Solving Kinetic Friction: Baseball Player Sliding Into Third Base

[Cassie]

I would really appreciate it if anyone could help me with this problem.

A baseball player slides into third base with an initial speed of 8.50 m/s. If the coefficient of kinetic friction between the player and the ground is 0.44, how far does the player slide before coming to rest?

I'm not sure if the solution is simpler then my attempts, but I can say I am quite confused.

 

[hotvette]

I'm not sure if the solution is simpler then my attempts, but I can say I am quite confused.

Welcome to PF!

Tell us details of your attempts and specifics about your confusion.

 

[Cassie]

I have drawn the free body diagram, but my professor was very vague on most of this chapter. I am confused how to solve for acceleration or time using kinematics without one or the other. I am sure there is some connection I am failing to make.

 

[bob1182006]

the solution doesn't seem that complicated.
Usually when you solve for acceleration you use the FBD and use Newton's second law to find the acceleration, the mass of the baseball player isn't known so you will get a formula with mass as a variable.

when you get all of the Forces acting on the baseball player you can solve for acceleration and then go from there.

 

[Cassie]

Sadly, I tend to over complicate and over analyze my physics work.

There is the normal force and weight as well as the frictional force.

 

[johnsonandrew]

Are you familiar with the acceleration formula Vf^2 = Vi^2 + 2 * a * d ?
Vf = final velocity, Vi= initial velocity, a=acceleration, d= distance

 

[Cassie]

Are you familiar with the acceleration formula V_{}final

No, I don't think so. Its possible that I just might not recognize it though.

 

[Cassie]

Yes I am, but I don't understand how I would use that formula.

 

[Cassie]

No, I don't think so. Its possible that I just might not recognize it though.

edit: scratch this. posted it too soon.

 

[johnsonandrew]

Alright, you are not given mass or weight by any chance are you? I believe you'll have to use this formula if you are not

 

[hotvette]

Suggest you do the following. List the magnitude and direction of each force and we'll tell you if you are on the right path. By the way, once you know the magnitude/direction of the forces, writing and integrating the F=ma equations is usually simple. Try it by using +x as the direction the player is sliding and +y as up. As was mentioned in the rapid succession of replies, you must have been given the mass of the runner, no?

 

[johnsonandrew]

Yeah hotvette knows what he's talking about. What would be the force acting against friction in the +x direction though hotvette? Like bob118 said, if you are not given mass I think you'll have to find acceleration, then plug that into the formula I gave you... But I don't know how you would find the sum of forces ( what is +x force equal to?) . the sum of forces, x, would be the force in the +x direction minus Ffriction...

 

[hotvette]

What would be the force acting against friction in the +x direction though hotvette?

That's what I'm trying to get Cassie to do. List the forces involved.

 

[Cassie]

Alright, you are not given mass or weight by any chance are you? I believe you'll have to use this formula if you are not

I am not given mass. Just initial speed and the kinetic friction coefficient.

 

[Cassie]

That's what I'm trying to get Cassie to do. List the forces involved.

I did list them. Normal force and weight in the y-direction, and friction force in the x-direction.

 

[splac6996]

Recall that you can use your force equations in the x direction and the y direction. Do some substituting between them and you will get an acceleration. Once you have that use your kinematic equation to solve for the change in postion. (delta s)

 

[johnsonandrew]

One more, no? m*a in the positive x direction or no?


edit-- nevermind I'm confused

 

[splac6996]

ma is in the positive direction because your friction goes in the opposite direction so he comes to a stop

 

[johnsonandrew]

How can you find force of friction without normal force, mass, or weight?

splac6996: Then wouldn't you get Fxtotal = ma- Ff ?

(taking forward as positive)

 

[bob1182006]

Mass will just be a variable in this answer so it won't be numerical it will be a formula.

and Cassie you should write out Newton's second law now that you know the forces acting on the body. with respect to your axis's and which way is positive.

 

[splac6996]

Fx=Ff=ma
Fy=N-mg=0
N=mg
Ff=coef*N so
coef*mg=ma

 

[johnsonandrew]

Fx=Ff=ma
Fy=N-mg=0
N=mg
Ff=coef*N so
coef*mg=ma

ohhh! so
0.44 * g= a
0.44(9.81)= a = 4.32 m/s/s

then acceleration formula says vf^2= vi^2 + 2ad
so 8.5^2 = 0^2 +2(4.32)(d)
72.25 = 8.64d
d= 8.36 m is that correct?

 

[bob1182006]

ohhh! so
0.44 * g= a
0.44(9.81)= a = 4.32 m/s/s

then acceleration formula says vf^2= vi^2 + 2ad
so 8.5^2 = 0^2 +2(4.32)(d)
72.25 = 8.64d
d= 8.36 m is that correct?

yes but you should've let Cassie do it or done it though pm's :s.

Also your signs are wrong, gravity (if we're going negative y is down) *.44 will give a -acceleration which opposes the motion to the right (+x side of the x axis).
and the velocity given is initial not final.

 

[Cassie]

I contacted my TA about the problem. The formula was simply V^2 / 2*g*coef=s
Thank you for all your help!

 

[splac6996]

that should be right let me know how it turns out

 

[bob1182006]

I contacted my TA about the problem. The formula was simply V^2 / 2*g*coef=s
Thank you for all your help!

yep vf^2=2as
v^2/(2a)=s
a=g*Uk
v^2/(2*g*uk)=s

 

[johnsonandrew]

yes but you should've let Cassie do it or done it though pm's :s.

Also your signs are wrong, gravity (if we're going negative y is down) *.44 will give a -acceleration which opposes the motion to the right (+x side of the x axis).
and the velocity given is initial not final.

Oh oops. Alright I see