A 90.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.511. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.91 s, what is his initial speed?

Question

anonymous · Answer

The frictional force is simply$$F_f = \mu_k m g$$

His initial speed can be determined from Impulse$$I = \int\limits F dt$$From Newton's Second Law$$F = {dp \over dt}$$where p is momentum. Therefore,$$I = \int\limits {dp \over dt} dt = \int\limits dp = \Delta p = m(v_f - v_i)$$where $v_f$ is the final velocity and $v_i$ is the initial velocity. We can find the initial velocity of the player as$$F \Delta t = mv_i \rightarrow \mu_k m g \Delta t = m v_i \rightarrow \mu_k g \Delta t = v_i$$