Frictional Force 2: Find Coefficient of Kinetic Friction on Ice

[Peterson]

INTRODUCTION:
This is a problem from my Introduction to Physical Science class using "Conceptual Physics" 10th Ed.by Paul G. Hewitt

EXACT PROBLEM:
"A hockey puck slides 20.0m across a frozen pond in a time of 8.0 seconds before coming to a rest."

PROBLEMS FACED:
Find the coefficient of kinetic friction between the ice and the puck.

MY THOUGHTS:
I don't know what to do here really. I am not sure which formulas to use.

 

[fizzzzzzzzzzzy]

you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation.

 

[Peterson]

you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation.

That sounds like a plan! So I need to use one of these friction equations?
https://www.physicsforums.com/showpost.php?p=920984&postcount=4

 

[Peterson]

you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation.

so how do I rewrite this to solve for (de)acceleration?

Does this look correct?
a = (x-x0-v0t)/(.5t)

Do the X variables cancel out?

 

[hage567]

"you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation."


I don't think this the proper equation to use, at least not by itself. V is this equation represents the initial velocity, which we don't know. So you can start by getting an expression for the initial velocity (Vo, by using another kinematic equation), and then put that into this equation for V.

If you bring arrange the equation so (X-Xo) = Vo + 0.5at^2, you can see that the (X-Xo) is just the displacement of the puck (20 - 0 = 20 m).

 

[Peterson]

"you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation."


I don't think this the proper equation to use, at least not by itself. V is this equation represents the initial velocity, which we don't know. So you can start by getting an expression for the initial velocity (Vo, by using another kinematic equation), and then put that into this equation for V.

If you bring arrange the equation so (X-Xo) = Vo + 0.5at^2, you can see that the (X-Xo) is just the displacement of the puck (20 - 0 = 20 m).

how can I figure out the velocity if I don't know the acceleration?

 

[hage567]

You can get an equation for Vo in terms of a by using one equation. Then you put that into the other equation you have with Vo and a, and you will only have "a" left over to solve for. It is a two equations, two unknowns kind of thing.

 

[Peterson]

You can get an equation for Vo by using one equation. Then you put that into the other equation you have with Vo and a, and you will only have "a" left over to solve for. It is a two equations, two unknowns kind of thing.

To me, this is like telling a lost person to just go North, when the lost person doesn't even know which direction north is.

I need a point of reference and a starting point. I know the forum rules about how you aren't supposed to do my work, but how are you helping me if you only give me a small fraction of advice every time?

 

[gabee]

Well, that's because we want you to try to work it out on your own. Unfortunately people who are lost see these pieces of advice as cryptic hints. But, you should probably know these two equations:

x = x_0 + v_0t + \frac{1}{2}gt^2

v = v_0 + at

Then think about what the problem gives you. The problem says the puck is initially traveling at some speed v0 (which you don't know) and it is decelerating because of friction. It travels a certain distance from the point x0 = 0 to x = 20 in t = 8 seconds. See if you can do it from there!

 

[Peterson]

Well, that's because we want you to try to work it out on your own. Unfortunately people who are lost see these pieces of advice as cryptic hints. But, you should probably know these two equations:

x = x_0 + v_0t + \frac{1}{2}gt^2

v = v_0 + at

Then think about what the problem gives you. The problem says the puck is initially traveling at some speed v0 (which you don't know) and it is decelerating because of friction. It travels a certain distance from the point x0 = 0 to x = 20 in t = 8 seconds. See if you can do it from there!

I am familiar with these equations, but why did you replace variable a with variable g? Gravity, I take it?

 

[gabee]

Oh crap, yes. That's just a typo, I'm so used to using g there it just slipped out :X But yes, the same equation, just use a instead of g.

 

[Peterson]

Oh crap, yes. That's just a typo, I'm so used to using g there it just slipped out :X But yes, the same equation, just use a instead of g.

so does that mean it is:

x = x0 + v0t + (1/2)gt^2
20 = 0 + v0(8) + (1/2)(9.8)(8^2)
20 = v0(8) + 313.6
-293.6/8 = v0
v0 = 36.7 m/s

 

[gabee]

No, it looks like this typo is causing more trouble than it was worth :X Your acceleration is not g, you aren't even considering gravity in this problem. The equations I meant to write are:x = x_0 + v_0t + \frac{1}{2}at^2

v = v_0 + at

 

[Peterson]

No, it looks like this typo is causing more trouble than it was worth :X Your acceleration is not g, you aren't even considering gravity in this problem. The equations I meant to write are:


x = x_0 + v_0t + \frac{1}{2}at^2

v = v_0 + at

superb. idk then.

 

[gabee]

Alright, we have those two equations, and the problem told us:

x0 = 0 m
x = 20.0 m
t = 8.0 s
v = 0 m/s (if the object comes to rest, it doesn't have velocity!)
v0 = ?
a = ?

