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My Physics Homework Cabin

  1. Mar 28, 2004 #1
    Hey y'all! My friend told me about this special forum site in which you can get help with your homework and I think that's a really useful tool. Anyways, I am currently enrolled in Physics 30 and I need some help. So instead of posting a new thread everyday, this is the name I gave to my thread so I wouldn't have to start a new one everytime, and I can just stick to this one. Anyways, here's my question:

    A 150N force is pulling a 50.0kg box along a horizontal surface. The force acts at an angle of 25.0 degrees. If this force acts through a displacement of 12.0m, and the coefficient of friction is 0.250, what is the speed of the box, assuming it started from rest?

    If anyone can help me, that'd be great. Oh yeah, and I'm a girl....not a guy....or maybe I shouldn't have chosen this name....
     
    Last edited: Mar 28, 2004
  2. jcsd
  3. Mar 28, 2004 #2

    Janitor

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    Are you told (or shown with a diagram) whether the 25 degree angle is helping to pin the mass against the horizontal surface, or helping to lift it from the surface? I.e., is the force tilted down that many degrees, or up that many degrees? It seems to me like it would matter in a friction problem such as this.
     
  4. Mar 28, 2004 #3

    ShawnD

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    In the future, try not to use blue writing. It makes it harder to read.

    Looks like an energy problem. Write an equation to show the energies at work (energy in vs. energy out)

    applied work = friction + kinetic energy

    [tex]F\cos(\theta)d = \mu(mg - F\sin(\theta))d + \frac{1}{2}mv^2[/tex]

    [tex]150\cos(25)(12) = (0.25)((50)(9.8) - (150)\sin(25))(12) + \frac{1}{2}(50)v^2[/tex]

    Just solve for V. Now that wasn't so hard :D
     
  5. Mar 28, 2004 #4
    I still don't really get it....can you show me step by step of how to arrive at it?
     
  6. Mar 28, 2004 #5

    ShawnD

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    Applied Work:
    When calculating work, force and direction have to be parallel to each other. A person applies a force on the box (150N) but since you only want the horizontal component of that force, multiply it by cos(25). d is simply 12.

    Frictional Work
    The force of friction is uN where u is the firction coefficient and N is the force between the surfaces. In the equation, I substituted N with (mg - Fsin(25)). mg is the force due to gravity (down). Fsin(25) is the vertical component of the force being applied on the box (up). d is still 12.

    Kinetic Energy
    The formula for kinetic energy is (1/2)mv^2.
     
    Last edited: Mar 28, 2004
  7. Mar 28, 2004 #6

    Doc Al

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    Two points:
    (1) The official--and wise--policy here is for questioners to attempt to solve the problem themselves, showing their work and where they got stuck, before asking for help. Read Tom's sticky post Read This Before Posting at the top of the forum.

    Of course, sharp guys like ShawnD are all too eager to help! :smile:

    (2) You'll get better and quicker help if you start a new thread with each question using a meaningful title. Don't just keep adding to this thread--unless you need additional help on this exact problem.
     
  8. Mar 28, 2004 #7

    Janitor

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    Note

    Shawn is taking the 25 degree force to be pointing above the horizontal. If you take it to be 25 degrees below the horizontal, you will want to add F sin theta to mg instead of subtracting it from mg.
     
  9. Mar 28, 2004 #8

    ShawnD

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    True, but don't do that unless it specifies that the force is downward. If you are given an angle, you assume it follows the standard rules where counter-clockwise is positive.
     
  10. Mar 28, 2004 #9
    Anyways....for your info:

    I already tried doing the question by myself, and I got stuck cause I wasn't sure what to do with friction....like, how to implement it into the question....
     
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