Answer:
The efficiency of the machine  is  [tex]\eta =[/tex]12%
Explanation:
From the question we are told that
  The time available to climb the stairs is  [tex]t = 30 \ minutes[/tex]
  The rate at which the stairs is  climbed is  [tex]v = 85 \ steps /minute[/tex]
   The height of each step is  [tex]h = 8.00 \ inch = 8 * (0.0254 \frac{m}{inh} ) = 0.2032 \ m[/tex]
   The mass of the person is  [tex]m = 150 lb[/tex]
   The amount of calories burned is  [tex]E = 690 kcal = 690 *1000 cal = 690000 * (4.186 J/cal) = 2888340 J[/tex]
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Generally the workdone is taking a step is mathematically represented as
    [tex]W = mgh[/tex]
Here  g (acceleration due to gravity is  [tex]4.448\ N/lb[/tex])
substituting values
     [tex]W = 150 * 4.44 * 0.2032[/tex]
    [tex]W = 135.58 \ J[/tex]
Now the total workdone during the course of the workout is mathematically represented as
    [tex]W_T = W * v * t[/tex]
substituting values
    [tex]W_T = 135.58 * 85 * 30[/tex]
    [tex]W_T = 345716.4 \ J[/tex]
 The efficiency of the machine is mathematically represented as
       [tex]\eta = \frac{W_T}{E} * \frac{100}{1}[/tex]
substituting values Â
       [tex]\eta = \frac{34716.4}{2888340} * \frac{100}{1}[/tex]
      [tex]\eta =[/tex]12%
