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 17 minutes.

A.) Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m2.

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

Answer 1

Part A.

Drag force is given by:

Fd = (1/2)*rho*A*Cd*V^2

rho = density of air = 1.2 kg/m^3

A = Cross-sectional area = 0.72 m^2

Cd = Drag coefficient of everyday moving objects = 1/2

V = Speed of runner = distance/time = 5000 m/(17*60 sec) = 4.90 m/sec

Using these values:

Fd = (1/2)*1.2*(1/2)*0.72*4.90^2 = 5.186 N

Fd = 5.2 N = Drag Force

Part B.

Ratio of drag force and weight will be:

Ratio = Fd/W

W = Weight of runner = m*g = 60*9.81 = 588.6 N

Ratio = 5.2/588.6 = 0.0088 = 8.8*10^-3

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