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natasha.aries Group Title

Hiii can someone help me with a Physics problem? Please and thank ya! :)

  • 2 years ago
  • 2 years ago

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  1. natasha.aries Group Title
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    • 2 years ago
  2. furnessj Group Title
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    oh boy, I looked at it! too late to get into though. you should get good help with that one, though it is tricky again..

    • 2 years ago
  3. ivanmlerner Group Title
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    OK, the work done is basically the force thats is influencing the movement times the distance, So we need to find the projection of the force in the direction of the movement, and it must be equal to the force of gravity summed with the friction, because it cannot have any aceleration for the speed to remain constant, and force=ma

    • 2 years ago
  4. ivanmlerner Group Title
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    Considering that, we get:\[-mg-\mu N+Fsin \theta=0\]

    • 2 years ago
  5. ivanmlerner Group Title
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    The normal force is the force balancing the force thats pulling the block towards the wall so:\[N=Fcos \theta\]

    • 2 years ago
  6. ivanmlerner Group Title
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    So the work done is given by:\[W=\Delta xFsin \theta=(mg+\mu Fcos \theta)\Delta x\]

    • 2 years ago
  7. natasha.aries Group Title
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    how do we find the change in x?

    • 2 years ago
  8. ivanmlerner Group Title
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    Now, whats missing is F. To find it we use the fact that the total force on the object must be zero. \[Fsin \theta=mg+\mu N=mg+\mu Fcos \theta\]Isolating F we get:\[Fsin \theta-F \mu \cos \theta =mg\]\[F=\frac{mg}{\sin \theta-\mu \cos \theta}\]Plug that into the formula for the work, and you have all the numbers now to calculate it. The change in x is given, it is 3m

    • 2 years ago
  9. natasha.aries Group Title
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    so instead of N should i plug in Fcosθ

    • 2 years ago
  10. ivanmlerner Group Title
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    This is the work done by F, now to get the work done by gravity it is easy, we have the change in x and we also have the magnitude of the force, the only tricky part is that the work must be negative, because the force is going against the movement and not favoring it.

    • 2 years ago
  11. natasha.aries Group Title
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    and what equation would we use for that?

    • 2 years ago
  12. ivanmlerner Group Title
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    Yes, for the first and\[W_G=-\Delta x mg\]for the second. Are you understanding what I'm doing here? Because this formula should be natural for you now.

    • 2 years ago
  13. natasha.aries Group Title
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    a little, not really because i havent used it before

    • 2 years ago
  14. natasha.aries Group Title
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    or at least i dont believe i have

    • 2 years ago
  15. furnessj Group Title
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    It is just gravitational potential in disguise natasha! (GPE=mgh) the work done is just the potential energy 'gained' (but negative)

    • 2 years ago
  16. natasha.aries Group Title
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    i understand the first part though, not really how you got the equation but how it makes sense

    • 2 years ago
  17. natasha.aries Group Title
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    but then where is the h @furnessj

    • 2 years ago
  18. ivanmlerner Group Title
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    Yes, thats it, the h is x, just a diferent name.

    • 2 years ago
  19. furnessj Group Title
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    h is the height gained, or just x here, anyway, leaving Ivan to it!

    • 2 years ago
  20. natasha.aries Group Title
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    oh okay, so why cant we keep it as h?

    • 2 years ago
  21. natasha.aries Group Title
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    oh okay!

    • 2 years ago
  22. ivanmlerner Group Title
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    If you understood everything so far, the only thing left is the normal force, wich you already have the formula for, and all the numbers involved. Just as a complementary comment for what furnessj said, the work done is also the change in the kinectic energy, and that is always true. However, the work is minus the change in the potential energy related to that force only if it is conservative, friction for example does work, but has no potential energy concept related to it.

    • 2 years ago
  23. ivanmlerner Group Title
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    Any doubts?

    • 2 years ago
  24. natasha.aries Group Title
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    no, but how can i find the normal force formula?

    • 2 years ago
  25. natasha.aries Group Title
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    @ivanmlerner ?

    • 2 years ago
  26. ivanmlerner Group Title
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    Oh, the normal force is the force that balances the force thats pulling the object towards the wall, so that it keeps still and don't "enter" the wall. It is the 3rd law of newton, that any force must have an opposite force balancing it. In this case what is pulling the object to the wall is the horizontal component of F, so its magnitude is Fcos theta

    • 2 years ago
  27. natasha.aries Group Title
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    Oh, it would be that simple?

    • 2 years ago
  28. ivanmlerner Group Title
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    Yes, that simple

    • 2 years ago
  29. gerryliyana Group Title
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    of course @natasha.aries , that simple :)

    • 2 years ago
  30. natasha.aries Group Title
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    haha! okay thanks! can i try plugging it in a few minutes, and would you be able to tell me if its right?

    • 2 years ago
  31. gerryliyana Group Title
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    ok... ur good learner .,

    • 2 years ago
  32. natasha.aries Group Title
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    haha thank you!

    • 2 years ago
  33. ivanmlerner Group Title
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    To give a medal for asking do I just click on best response by the asker?

    • 2 years ago
  34. natasha.aries Group Title
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    Yes!

    • 2 years ago
  35. natasha.aries Group Title
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    and when i did the first part i got 204 does that sound about right?

    • 2 years ago
  36. ivanmlerner Group Title
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    I got approximately 312 using g=10.

    • 2 years ago
  37. natasha.aries Group Title
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    the mass would be 5 correct?

