Question

The energy content of food is conventionally measured in Calories rather than joules. One Calorie in nutrition is equal to 4184 J. Metabolizing 1 g of fat can release 9.00 Calories. A 75.7 kg student decides to try to lose weight by exercising. He plans to run up and down the stairs in a football stadium as fast as he can and as many times as necessary. To evaluate the program, suppose he runs up a flight of 100 steps, each 0.150 m high, in 59.9 s. For simplicity, ignore the energy he uses in coming down (which is small). Part a (1 points) What is his average power output, in watts, as he runs up the stairs? Note that the power output is the rate of mechanical work generated, not the rate at which energy is consumed. Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23,-2, 1e6, 5.23e-8 Enter answer here W CHECK ANSWER 0 of 4 attempts used (1 points) Part b A typical efficiency for human muscles is 20.0%. For instance, when your body metabolizes 100 J of chemical energy it converts this into 20 J of mechanical work (here, climbing stairs), while the remainder is lost to thermal energy. Assuming all the energy consumed during exercise comes from burning fat, how many times must he run up the flight of stairs to burn 1.00 kg of fat? [for simplicity assume his total mass remains constant during this process] Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 flights of Enter answer here stairs CHECK 0 of 4 attempts used ANSWER

Answer 1

Answer:

Part a)

$P = 186 W$

Part B)

$N = 676 times$

Explanation:

Part a)

Average power is rate of energy consumption

So it is given as

$P = \frac{mgh}{t}$

$P = \frac{100(75.7)(9.81)(0.150)}{59.9}$

$P = 186 W$

Part b)

If the efficiency is given as

$\eta = 20$%

so we know that 20% of total energy consumed is converted into output work

So total power consumed here will be given as

$P = \frac{186}{0.20}$

$P = 930 W$

Energy consumed by him

$E = power \times time$

$E = 930 \times 59.9$

$E = 55707 J$

now we know that 1 g fat will release 9 Calories energy

So we have

$1 g = 9 \times 4184 J$

$1 g = 37656 J$

so to consume 1000 g of fat we need to release  total energy

$E = 37656 \times 10^3 J$

so total number of times

$N = \frac{37656 \times 10^3}{55707}$

$N = 676 times$