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Work & Energy Questions

  1. Oct 20, 2003 #1
    1) An 80-N box is pulled 20m up a 30 degree incline by an applied force of 100 N that points upward, parallel to the incline. If the coefficient of kinetic friction between box and incline is 0.220, calculate the change in the kinetic energy of the box.

    I made a triangle sketch, obviously, 30 degrees in the lower left corner. One of my big problems is I'm not sure what they mean by "80-N box." Does that mean it's a force of 80 N straight down, or is that like a force perpendicular against the incline? I used muK = Ff-Fn to find an Ff of 80.220 N. Then I did something with FcosΘ=work, and somehow came up with 86.6 J. I really don't know what I'm doing, and I have no correct answer listed anywhere, so I have no direction. Help!

    2)A 70kg diver steps off a 10m tower and drops, from rest, straight down into the water. If he comes to rest 5.0m below the surface, determine the average resistance force exerted on him by the water.

    I used vf^2=vi^2 + 2ad to get his initial velocity at the water (I got 14 m/s) but after that I'm clueless as to what to do next.

    Any help you can provide me would be great, I'm going to re-read the chapter. I've been sick from school for a few days, so I'm really behind and lost here.
     
  2. jcsd
  3. Oct 20, 2003 #2

    HallsofIvy

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    The phrase "80 N box" is talking about its weight. Yes, that means the force is directed downward. I'm not sure what you mean by "Ff" but you cannot simply add 80 N and the COEFFICIENT of kinetic friction. The friction force is the coefficient of kinetic friction times the normal force.

    You will need to set up the "force diagram" and use trigonometry to determine the components of weight along the incline and normal to the incline.
     
  4. Oct 20, 2003 #3
    Ok, i think that I have an answer for you. On your question of what the 80 N force is, it is the force due to gravity. 80 N is equal to mg (mass x gravity), (so the mass equals 80/9.8) and the force due to gravity is directly, completely, straight downward. Now the normal force is different. The normal force of an object is the support force that acts perpendicular to the surface on which the object rests. Now since there is a 30 degree incline, the normal force does not just equal mg (80 N), it equals mgcos(angle), where the (angle) is equal to 30 degrees.

    After figuring this out, you use the work-energy theorm, which goes as follows . . .

    The net work done on an object = the change in kinetic energy, or
    W = change in K

    So then you look at all of the work being done on the object. There is the work done by the pulling force (Fd), the work done by friction, (muFn), and the work done by gravity, (mgh). Now the work done by friction and gravity are negative, becasue they oppose the direction of motion of the block. Set the formula up as follows

    Wp - Wf - Wg = K
    Fd - mu(Fn) - mgh = K

    mu(Fn) is equal to mu(mgcos(angle), where Fn = mgcos(angle), which is from earlier.

    Remeber your trig and realize that h equals the hypotenuse times sin30, SOH.

    I got 895 J, tell me what you get
     
  5. Oct 20, 2003 #4
    The only thing I can't figure out in all that is how the force of gravity became MGH all of a sudden instead of just MG. What does the H stand for? Other than that, I'm good. Thanks!
     
  6. Oct 20, 2003 #5
    Now moving on to your second question, i think i have another answer. The velocity when the boy hits the water is 14 m/s, like you said. Then you just resort back to the work-energy theorm like you did earlier.

    You look at the situation, and see what is doing work on the boy. There is no pushing force, so the work due to gravity (Wg), and the work due to the water resistance are the only works present. The work due to the water resistance (Fd) is similar to friction, in that it resists motion. You know that work = Fd, so that is the water resistance work, and the work due to gravity = mgh.

    Net Work = Change in K, where final K = 0
    Wg + Fd = .5mvf^2 - .5mvo^2
    mgh + Fd = (.5)(m)(0) - (.5)(m)(14^2)

    I think that becuase since Wg and Fd are both negative work, they can be added, but i could be wrong

    Solve for F

    I dont know whether to use positive or negative 5 meters, so sorry about that. Plug and Chug. Good Luck catching up.
     
  7. Oct 20, 2003 #6
    the force due to gravity is equal to mg
    the work due to gravity is mgh

    W = Fd Fg = mg d = h

    Wg = mgh
     
  8. Oct 21, 2003 #7
    For the second problem, i was a little wrong. Since the diver is traveling down, in the direction of gravity, the work due to gravity is acting with the direction of motion, so it is positive, and the work due to the water is still negative becuase it slows the diver down. Sorry.

    Also, the problem could be set up a different way. The diver has zero kinetic and all potential energy (mgh) at the top of the tower, and zero potential and zero kinetic at the bottom (v = 0, and the arbitrary zero of the problem can be determined to be where he stops, or a 5 m below the surface, so h = 0). So in this case, all potential (work due to gravity) turns into all water resistance (not taking into consideration the splash created by the diver).

    Wg (or U) = Fd (or Q)
     
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