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Kinetic energy, Potential energy and change in water pressure

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm not particularly confident with my calculations and I would appreciate any feed back.

    A) What is the kinetic energy of an object with:
    mass = 5lbm
    velocity = 200 ft/sec

    B) What is the potential energy of an object with:
    mass = 5lbm
    height = 200ft
    g = 32.2 ft/sec2

    C) What is the pressure difference between the top and bottom of a water column that is 15ft high and gravity is 32.2 ft/sec2

    2. Relevant equations
    KE = 1/2mv2/gc

    PE = mgh/gc


    3. The attempt at a solution

    A)
    KE = 1/2(5 lbm)(200 ft/sec)2/32.17 ft-lbm/lbf-sec2
    KE = 3.11x103 lbf-ft

    B)
    PE = (5 lbm)(32.2 ft/sec2)(200ft)/32.17 ft-lbm/lbf-sec2
    PE = 1001 lbf-ft

    C) I'm not quite sure where to start with this one and I would appreciate an guidance.
     
  2. jcsd
  3. Jan 24, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Northbysouth! :smile:
    yes, they look fine to me, except:

    i] i don't understand the pound system :redface:, so i can;t confirm whether you need to divide by g !
    ii] if the question asks you to take g as 32.2, then shouldn't you use the same figure instead of 32.17 ? :confused:
    you'll need to find the density of water (mass per cubic foot) …

    pressure = weight/area = density*height*area/area = density*height :smile:
     
  4. Jan 24, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since the problem says that the mass (rather than weight) is "5 lbm", no, you shouldn't divide by g. "1 lbm" or "one pound mass" is the mass of an object that has, in standard gravity, a weight of 1 lb. The division by g has already been done.

     
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