A baseball player is running to second base at 5.03 m/s. When he is 4.80 m from the plate he goes into a slide. The coefficient of kinetic friction between the player and the ground is 0.180, and the
Coefficient of static friction is 3.14. What is his speed when he reaches the plate?
A) 2.89 m/s
B) 1.96 m/s
C) 2.56 m/s
D) 4.47 m/s
E) He stops before reaching the plate.

Question

Quizplus · Accepted Answer

Explanation:The force of kinetic friction is:$$f_k=\mu_k n=\mu_k mg$$The work done by the force of kinetic friction is:$$W_f=-\mu_k mgd=-\mu_k mg(4.80	ext{ m})$$The change in kinetic energy is:$$\Delta K=K_f-K_i=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy:$$W_{net}=\Delta K$$Therefore,$$-\mu_k mgd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$Solving for $v_f$, we get:$$v_f=\sqrt{v_i^2-2\mu_k gd}=\sqrt{(5.03	ext{ m/s})^2-2(0.180)(9.81	ext{ m/s}^2)(4.80	ext{ m})}$$$$v_f=2.89	ext{ m/s}$$

A Baseball Player Is Running to Second Base at 5