If 10% of an 80-kg rock climber's total energy expenditure goes into the gravitational energy change when climbing a 70-m vertical slope, what is the climber's average metabolic rate (in Watts) during the climb if it takes her 11 min to complete the climb?
oubaas
power P = 70*80*9.806/(0.1*60*11) = 830 watt (quite a number, being more than 1HP) )
Andrew Smith
There are FOUR things being tested. 1 Do you recognize that in SI you must use seconds for time.? 2. Do you know that power = energy / time? 3. Do you recognize that gravitational energy = mgh. 4. IF 10 % goes into gravity how do you find 100% when you know 10%?
OK so metabolic energy =mgh *100/10
Power = metabolic energy / time = mgh * 100/10 / (11*60) = 80*9.8*70 *10/(11*60) ~=830W Which is a very realistic figure.
Ash
Gravitational energy change = mgh = (80 kg)(9.8 m/s²)(70 m) = 54880 J
If gravitational energy change of 54880 J is 10%, then total energy expended will be...
Total energy = (100%/10%)*54880 J = 548800 J
Average metabolic rate = Total energy/time in sec = (548800 J)/[(11min * 60)s] ≈ 832 W
billrussell42
energy to climb 70 m, E = mgh = 80•9.8•70 = 54880 J
total energy expended would be 10 times that or 548800 J
power in watts = energy in joules / time in seconds
11 min = 660 s
P = 548800 J / 660 s = 832 J/s or 832 watts
Ruel The Midianite
What is the energy needed to move an 80 kg mass upwards by 70 m in the Earth's gravitational field.?
The question says that that amount of energy is only 10% of the climber's energy over an 11 minute period, so the total energy in 11 minutes is 10X as much. The metabolic rate is the average energy expenditure per unit time.