Question

Suppose you spend 30.0 minutes on a stair-climbing machine, climbing at a rate of 65 steps per minute, with each step 8.00 inches high. If you weigh 150 lb and the machine reports that 650 kcal have been burned at the end of the workout, what efficiency is the machine using in obtaining this result?

Answer 1

Answer: The machine is using an efficiency of 9.9%

Explanation:

Efficiency = (work output/work input)*100%

Work input = 650Kcal = 650*4184J = 2719600J

work output = Force * distance

force = mass *acceleration , where mass = 150lb or 68.04kg, a = 10ms⁻²

force = 68.04*10 = 680.4N

distance = velocity * time (in seconds)

velocity = 65 steps/min * (1/60) = 1.083 steps/s

multiply 1.083 steps/s by the height of each step ( in meters) to obtain velocity in units of m/s.

1.083 * 0.2032m = 0.220m/s

Note: 8.00 inches = 0.2032m

distance moved = 0.220*1800s = 396m

Note: 30 minutes = 1800s

Therefore, work output = 680.4*396 = 269438.4J

Efficiency = (269438.4J/2719600J)*100%

Efficiency = 9.9%