In this case, the net force acting on the runner is the force of friction between her shoes and the pavement, which is given by:
$$F_f=\mu_k n$$
where:
* $$F_f$$ is the force of friction
* μk is the coefficient of kinetic friction
* n is the normal force
The normal force is equal to the weight of the runner, which is given by:
$$n=mg$$
where:
* m is the mass of the runner
* g is the acceleration due to gravity
Combining these equations, we get:
$$F_f=\mu_k mg$$
and
$$a=\frac{F_f}{m}=\frac{\mu_k mg}{m}=\mu_k g$$
Substituting the given values, we get:
$$a=(0.72)(9.8 m/s^2)=7.06 m/s^2$$
Therefore, the greatest acceleration a runner can muster if friction between her shoes and pavement 72 percent of weight is \( 7.06 \ m/s^2 \).
