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   "source": [
    "# Chapter 2 Fundamental Concepts"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:2.1 Pg:34"
   ]
  },
  {
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   "execution_count": 1,
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   "outputs": [
    {
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     "text": [
      "Potential energy = 3217.39 ft-lbm\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "z=100 #ft\n",
    "m=32.1739 #lbm\n",
    "#calculations\n",
    "PE=m*z\n",
    "#results\n",
    "print \"Potential energy = %.2f ft-lbm\"%(PE)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:2.2 Pg:35"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Absolute energy of this mixture = 5.40e+17 ft-lbf\n",
      "\n",
      " In case b, there is no change in mass\n",
      "\n",
      " Change in mass = -3.15e-09 lbm\n",
      "The answers are a bit different due to rounding off error in textbook.\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "m0=18.016 #lbm\n",
    "gc=32.1739 #lbm ft/lbf sec**2\n",
    "c=186000*5280\n",
    "dU=94.4*10**6 #ft-lbf\n",
    "#calculations\n",
    "U=m0/gc *c**2\n",
    "dm= -dU*gc/c**2\n",
    "#results\n",
    "print \"Absolute energy of this mixture = %.2e ft-lbf\"%(U)\n",
    "print \"\\n In case b, there is no change in mass\"\n",
    "print \"\\n Change in mass = %.2e lbm\"%(dm)\n",
    "print \"The answers are a bit different due to rounding off error in textbook.\""
   ]
  }
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