Using energy considerations, calculate the average force a 60.0-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 headwind that exerts an average force of 30.0 N against him.

Work energy theorem: According to the work-energy theorem, the work done of the body equals the change in kinetic energy of the body.

Mathematically,

$W={\text{KE}}_f-{\text{KE}}_i$

The work done by nonconservative force is,

$W_{net}=W_c+W_{nc}$ (1.1)

Here,$W_{net}$is the total work done,$W_c$is the work done by all conservative forces, and$W_{nc}$is the work done by all nonconservative forces.

The work-energy theorem states that the total amount of work done is equal to the total amount of energy used.

$W_{net}=\Delta \text{KE}$ (1.2)

Here,$\Delta \text{KE}$is the change in kinetic energy.

The work done by all conservative forces is,

$W_c=Fd$ (1.3)

Here, F the average force exerted by the sprinter and d is the distance.

The work done by all nonconservative forces is,

$W_{nc}=-fd$ (1.4)

Here, f is the opposing force.

We obtain from equations (1.1), (1.2), (1.3), and (1.4),

$\begin{array}{rcl}\Delta \text{KE}& =& Fd-fd\\ \frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2& =& Fd-fd\end{array}$ (1.5)

Here, m denotes mass, v denotes final velocity, and u denotes initial velocity.

Rearranging equation (1.5) in order to get an expression for the average force exerted by the sprinter,

$F=\frac{m\left(v^2-u^2\right)}{2d}+f$

Putting 60 kg for m, 8m/s for v, 2m/s for u, 25m for d, and 30 N for f,

localid="1668678896333" $\begin{array}{rcl}F& =& \frac{\left(60\text{kg}\right)\times \left[{\left(8\text{m}/\text{s}\right)}^2-{\left(2\text{m}/\text{s}\right)}^2\right]}{2\times \left(25\text{m}\right)}+\left(30\text{N}\right)\\ & =& 102\text{N}\end{array}$

Therefore, the required average force exerted by the sprinter is 102 N.