Question

For sprinters running at 6 m/sm/s around a curved track of radius 26 mm, how much greater (as a percentage) is the average total force on their feet compared to when they are running in a straight line

Answer 1

Answer:

The percentage is   $k =$1.02%

Explanation:

From the question we are told that

   The speed of the sprinter is  $v_s = 6 \ m/s$

    The radius of the curve is  $r = 26 m$

   

The centripetal acceleration at the curve is mathematically represented as

     $a = \frac{v^2}{r}$

substituting values

    $a = \frac{ 6^2}{26}$

    $a = 1.385 \ m/s^2$

Now the force acting on the sprinter around the curve are

      $F_s$ which represents centripetal force which  is mathematically evaluated as

     $F_s = ma$

     $F_s =1.385 * m$

   and  $F_v$ which is the centrifugal force which is generally represented as

          $F_v = mg$

           $F_v = 9.8 * m$

Now the resultant force which is the force acting on the sprinter when running around the curve is mathematically represented as

       $F_r = m \sqrt{ a^ 2 + g^2}$

substituting values

     $F_r = m \sqrt{ 1.385^ 2 + 9.8^2}$

       $F_r = 9.9 * m$

the average total force on their feet compared to when they are running in a straight line is mathematically evaluated as

     $k = \frac{F_r - F_v}{ F_v} * 100$

     $k = \frac{9.9 *m - 9.8 * m }{ 9.8 * m } * 100$

    $k =$1.02%