Question

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 19minutes.
Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m2 and the density of air is 1.2 kg/m3.

What is this force as a fraction of the runner's weight?

Answer 1

Answer:

Part A.1) The drag force on the runner is 8.33N .

Part A.2 ) The fraction of the drag force to the runner’s weight is 0.014 .

Explanation:

Drag force acts opposite to the direction of the of motion of an object or body. It tends to oppose the motion of the body. The drag force on the runner acts to oppose the motion of runner while he runs. This drag force depends on the speed of runner.

Solution:

Consider the drag coefficient for the runner is approximately 11 .

Write the expression for the speed of the runner.

v=  d/t

Substitute  5 km for d, and 19 min. for t;

V=$\frac{5*1000}{19}$

 V=4.39 m/s

Write the expression for the drag force on the runner.

​ $F_{d}$ =  0.5ρAv² $C_{d}$

subtitute A=0.72 m², ρ=1.2 kg/m³ ,   $C_{d}$  =1 and v= 4.39 m/s

  $F_{d}$ = (0.5)*1.2*0.72* 4.39*1

  $F_{d}$ = 8.33 N

Part A.1

The drag force on the runner is 8.33N .

(A.2)

Write the expression for the weight of the runner

     w = mg

Here, w is the weight of the runner, g is acceleration due to the gravity, and mm is the mass of the runner.

     m=60 kg ; 9.81m/s² for g

      w= ( 60*9.81 )

      w=588.6 N

The fraction of the drag force to the runner’s weight can be calculated as,

f=   ($F_{d}$ )/ (w)

Here, ff is the drag force as a fraction of runner’s weight.​    

            f= (8.33) / (588.6)

            f=0.014

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