Statuc and kinetic frictional forces

[pookisantoki]

A 103kg baseball player slides into second base. Then coefficient of kinetic friction between the player and the ground is 0.607 a.) what is the magntiude of tfictional force? b.) if the player comes to rest after 2.22sec, what is his intial speed?

a.) Fs max=Ms*Fn
Ms=.607
Fn=103
0.607*103=62.521
but it's wrong
b.) I thought could use V=Vo+At
Vo=?
V=O
t=2.22
A=)Fk/m

so 0=Vo+(Fk/m)*2.22
Is this the right set up?

 

[206PiruBlood]

Remember that frictional force is equal to the normal force times the coefficient, not the mass.

Your method for part B seems fine, you just need to adjust your acceleration after recalculating the force of friction.

 

[pookisantoki]

0=Vo+(Fk/m)*2.22
for part b
and I set it up where
v0= -(fK/m)/2.22
Fk is .607
and m is 103
so -(.607/103)/2.22=-.002655
but its wrong what am I doing wrong. thank you

 

[rl.bhat]

Normal force Fn = m*g

 

[pookisantoki]

So i redid this problem by doing this:
Since fk=mk*Fn and Fn=m*g so (mk*m*g)/m but the m gets canceled out so...
Vo=-(Mk*g)/2.22
=(.607*9.8)/2.22
=-2.6795
But it's still wrong what am i doing wrong?