ramp), then you can add it into the net work, and use the work-energy theorem as you were doing.

    EDIT: Sorry, I forgot seeing that you found the frictional force already, and the work. I think the only problem is the value you used for displacement: you just did the x-component of the displacement, but remember (like I mentioned above) that friction works along the entire distance along the ramp.
     
  4. Feb 8, 2014 #3

    didnt i do that in this step μk*FN*d*cosθ?
     
  5. Feb 8, 2014 #4
    So in other words,

    I would use the hypotenuse instead of the 5.5 m up which was given? I tried that and it didn't work also.
     
  6. Feb 8, 2014 #5
    Yes, you would use the hypotenuse, but, I think the 5.5m given in the problem is the hypotenuse. The vertical distance would be the 5.5m then times the sine of the incline. I see how you would think it refers to the vertical distance from the wording, but the key word here is "along", to indicate the distance is the hypotenuse -- or the distance along the sloped ramp instead of just the vertical distance.
     
  7. Feb 9, 2014 #6
    Okay. So here are my calculations:

    Fn = mgcos(θ) +Fpsin(θ)

    =(18 kg)(9.8 m/s2)(cos(32)) + 150sin(32) = 229.08 N


    Then,

    WFr = (μk)(Fn)(d)(cos(θ))

    = (0.14)(229.08)(5.5)(cos(32)) = 149.6

    Finally,

    Wnet=0.5mv2 OR v=√(2W)/m= 6.1 m/s


    Are my calculations wrong? Or did I use a wrong number? The answer I got is not correct.
     
  8. Feb 9, 2014 #7
    The problem is the displacement you use to calculate work done by friction. You use [itex]dcos\theta[/itex], but that's just the x-displacement. You have to use the entire displacement along the slope, since friction acts all along that distance. Look at my other post for reference.
     
  9. Feb 9, 2014 #8
    So would WFr = (μk)(Fn)(dx)cos(θ))+μk)(Fn)dy(sin(θ))?
     
  10. Feb 9, 2014 #9
    Close, but now you're doing it in two separate parts -- one for the x displacement and one for the y displacement, but friction doesn't work in both directions, it only works along the path of motion. Just look right at the problem's givens. How far is the block pushed along the incline?
     
  11. Feb 9, 2014 #10
    It's pushed 5.5 m, right?
     
  12. Feb 9, 2014 #11
    Yes, that's the distance along the incline, so that's the distance along which friction acts.
     
  13. Feb 9, 2014 #12
    but i did that in my original post and the answer was wrong.
     
  14. Feb 9, 2014 #13
    No, you're using [itex]dcos\theta[/itex]. It needs to be just [itex]d[/itex].
     
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