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

`v=6.65m/sec`

Step-by-step 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`