Question

A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.646. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.36 s, what is his initial speed?

Answer 1

628.022466 N

8.61 m/s

Step-by-step explanation:

m = Mass

`\mu` = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

`F_f=\mu mg\\\Rightarrow F_f=0.646* 99.1* 9.81\\\Rightarrow F_f=628.022466\ N`

Magnitude of frictional force is 628.022466 N

`F=ma\\\Rightarrow a=(F_f)/(m)\\\Rightarrow a=(628.022466)/(99.1)\\\Rightarrow a=6.33726\ m/s^2`

`v=u+at\\\Rightarrow 0=u-6.33726* 1.36\\\Rightarrow u=8.61\ m/s`

Initial speed of the player is 8.61 m/s