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Running up some stairs, kiloWatts

  1. Nov 6, 2004 #1
    Ok, so I'm supposed to run up some stairs and then calculate how much effect (watts) I used.

    This is how I did it:

    Stairs was 2.52 meters high and 3.5 meters long, my mass is 72 kilos and I ran up the stairs in 2.72 seconds.

    Energy for moving up 2.52 meters:

    [tex]E_{p} = mgh = 72*9.82*2.52 = 1781.7408[/tex]

    [tex]P_{1} = \frac{E_{p}}{t} = \frac{(\frac{72*9.82*2.52}{2.72})}{2.72} \approx 0.66 kW[/tex]

    Energy: for moving the 3.5 meters horizontally:

    [tex]v = \frac{3.5}{2.72}[/tex]

    [tex]E_{k} = \frac{mv^2}{2} = \frac{72(\frac{3.5}{2.72})^2}{2.72}[/tex]

    [tex]P_{2} = \frac{E_{k}}{t} \approx 0.22 kW[/tex]

    Alltogether: [tex]P_{1} + P_{2} = 0.66 + 0.22 = \underline{0.88 kW}[/tex]

    I'm not supposed to think about air friction etc.
    Question is ofcourse: Am I anywhere close to answering this in a correct way?
  2. jcsd
  3. Nov 6, 2004 #2
    For the first part, why did you divide by 2.72 twice? [itex]P=\frac{\Delta{E}}{\Delta{t}}[/itex]. The second part is tricky. If you ignore non-conservative forces such as friction, the only work you do in the horizontal direction is that which is required for you to accelerate to a certain velocity. Once you reach a given horizontal velocity, because there are no frictional forces (in theory), you do not have to do any more work to continue moving at that velocity. Therefore, you must make some assumption for this part. You can assume that you accelerated at a constant rate in the horizontal direction, and use [itex]\Delta{d}=v_{1}\Delta{t}+\frac{1}{2}a(\Delta{t})^2[/itex] to find a, [itex]F_{net}=ma[/itex] to find net force, then sub that into [itex]W=Fd\cos{\theta}[/itex], where theta would be 0. I get about 88W for just the horizontal motion.
  4. Nov 6, 2004 #3
    Looking at the math a second time I see I wrote in a number of mistakes, sorry - first time using this LaTeX thing :)

    this is how the first P was supposed to be (not dividing with 2.72 twice)

    [tex]P_{1} = \frac{E_{p}}{t} = \frac{72*9.82*2.52}{2.72} \approx 0.66 kW[/tex]

    And the second E was supposed to be like this (dividing with 2, not 2.72):

    [tex]E_{k} = \frac{mv^2}{2} = \frac{72(\frac{3.5}{2.72})^2}{2}[/tex]

    Also this very basic class I'm taking doesn't deal with cosinus etc yet so i expected the answer to be pretty much basic and simple (like [tex]mgh[/tex] and [tex]\frac{mv^2}{2}[/tex])
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