time: 5.68 sec, stairs: 12, height: .1778m 2. Running up the stairs (5 pts) . Ti

Question

time: 5.68 sec, stairs: 12, height: .1778m

2. Running up the stairs (5 pts). Time how long it takes for you (or a friend) to run as fast as possible up a staircase. Measure the height of one step and count the steps to find how high you climbed.

a) How much mechanical energy did you produce to climb the stairs?
b) What was the average mechanical power produced during your climb?

c) Estimate the muscle mass in your upper legs (quadriceps, hamstrings, and buttocks). (Recall that biological tissue has about the same density as water.) If each kilogram of muscle can produce about 100 watts of mechanical power, are these muscles sufficient to provide the power required for you to run up the stairs?

d) If you are 25% efficient at converting food energy to mechanical energy, and there are about 4200 joules in one food Calorie, how many Calories did you consume during your climb?

e) The food energy that was not converted into mechanical energy was mainly turned into heat. If your body were completely insulated (for instance, if you were wearing heavy winter clothes), how much would your body temperature increase as a result of running up the stairs? Assume that your entire body has a specific heat equal to that of water, C = 4.18 J/g·°C

Explanation / Answer

time: 5.68 sec,

stairs: 12,

Assuming mass = 60 kg

height of each stair: 0.1778m

So total height = 12 *0.1778 = 2.1336 m

a) Total mechanical energy produced = increase in potential energy = mgh = 60*9.81*2.16 = 1255.8 J

b) Average mechanical power = energy / time = 1255.8/5.68 = 221.1 W

c) Mass of quadriceps, hamstrings, and buttocks is around 8 Kg (rough estimate)

So the power it can generate = 800 W, which is sufficient.

d) Total energy required to climb = 1255.8 J

So total energy consumed = 1255.8 / 0.25 = 5023.2 J = 1.196 food calorie

e) Amount of heat = 5023.2-1255.8=3767.4 J

Let the temperature rise be T

mass = 60 kg

C = 4180 J/kg°C

So 3767.4 = 60 * 4180 *T

=>Rise in temperature, T = 0.015 °C

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