Question

Scenario 2: Use the following information to answer questions 3 and 4:

Answer 1

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

$\dfrac{MET*weight (kg)*3.5}{200}$

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the  standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = $\dfrac{4.3*75 (kg)*3.5}{200}$

= 5.644

Therefore, for walking for 52 mins; Jim  burns approximately 293.475 kcal which is nearest to 300 kcal.

4.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

$V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7$

where ;

$V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = 0.0012$

∴$0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min$

Converting to watts;

Since;  6.118kg-m/min is =  1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

≅ 47 watts