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Help with this simple WORK problem

  1. Mar 30, 2006 #1
    Hi everyone. I just found this forum and it looks like a good place to get some help on my physics homework. Anyways, this year, I have bad luck. I have a REALLY bad physics teacher (I'm not bashing on teachers here, but it's true). He doesn't do anything; we haven't learned ANYTHING from his class, he is lazy and disorganized, and rambles on and on about things not related to physics, and can't even solve simple problems. Thanks for letting me rant, I'm really frustrated with this teacher. He gives us some work and expects us to do it without any help.

    Here's a problem I can't figure out:

    You slide a crate up a ramp at an angle of 30.0° by exerting a 230 N force parallel to the ramp. The crate moves at constant speed. The coefficient of friction is 0.28. How much work have you done on the crate when it is raised a vertical distance of 1.10 m?

    I'm sure it's pretty simple to you guys, but I don't know how to do it. Can someone provide me with an answer or even an explanation along with the answer? I would say I'm a pretty smart guy, so I can figure it out if I get an answer, but an brief explanation would be nice as well.

  2. jcsd
  3. Mar 30, 2006 #2


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    work is force multiplied by distance moved that should be enough for you to have a go.
  4. Mar 30, 2006 #3
    I tried that, and I can't get the answer because there is the coefficient of friction, the angle, etc. etc. that I am not familiar with.
  5. Mar 30, 2006 #4
    Okay so I find the distance of the hypotenuse of the triangle by the sin formula thing and just multiply by the force? Is it that simple?
  6. Mar 30, 2006 #5
    Do I have to use the coefficient of friction to solve this problem?
  7. Mar 30, 2006 #6
    Okay so I tried solving it but it didn't work again.

    So here is what I did:

    I solved for distance using sin(30)=1.10/d
    Then I plugged in d in the formula and I plugged in the f which is 230 N according to the problem and I got 104.55 but the answer doesn't match the book. Gah! :(
  8. Mar 30, 2006 #7
    How does the coefficient of friction fit into this? Sorry for all the posts.
  9. Mar 30, 2006 #8
    well i did what you did apparently and i didn't get 104.55. i got 506. you don't need to use the coeff of friction unless your finding work due to friction, or total work.
  10. Apr 1, 2006 #9
    You are moving at a constant speed so you are not accelerating, so your Force applied is going to be equal to the Force Parallel + Force of Friction.
  11. Apr 1, 2006 #10
    The only work done is by gravity
  12. Apr 1, 2006 #11


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    You found the right distance d, but your mistake is that you used that the work is the force divided by the distance. This is the mistake. The right equation is work = force times distance (when the distance is in the same direction as the motion!!!). So 230N times 2.20 meters = 506 Joules (a Joule is a different name for a Newton-meter).

  13. Apr 2, 2006 #12


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    I am a physics teacher and I take absolutely no offense! There *are* some bad teachers!!
    That is terrible! What the heck does he spend time talking about?!
    I already shpwed how to get the correct answer but here is a brief overview. In general, the work energy theorem says that the change of kinetic energy is equal to the work done by all the forces:[itex] {1 \over 2} m v_f^2 - {1\over 2} m v_i^2 = sum~of~the~work~done~by~all~the~forces[/itex] .

    In your example, there are 4 forces acting: gravity, the normal force, a friction force and you. In general, the work done by a force is given by [itex] F d cos(\theta) [/itex] where the angle is measured between the direction of the force and the direction of the motion. And d is the distance traveled. (I am assuming motion along a straight line and constant force, otherwise you need to do an integral).

    In your example, the speed is constant so the difference of kinetic energy is zero so the sum of all the forces must be zero. But you don't even need to know that to answer the question. If they only want the work done by you, just use [itex] F_{you} d cos(\theta) [/itex]. Since in that example the force you are applying is in the same direction as the motion, the angle is zero and you end up with simply Work done by you = F times d.

    Hope this helps

    Last edited: Apr 2, 2006
  14. Apr 2, 2006 #13
    Thanks for the help guys it was useful.

    nrqed, my physics teacher keeps on talking about the war in Iraq and how he used to play sports when he was younger and how he broke his leg going up a hill etc etc etc. He is easily distracted and constantly relies on a computer to give us lectures.
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