Question

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?

Answer 1

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction  $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

 $F=\mu mg$

 $F=0.555*97.6*9.8$

 $F=531.388N$

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 $a_f=-\mu g$

 $a_f=-0.555 *9.81$

 $a_f=-54455m/sec^2$

Therefore

 $v=u-at$

 $v=0+5.45*1.22$

 $v=6.65m/sec$