Solve Distance: Baseball Player Sliding w/ 7.9 m/s & 0.41 Coefficient

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Can anyone help me to solve for the distance here ?? I'm lost ...

A baseball player slides with an initial speed of 7.9 m/s. If the coefficient of kinetic friction between the player and the ground is 0.41, how far does thet player have to slide before coming to rest ?
 
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Can you calculate the magnitude of the deceleration (Hint: make a free body diagram and use Newton's second law; the mass of the player will cancel out)?

(You can also use the work-energy theorem if you have covered that in class.)
 
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can anyone expand on this explanation ? I am still lost
 
u could calculate force that will decelerate the baseball player using the coefficient of kinetic friciton

[tex]\mu_k = \frac{F_{fr}}{F_{N}}[/tex]figure out the deceleration from that force...
[tex]\frac{F}{m_{player}}=a_{player}[/tex]

then use a kinematic equation to solve for distance
probably

[tex]{v_0^2} = {v_1^2} + 2ax[/tex]
 
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The only force acting on the player is friction, so Newton's second law gives:

[tex]F_{net} = ma = f = -\mu mg[/tex]

The mass cancels out when you solve for a:

[tex]a = -\mu g[/tex]

Now that you have the magnitude of the acceleration, you can use the kinematic equation suggested by teken894, which should be written:

[tex]v^2 = v_o^2 + 2a\Delta x[/tex]
 
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jamesrc said:
Now that you have the magnitude of the acceleration, you can use the kinematic equation suggested by teken894, which should be written:

[tex]v = v_o^2 + 2a\Delta x[/tex]
Nope, it shouldn't be written like that. It should instead read:
[tex]v ^ 2 = v_0 ^ 2 + 2ad[/tex]
Note that it's v2 not v. :)