    • 2 years ago
  38. ivanmlerner Group Title
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    yes

    • 2 years ago
  39. natasha.aries Group Title
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    idk what i did wrong :/

    • 2 years ago
  40. ivanmlerner Group Title
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    Are you talking about part (a)?

    • 2 years ago
  41. natasha.aries Group Title
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    yes

    • 2 years ago
  42. ivanmlerner Group Title
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    Did you use a calculator? If you did, check for missing parentesis. Also, did you use w=xFsin theta? with that formula for F?

    • 2 years ago
  43. natasha.aries Group Title
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    yes, it would be (5)(9.81) / (3sin30-.3cos30) right?

    • 2 years ago
  44. ivanmlerner Group Title
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    That would be F, but you got an extra 3 in the denominator in front of sin(30)

    • 2 years ago
  45. ivanmlerner Group Title
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    You still need to multiply that by 3, and by sin(30)

    • 2 years ago
  46. natasha.aries Group Title
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    wait, im confused. so it would be (5)(9.81) / (sin30-.3cos30) but why do we need to multiply it by 3sin30 ?

    • 2 years ago
  47. natasha.aries Group Title
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    anyone?

    • 2 years ago
  48. ivanmlerner Group Title
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    Sorry, I had to go in a rush, look at the formula for work on the begining of my posts:\[W=\Delta xFsin \theta\]So the one you are using is the one for F alone, and 3sin(30) is the rest: change in x=3 and sin theta=sin(30). Take a look again at the thinking I made to get to the first answer.

    • 2 years ago
  49. natasha.aries Group Title
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    its all good! and now i get 306 :)

    • 2 years ago
  50. natasha.aries Group Title
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    and so for the last one it would be 306(cos)30?

    • 2 years ago
  51. ivanmlerner Group Title
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    Using 9.81 as g that is correct, and for the second thats almost it.

    • 2 years ago
  52. ivanmlerner Group Title
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    306 is correct, but this is the work done by the force.

    • 2 years ago
  53. ivanmlerner Group Title
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    On the last it is asking for the normal, wich is a force, not work.

    • 2 years ago
  54. ivanmlerner Group Title
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    It is given by Fcos(30) and F is what you were calculating before.

    • 2 years ago
  55. ivanmlerner Group Title
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    F= that 204 you were calculating.

    • 2 years ago
  56. natasha.aries Group Title
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    but Fcos(30) is the normal force?

    • 2 years ago
  57. natasha.aries Group Title
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    i got 204 by mistake though

    • 2 years ago
  58. ivanmlerner Group Title
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    Yes, read what I wrote on the begining. You got 204 calculating the force alone, you had forgotten the other terms but here the force is exacly what you need, take a look at the begining and try to understand that, you don't seem to have understood it well.

    • 2 years ago
  59. natasha.aries Group Title
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    i tried it again, and still didnt get that :/

    • 2 years ago
  60. natasha.aries Group Title
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    im sorry :(

    • 2 years ago
  61. ivanmlerner Group Title
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    Wait, I said 204 is the force, not the answer, the answer is 204cos(30)

    • 2 years ago
  62. natasha.aries Group Title
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    but i dont get 204 at all

    • 2 years ago
  63. ivanmlerner Group Title
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    The formulas are:\[N=Fcos \theta\]\[F=\frac{mg}{\sin \theta-\mu \cos \theta}\]

    • 2 years ago
  64. ivanmlerner Group Title
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    The explanation for them is in the begining.

    • 2 years ago
  65. natasha.aries Group Title
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    i did that

    • 2 years ago
  66. ivanmlerner Group Title
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    Where you using that?

    • 2 years ago
  67. natasha.aries Group Title
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    yes

    • 2 years ago
  68. ivanmlerner Group Title
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    Show the numbers again.

    • 2 years ago
  69. natasha.aries Group Title
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    5)(9.81) / (sin30-.3cos30)

    • 2 years ago
  70. ivanmlerner Group Title
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    And you need to multiply all of that by cos(30)

    • 2 years ago
  71. natasha.aries Group Title
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    i finally got it! thats what ive been doing! i just separated it a few more times now. thank you so much. so the first one you would multiply it by 3sin30 and the last one cos30?

    • 2 years ago
  72. ivanmlerner Group Title
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    Yes, but remember what those numbers mean.

    • 2 years ago
  73. natasha.aries Group Title
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    i will, im just unsure with how u knew which equations to use

    • 2 years ago
  74. ivanmlerner Group Title
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    I didn't I foun out wich one I needed to use, and I posted wverything I did, project the force to find the work, see what force is causing the normal, do the math and you get to the formulas.

    • 2 years ago
  75. natasha.aries Group Title
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    when i checked the answers, the right answers were 310J... -150J...180 N

    • 2 years ago
  76. natasha.aries Group Title
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    310 for a -150 for b and 180 for c

    • 2 years ago
  77. natasha.aries Group Title
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    i got a similar answer for the first 2 parts, but not for the third!

    • 2 years ago
  78. ivanmlerner Group Title
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    Remember that on the last one you only need to multiply the force F (204N) by cos(30) (sqrt(3)/2) because we need the projection of the force at the direction perpendicular to the wall. If you use those numbers you'll get the answer.

    • 2 years ago
  79. natasha.aries Group Title
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    why sqrt of 3/2 ?

    • 2 years ago
  80. natasha.aries Group Title
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    i think you just mean cos30 because those two are the same things :)

    • 2 years ago
  81. ivanmlerner Group Title
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    What?? wait, thats what i meant:\[\cos 30°=\frac{\sqrt{3}}{2}\]

    • 2 years ago
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is replying to Can someone tell me what button the professor is hitting...

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