When you ride a bicycle at constant speed nearly all the...

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.42 m2 and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90. Metabolic power is the rate at which your body "burns" fuel to power your activities. For many activities, your body is roughly 25% efficient at converting the chemical energy of food into mechanical energy. What is the cyclist's metabolic power while cycling at 7.3 m/s? The food calorie is equivalent to 4190 J. How many calories does the cyclist burn if he rides over level ground at 7.3 m/s for 1.6 h ?

Transcript

00:01 In this question, we have to find the metabolic power denoted by p.

00:09 And we have to find the number of calories burn.

00:14 So number of calories burn, let us denote the calories burn by c .b.

00:22 C .b.

00:23 Is calories burn.

00:25 And what we will do, like calories burn.

00:31 And we are given in the question, then there is some data which can be helpful like cross -sectional area is given 0 .42 meters square.

00:44 So this is given to us drag coefficient, let us denote the drag coefficient by cd.

00:52 So cd is drag coefficient.

00:56 It is given 0 .90.

01:02 And also it is given that metabolic power we have to find.

01:06 At velocity 7 .3 meter per second and calories are given calories given are 4190 jury so and time the velocity 7 .3 meter per second is same but we have to find the caldip one for the time let us denote by t so we we have to find 4 .1 .6.

01:49 So starting from the metabolic power, firstly we will calculate, like the metabolic power, p is d multiplies v.

02:06 So v is velocity and d is drag force.

02:12 So d is drag force.

02:14 This is the formula.

02:16 So d is drag force.

02:22 Direct force and drag force can be calculated by using the formula cd that is drag coefficient multiplies 1 over 2 multiplies row low is density of air and cross -sectional area and v square so row is density of air and its value is density of air values 1 .2 p .g per meter cube.

03:10 Now using the information given in the question we will put all the data in this formula and will calculate the drag force.

03:20 So it will be 0 .90 multiplies 0 .5 multiply's 1 .2 multiply's 0 .4 to multiply 0 .4 to multiply 7 .3 square.

03:44 Solving this we will get 12 .086 newton as it is a force.

03:53 So drag force came to be 12 .086 newton...

Question

Please give Ace some feedback