A baseball player slides into third base with an initial speed of 8.55 m/s. If the coefficient of kinetic friction between the player and the ground is 0.45, how far does the player slide before coming to rest?

Question

anonymous · Answer

If i were you i think id divide but idk

queelius · Answer

Assuming normal force between person and ground is weight = mg, then:

Let u = 0.45

net force F = -umg; F = ma, so a = -umg / m = -ug. So, acceleration is constant => v(t) = -ugt. So, what do you think the next step is?

queelius · Answer

Err, correction. v(t) = 8.55 - ugt

shane_b · Answer

$$Vi = 8.55 m/s$$$$F_f=U_k mg=0.45m(9.8 m/s^2)$$Calculate the acceleration (which will be negative since it's due to the force of friction):
$$a = \frac{F}{m}$$$$a = \frac{(0.45)(9.8m/s^2) m}{m}$$$$a = 4.41 m/s^2$$
Time to stop:$$t=\frac{8.55m/s}{4.41 m/s^2}=1.94s$$
Now see how far they go in 1.94s with an acceleration of -4.41m/s^2:$$x = V_0t+\frac{1}{2}at^2$$$$x = (8.55m/s)(1.94s)+\frac{1}{2}(-4.41m/s^2)(1.94s)^2=8.29m$$Did it in a hurry...hopefully there are no silly mistakes.

anonymous · Answer

omg shane your a genuis. :o