# Force Applications

1. Feb 8, 2006

### Huskies213

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 comming to rest ?

2. Feb 8, 2006

### jamesrc

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.)

3. Feb 8, 2006

### Huskies213

Re

can anyone expand on this explanation ? im still lost

4. Feb 8, 2006

### teken894

u could calculate force that will decelerate the baseball player using the coefficient of kinetic friciton

$$\mu_k = \frac{F_{fr}}{F_{N}}$$

figure out the deceleration from that force...
$$\frac{F}{m_{player}}=a_{player}$$

then use a kinematic equation to solve for distance
probably

$${v_0^2} = {v_1^2} + 2ax$$

Last edited: Feb 8, 2006
5. Feb 9, 2006

### jamesrc

The only force acting on the player is friction, so Newton's second law gives:

$$F_{net} = ma = f = -\mu mg$$

The mass cancels out when you solve for a:

$$a = -\mu g$$

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

$$v^2 = v_o^2 + 2a\Delta x$$

Last edited: Feb 9, 2006
6. Feb 9, 2006

### VietDao29

Nope, it shouldn't be written like that. It should instead read:
$$v ^ 2 = v_0 ^ 2 + 2ad$$
Note that it's v2 not v. :)

7. Feb 9, 2006

### jamesrc

Oops. Sorry about the typo. Fixed now.