Question

A baseball player slides into third base with an initial speed of 4.0 m/s . if the coefficient of kinetic friction between the player and the ground is 0.40, how far does the player slide before coming to rest?

Answer 1

The work done by the friction force to stop the player is equal to his loss of kinetic energy:
$W= \Delta K$

The work done by the friction force is the magnitude of the force $\mu m g$ times the distance covered by the player, d:
$W= \mu mg d$

The loss in kinetic energy is simply equal to the initial kinetic energy of the player, since the final kinetic energy is zero (the player comes to rest):
$\Delta K=K_i = \frac{1}{2}mv^2$

Substituting into the first equation, we get:
$\mu m g d= \frac{1}{2}mv^2$
from which we find d, the distance covered by the player:
$d= \frac{v^2}{2 \mu g}= \frac{(4.0 m/s)^2}{2 \cdot 0.4 \cdot 9.81 m/s^2}=2.04 m$