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

The percentage is
`k =`1.02%

Step-by-step 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 = (v^2)/(r)`

substituting values

`a = ( 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 √( a^ 2 + g^2)`

substituting values

`F_r = m √( 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 = (F_r - F_v)/( F_v) * 100`

`k = (9.9 *m - 9.8 * m )/( 9.8 * m ) * 100`

`k =`1.02%