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:
$d= \frac{v^2}{2 \mu g}= \frac{(4.0 m/s)^2}{2 \cdot 0.4 \cdot 9.81 m/s^2}=2.04 m$

Answer 2

The player will slide a distance of 2.04 m before coming to rest.

Given data:

The initial speed of player is, u = 4.0 m/s.

The coefficient of kinetic friction is, $\mu = 0.40$.

Apply the work-energy principle as,

Work done by frictional force = Kinetic energy change

$W = \dfrac{1}{2}mv^{2} - \dfrac{1}{2}mu^{2}\\-f \times d = \dfrac{1}{2}mv^{2} - \dfrac{1}{2}mu^{2}\\-\mu mg \times d = \dfrac{1}{2}m(0^{2} )- \dfrac{1}{2}mu^{2}\\-\mu g \times d = - \dfrac{1}{2}u^{2}\\-(0.40) \times 9.8 \times d = - \dfrac{1}{2} \times 4^{2}\\d = 2.04 \;\rm m$

Thus, the distance slide by the player before coming to rest is 2.04 m.

Learn more about the work-energy principle here:

https://brainly.com/question/15442553?referrer=searchResults