Question

Suppose you spend 30.0 minutes on a stair-climbing machine, climbing at a rate of 85 steps per minute, with each step 8.00 inches high. If you weigh 150 lb and the machine reports that 690 kcal have been burned at the end of the workout, what efficiency is the machine using in obtaining this result?

Answer 1

The efficiency of the machine is
`\eta =`12%

Step-by-step explanation:

From the question we are told that

The time available to climb the stairs is
`t = 30 \ minutes`

The rate at which the stairs is climbed is
`v = 85 \ steps /minute`

The height of each step is
`h = 8.00 \ inch = 8 * (0.0254 (m)/(inh) ) = 0.2032 \ m`

The mass of the person is
`m = 150 lb`

The amount of calories burned is
`E = 690 kcal = 690 *1000 cal = 690000 * (4.186 J/cal) = 2888340 J`

Generally the workdone is taking a step is mathematically represented as

`W = mgh`

Here g (acceleration due to gravity is
`4.448\ N/lb`)

substituting values

`W = 150 * 4.44 * 0.2032`

`W = 135.58 \ J`

Now the total workdone during the course of the workout is mathematically represented as

`W_T = W * v * t`

substituting values

`W_T = 135.58 * 85 * 30`

`W_T = 345716.4 \ J`

The efficiency of the machine is mathematically represented as

`\eta = (W_T)/(E) * (100)/(1)`

substituting values

`\eta = (34716.4)/(2888340) * (100)/(1)`

`\eta =`12%