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natasha.aries

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

  • one year ago
  • one year ago

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  1. natasha.aries
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    • one year ago
  2. furnessj
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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..

    • one year ago
  3. ivanmlerner
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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

    • one year ago
  4. ivanmlerner
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    Considering that, we get:\[-mg-\mu N+Fsin \theta=0\]

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

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

    • one year ago
  7. natasha.aries
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    how do we find the change in x?

    • one year ago
  8. ivanmlerner
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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

    • one year ago
  9. natasha.aries
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    so instead of N should i plug in Fcosθ

    • one year ago
  10. ivanmlerner
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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.

    • one year ago
  11. natasha.aries
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    and what equation would we use for that?

    • one year ago
  12. ivanmlerner
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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.

    • one year ago
  13. natasha.aries
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    a little, not really because i havent used it before

    • one year ago
  14. natasha.aries
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    or at least i dont believe i have

    • one year ago
  15. furnessj
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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)

    • one year ago
  16. natasha.aries
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    i understand the first part though, not really how you got the equation but how it makes sense

    • one year ago
  17. natasha.aries
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    but then where is the h @furnessj

    • one year ago
  18. ivanmlerner
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    Yes, thats it, the h is x, just a diferent name.

    • one year ago
  19. furnessj
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    h is the height gained, or just x here, anyway, leaving Ivan to it!

    • one year ago
  20. natasha.aries
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    oh okay, so why cant we keep it as h?

    • one year ago
  21. natasha.aries
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    oh okay!

    • one year ago
  22. ivanmlerner
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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.

    • one year ago
  23. ivanmlerner
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    Any doubts?

    • one year ago
  24. natasha.aries
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    no, but how can i find the normal force formula?

    • one year ago
  25. natasha.aries
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    @ivanmlerner ?

    • one year ago
  26. ivanmlerner
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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

    • one year ago
  27. natasha.aries
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    Oh, it would be that simple?

    • one year ago
  28. ivanmlerner
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    Yes, that simple

    • one year ago
  29. gerryliyana
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    of course @natasha.aries , that simple :)

    • one year ago
  30. natasha.aries
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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?

    • one year ago
  31. gerryliyana
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    ok... ur good learner .,

    • one year ago
  32. natasha.aries
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    haha thank you!

    • one year ago
  33. ivanmlerner
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    To give a medal for asking do I just click on best response by the asker?

    • one year ago
  34. natasha.aries
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    Yes!

    • one year ago
  35. natasha.aries
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    and when i did the first part i got 204 does that sound about right?

    • one year ago
  36. ivanmlerner
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    I got approximately 312 using g=10.

    • one year ago
  37. natasha.aries
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    the mass would be 5 correct?

    • one year ago
  38. ivanmlerner
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    yes

    • one year ago
  39. natasha.aries
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    idk what i did wrong :/

    • one year ago
  40. ivanmlerner
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    Are you talking about part (a)?

    • one year ago
  41. natasha.aries
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    yes

    • one year ago
  42. ivanmlerner
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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?

    • one year ago
  43. natasha.aries
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    yes, it would be (5)(9.81) / (3sin30-.3cos30) right?

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

    • one year ago
  45. ivanmlerner
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    You still need to multiply that by 3, and by sin(30)

    • one year ago
  46. natasha.aries
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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 ?

    • one year ago
  47. natasha.aries
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    anyone?

    • one year ago
  48. ivanmlerner
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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.

    • one year ago
  49. natasha.aries
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    its all good! and now i get 306 :)

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

    • one year ago
  51. ivanmlerner
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    Using 9.81 as g that is correct, and for the second thats almost it.

    • one year ago
  52. ivanmlerner
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    306 is correct, but this is the work done by the force.

    • one year ago
  53. ivanmlerner
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    On the last it is asking for the normal, wich is a force, not work.

    • one year ago
  54. ivanmlerner
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    It is given by Fcos(30) and F is what you were calculating before.

    • one year ago
  55. ivanmlerner
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    F= that 204 you were calculating.

    • one year ago
  56. natasha.aries
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    but Fcos(30) is the normal force?

    • one year ago
  57. natasha.aries
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    i got 204 by mistake though

    • one year ago
  58. ivanmlerner
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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.

    • one year ago
  59. natasha.aries
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    i tried it again, and still didnt get that :/

    • one year ago
  60. natasha.aries
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    im sorry :(

    • one year ago
  61. ivanmlerner
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    Wait, I said 204 is the force, not the answer, the answer is 204cos(30)

    • one year ago
  62. natasha.aries
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    but i dont get 204 at all

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

    • one year ago
  64. ivanmlerner
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    The explanation for them is in the begining.

    • one year ago
  65. natasha.aries
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    i did that

    • one year ago
  66. ivanmlerner
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    Where you using that?

    • one year ago
  67. natasha.aries
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    yes

    • one year ago
  68. ivanmlerner
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    Show the numbers again.

    • one year ago
  69. natasha.aries
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    5)(9.81) / (sin30-.3cos30)

    • one year ago
  70. ivanmlerner
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    And you need to multiply all of that by cos(30)

    • one year ago
  71. natasha.aries
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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?

    • one year ago
  72. ivanmlerner
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    Yes, but remember what those numbers mean.

    • one year ago
  73. natasha.aries
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    i will, im just unsure with how u knew which equations to use

    • one year ago
  74. ivanmlerner
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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.

    • one year ago
  75. natasha.aries
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    when i checked the answers, the right answers were 310J... -150J...180 N

    • one year ago
  76. natasha.aries
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    310 for a -150 for b and 180 for c

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

    • one year ago
  78. ivanmlerner
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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.

    • one year ago
  79. natasha.aries
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    why sqrt of 3/2 ?

    • one year ago
  80. natasha.aries
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    i think you just mean cos30 because those two are the same things :)

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

    • one year ago
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