Question

using energy considerations, calculate the average force (in n) a 68.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a head - wind that exerts an average force of 30.0 n against him.

Explanation:

Step1: Calculate the change in kinetic energy

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$. The initial velocity $v_1 = 2.00\ m/s$ and the final velocity $v_2=8.00\ m/s$, and mass $m = 68.0\ kg$. The change in kinetic energy $\Delta K=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$.
$\Delta K=\frac{1}{2}\times68.0\times(8.00^{2}-2.00^{2})=\frac{1}{2}\times68.0\times(64 - 4)= \frac{1}{2}\times68.0\times60=2040\ J$.

Step2: Calculate the work - done by the head - wind

The work - done by a force $W = Fd$. The force of the head - wind $F_{wind}=30.0\ N$ and the distance $d = 25.0\ m$. So the work - done by the head - wind $W_{wind}=F_{wind}d=30.0\times25.0 = 750\ J$.

Step3: Calculate the work - done by the sprinter

Let the force exerted by the sprinter be $F$. The work - done by the sprinter $W_{sprinter}=Fd$. According to the work - energy theorem, the net work done on the sprinter is equal to the change in his kinetic energy. So $W_{sprinter}-W_{wind}=\Delta K$.
$Fd - 750=2040$. Since $d = 25.0\ m$, we have $25F-750 = 2040$.

Step4: Solve for the force exerted by the sprinter

First, add 750 to both sides of the equation: $25F=2040 + 750=2790$. Then divide both sides by 25: $F=\frac{2790}{25}=111.6\ N$.

Answer:

$111.6$