It is well known that runners run more slowly around a curved track than a straight one. One hypothesis to explain this is that the total force from the track on a runner's feet--the magnitude of the vector sum of the normal force (that has average value mg to counteract gravity) and the inward-directed friction force that causes the runner's centripetal acceleration--is greater when running around a curve than on a straight track. Runners compensate for this greater force by increasing the time their feet are in contact with the ground, which slows them down.
For sprinters running at 8 m/s around a curved track of radius 18 m,how much greater (as a percentage) is the average total force on their feet compared to when they are running in a straight line? Express your answer as a percent.
(Fcurved - Fstraight) / Fstraight

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aristocles aristocles · Answer 1 · 2019-11-08T02:18:18+01:00

Answer:

[tex]percentage = 6.27 %[/tex]

Explanation:

As we know that when runner is moving on straight track then the net force on his feet is given as

[tex]F_s = mg[/tex]

while when runner is moving on circular track then we have

[tex]F_c = \sqrt{(mg)^2 + (\frac{mv^2}{R})^2}[/tex]

[tex]F_c = m\sqrt{g^2 + \frac{v^4}{R^2}}[/tex]

[tex]F_c = m\sqrt{9.8^2 + \frac{8^4}{18^2}}[/tex]

[tex]F_c = m(10.42)[/tex]

now percentage change is given as

[tex]percentage = \frac{F_c - F_s}{F_s} \times 100[/tex]

[tex]percentage = \frac{m(10.42) - m(9.81)}{m(9.81)}\times 100[/tex]

[tex]percentage = 6.27 %[/tex]