Assuming that muscles have an efficiency of 22 percent for converting energy into work, how much energy (in kJ) is expended by an 80-kg person in climbing stairs through a vertical distance of 25 m.
2007-02-23 11:55:43 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Simplifying throughout, we have:
PE = mgh = 80*10*25 = ke; where m = mass of person, g ~ 10 m/sec^2 (actually 9.81 at Earth's surface), and h = height of the climb. The effective kinetic energy, ke = is converted to potential energy as the person walks up the stairs, but since r = 22 % = the efficiency, the person has to expend about 5 times more energy to get to PE.
ke = r KE, where the KE is the gross amount of energy the body expends to get an effective ke. Therefore, PE = mgh = r KE and KE = mgh/r ~ 80*10*25/.22, where KE is the gross energy used up due to inefficiencies. You can do the math.
Lessons learned: There is no energy expended in the horizontal run upstairs; only in the vertical rise overcoming the force of gravity. The effective kinetic energy is converted to potential energy when the person stops 25 m vertically up the ladder. The gross energy, due to inefficiencies is higher than the effective energy; so ke = r KE holds.
2007-02-23 12:46:51 · answer #1 · answered by oldprof 7 · 1⤊ 0⤋
Work done in climbing = mgh work it out
This is 22% of work done by muscles. So 100% of WD by muscles = mgh/.22
2007-02-23 21:29:52 · answer #2 · answered by hello 6 · 0⤊ 0⤋