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  • Anyone Good with Physics?

    I got these two problems wrong on a test today, and its been driving me nuts. I have pages of test work done out and I can't come up with anything right. Can anyone offer some insight?

    1. A skier is at the top of a slope inclined at 10.5 degrees. The μ of the slope is 0.0750, as is the μ of the flat ground at the end of the slope. The slope is 200m in length. Starting from rest at the top, what is his speed at the bottom of the hill and how far from the bottom of the hill will he finally come to a stop?


    2. A 700N man is bungee jumping from a bridge 36 meters above a river. The bungee cord is 25m long unstretched. In order to safely jump, he must stop his descent when he is 4m above the river. What is the 'k' (stretchability in N/m) value of the bungee cord?
    "63,000 bugs in the code, 63,000 bugs, you get 1 whacked with a service pack, now there's 63,005 bugs in the code."
    "Before you critisize someone, walk a mile in their shoes. That way, when you critisize them, you're a mile away and you have their shoes."
    Utopia Software - Current Software: Utopia News Pro (news management system)

  • #2
    I feel your pain man......the classes SUCK.
    Rebel Designs Studios

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    • #3
      1. Consider the energy of the skier throughout the course. At the top of the hill he has only potential gravitational energy, mgH. You can find H because you know the slope of the hill as well as it's length. At the bottom of the hill the skier has no potential energy, only kinetic energy. But he also lost some energy due to friction, so the equation looks like this: (initial energy - final energy = lost energy)
      mgH - (1/2)mV^2 = NμL
      Where L is the length of the hill, and N is the normal force acting on the skier, which equals mgCos(10.5). Solve this equation for v and you have the skier's speed at the bottom of the slope
      Now you want to find what distance the skier can cover with his inital energy, so we write a new equation: (again, initial energy - final energy = lost energy, except now the final energy equals zero because we want to know where he comes to a stop)
      mgH = NμD
      Solve for D and you have the distance the skier has travelled. BUT, remember that this distance includes the length of the slope, so you need to substract 200m from it to know how far from the bottom of the slope the skier will come to a halt.

      2. Again, use energy conservation:
      mgh - (1/2)kx^2 = 0
      Where:
      h = 32m, because he started 36 meters above water and we want him to stop 4 meters above.
      x = the prolongation of the bungee cord. We already know that when unstretched the cord is 25 meters in length, so the prolongation should be 32 - 25 = 7 meters.
      Solve for k.

      Hope I helped.
      Chen Avinadav
      Better to remain silent and be thought a fool than to speak out and remove all doubt.

      גם אני מאוכזב מסיקור תחרות לתור מוטור של NRG הרשת ע"י מעריב

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      • #4
        Oh and by the way, in the future you may want to post this kind of questions over at www.PhysicsForums.com, you'd probably get them answered a lot quicker over there.
        Chen Avinadav
        Better to remain silent and be thought a fool than to speak out and remove all doubt.

        גם אני מאוכזב מסיקור תחרות לתור מוטור של NRG הרשת ע"י מעריב

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        • #5
          Thanks.

          I registered two accounts and didn't get an email confirmation with either of them (also used two different email accounts). Grr.
          "63,000 bugs in the code, 63,000 bugs, you get 1 whacked with a service pack, now there's 63,005 bugs in the code."
          "Before you critisize someone, walk a mile in their shoes. That way, when you critisize them, you're a mile away and you have their shoes."
          Utopia Software - Current Software: Utopia News Pro (news management system)

          Comment


          • #6
            Originally posted by CeleronXT
            Thanks.

            I registered two accounts and didn't get an email confirmation with either of them (also used two different email accounts). Grr.
            PM me your username and I'll activate you manually.
            http://www.physicsforums.com

            Comment

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