1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Drilling was never too fun.

  1. Jul 31, 2007 #1
    Every morning at seven o' clock
    There's twenty terriers drilling on the rock.
    The boss comes around and he says, "Keep still
    And bear down heavy on the cast-iron drill

    And drill, ye terriers, drill." And drill, ye terriers, drill.
    It's work all day for sugar in your tea
    Down beyond the railway. And drill, ye terriers, drill.

    The foreman's name was John McAnn.
    By God, he was a blamed mean man.
    One day a premature blast went off
    And a mile in the air when big Jim Goff. And drill...

    Then when next payday came around
    Jim Goff a dollar short was found.
    When he asked what for, came this reply:
    "You were docked for the time you were up in the sky." And drill...

    What was Goff's hourly wage? State the assumptions you make in computing it.

    ----------
    Equations are the basic three of mechanics.

    If he went one mile to the sky, I assume he came back in one mile, making the total distance 2 miles, and a total displacement of zero. This problem is in a chapter that deals with motion in one dimension. I need to find the time that Goff was in the air. But I only know that a = -32.2 ft/s^2.

    Using v[final]^2 = v[initial]^2 + 2as to represent only the upward journey, I got the initial velocity, with v[final] = 0, a = -32.2, and s = 5280, to be the square root of 340032, or about 583.123 ft/s. Using the velocity as a proportion to 5280 feet, it took Goff about 9.055 seconds to get to his maximum height, meaning that 18.11 seconds were spent up in the air.

    For an hourly wage, I set up a another proportion; if one dollar is docked for about 18 seconds, then how many dollars would be docked for 3600 seconds?

    I got about 198 dollars.

    ...this seems wrong. I think I did the calculations right, but the thought of someone being paid 198 dollars an hour...for mining nonetheless. I know it has nothing to do with the logic of the problem, but, is this a good way to go about the problem? Is my train of thought correct?
     
  2. jcsd
  3. Jul 31, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your thinking is good, but you made an error here:
    Realize that 583 ft/s is the initial speed of the miner and his speed is not constant. To calculate the time it takes him to rise up that mile, you'll need his average speed. (How does that relate to the initial speed?)
     
  4. Jul 31, 2007 #3
    Hrm. So, could I just plug into the equation v[final] = v[initial] + at, and get that time, with v[inital] = 583 and v[final] = 0?
     
  5. Jul 31, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes: That's a perfectly fine way to calculate the time.
     
  6. Jul 31, 2007 #5
    Thank you! I got a [slightly] reasonable answer of 99.40 dollars.

    :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Drilling was never too fun.
  1. Power Drill Motors (Replies: 0)

Loading...