Question

A 170lb baseball player slides into 2nd base. the player was running 8.2 m/s and traveled 5.5 ft. What is the coefficient of kinetic friction between the player and the ground while he is sliding?

Answer 1

:)

Answer 2

You need to start by putting all your data in a common format.
Use the equation \[\sum F=F_{slide} - F_{fric}\]
F=ma, friction=k*natural force, natural force=mg

Answer 3

so the natural force would be 16.5?

Answer 4

What does the k stand for?

Answer 5

Frictional coefficient.

Answer 6

Would that be his weight in newtons?

Answer 7

A newton is defined as kg*m/s^2

Answer 8

Yes, so would I just transfer 170 lb to newtons?

Answer 9

Which would be 757.3 n

Answer 10

\[170 lbs \times \frac{1kg}{2.2046lbs}=77.11kg\]
\[77.11kg \times 9.8\frac{m}{s^2}=755.7N\]

Answer 11

Okay would the final equation be \[755.7N \div 1.68g(9.8m/s ^{2}\]

Answer 12

No, I don't know where you got 1.68g from.

Answer 13

Oops nevermind, Im confused then on what to put in for the grams :/

Answer 14

or the mass

Answer 15

mass is kg, force is newtons, velocity is m/s, acceleration is m/s^2 assuming you are using metric.

Answer 16

Yes I am using metric but how do I fill in the mass if I dont have it?

Answer 17

You do have the mass.

Answer 18

So I can use the baseball players weight for the newton and kg?

Answer 19

The baseball player's "weight" is just his mass times the force of gravity.

Answer 20

So will the equation be \[755.7N \div 76.9kg(9.8m/s ^{2)}\]

Answer 21

The sum of the forces will be the force of the slide - the force of friction, force is equal to mass times acceleration, so you will have something like sumF=ma - mu x Fn and then solve for mu.

Answer 22

If you draw a free body diagram you can see more clearly the forces at work.

Answer 23

Hmm okay, Im just confused on what the force of friction is and the force of slide

Answer 24

The force of friction is a force that causes an object to want to stay at rest. So if the ball player is sliding then the force of friction is what "pulls" the opposite direction to slow him down.

Answer 25

Okay but how would I find that

Answer 26

Okay, sorry for the delay, real life called.
So our knowns
\[v=8.2\frac{m}{s}, \space m=77.18kg, \space d=1.68m\] Since we have velocity and distance we can calculate time\[v=\frac{d}{t} \rightarrow t=\frac{d}{v} \rightarrow \frac{1.68m}{8.2\frac{m}{s^2}}=0.20s\] Now to find the force of friction, f=ma, we need to find his acceleration in the x direction (slowing down). \[a=\frac{v_f^2-v_i^2}{2\times (x_f-x_i)} \rightarrow \space \frac{0^2-8.2m^2}{2\times(1.68m-0)}=-20.38\]
So the force of friction is\[F_{fric}=ma \rightarrow\space 77.18kg \times -20.38\frac{m}{s^2}=-1572.6N\] And you know that \[F_{fric}=\mu F_{nat} \rightarrow\space \mu mg\]You are looking for mu so solve algebraically and get\[\frac{F_{fric}}{mg}=\mu \rightarrow\space \frac{1572.6N}{77.15kg \times 9.8\frac{m}{s^2}}=2.079\]

Answer 27

ok thank you so much :)