We have two equations and two unknowns, so we need to solve a system of equations! Go ahead and write those numbers into those equations and you'll probably see how to solve it. It's just algebra from here!

 

[Peterson]

Alright, we have those two equations, and the problem told us:

x0 = 0 m
x = 20.0 m
t = 8.0 s
v = 0 m/s (if the object comes to rest, it doesn't have velocity!)
v0 = ?
a = ?

We have two equations and two unknowns, so we need to solve a system of equations! Go ahead and write those numbers into those equations and you'll probably see how to solve it. It's just algebra from here!

Ok then:

V = v0 + at
0 = v0 + a8
v0 = -a8

x = x0 + v0t +(.5)a(t^2)
20 = 0 +(-a8) + (.5)a(8^2)
20 = -a64 + a32
.625 = -a64 +a
.0097 = -a +a
.0097 = 0 ?

well now, maybe that algebra isn't so simple after all...

 

[Peach]

Try to do it without the numbers first, solve for it in terms of variables only. All the numbers is confusing you. When you have the eqn already, then plug in the numbers.

 

[gabee]

Ok then:

V = v0 + at
0 = v0 + a8
v0 = -a8

x = x0 + v0t +(.5)a(t^2)
20 = 0 +(-a8) + (.5)a(8^2)
20 = -a64 + a32

So far, so good...

.625 = -a64 +a

Here is where you made a mistake. If 20 = -a*64 + a*32, isn't this the same as
20 = -64*a + 32*a
20 = -32*a

 

[Peterson]

So far, so good...Here is where you made a mistake. If 20 = -a*64 + a*32, isn't this the same as
20 = -64*a + 32*a
20 = -32*a

OHHHHHH! Stupid mistake there!

Ok then:

V = v0 + at
0 = v0 + a8
v0 = -a8

x = x0 + v0t +(.5)a(t^2)
20 = 0 +(-a8) + (.5)a(8^2)
20 = -a64 + a32
20 = -32a
a = -.625

v0 = -a8
v0 = .625(8)
v0 = 5

so here's all of the variables that we know now:
x0 = 0 m
x = 20.0 m
t = 8.0 s
v = 0
v0 = 5
a = -.625

 

[gabee]

Alright! So we have our acceleration on the puck. Now, remember Newton's second law, F = ma. We can ignore forces in the y-direction because they are mg and N, and they cancel exactly in this case. So, we are only concerned about forces along the x-axis. And that force acting on the puck is...?

 

[Peterson]

Alright! So we have our acceleration on the puck. Now, remember Newton's second law, F = ma. We can ignore forces in the y-direction because they are mg and N, and they cancel exactly in this case. So, we are only concerned about forces along the x-axis. And that force acting on the puck is...?

Friction?
Ffrictional = (Muk)(Fn) = (Muk)mg ?

 

[gabee]

That's exactly right. Now, F = ma, and we have the only force acting on our puck (friction = mu*m*g), we have the mass, and we have the acceleration...let's plug it all in.

 

[Peterson]

That's exactly right. Now, F = ma, and we have the only force acting on our puck (friction = mu*m*g), we have the mass, and we have the acceleration...let's plug it all in.

Force = ma
Force = (20)(5)
Force = 100N

I don't see where Force plugs into (friction = mu*m*g).

 

[gabee]

Not quite. When you know the forces which are acting on the object, you can add up the forces (to obtain the NET force acting on an object) and plug them in for the F in F=ma.

Just as a sidenote, unrelated to this problem, the way you might do this is the following: Say you have a force A pulling an object to the right, and a force B pulling the object to the left. If you say the +x direction is to the right, we can say that the NET force F = A - B = ma. Then, if you know A and B and m, you can figure out how the object will accelerate, or if you know m and a and B, you can figure out A, and so on.

So, in this case, since there is only one force acting on the object, friction, that friction formula is your F in F = ma.

 

[Peterson]

Not quite. When you know the forces which are acting on the object, you can add up the forces (to obtain the NET force acting on an object) and plug them in for the F in F=ma.

Just as a sidenote, unrelated to this problem, the way you might do this is the following: Say you have a force A pulling an object to the right, and a force B pulling the object to the left. If you say the +x direction is to the right, we can say that the NET force F = A - B = ma. Then, if you know A and B and m, you can figure out how the object will accelerate, or if you know m and a and B, you can figure out A, and so on.

So, in this case, since there is only one force acting on the object, friction, take the formula for the force of friction and put it in for F in F = ma.

I realized my 20 is 20 METERS.

(mu)mg = ma
(mu)m(9.8) = m(5)
(mu) = [(m)(9.8)]/[(m)(5)]

if this is correct, I still don't have a mass of the hockey puck.

 

[gabee]

First, where did 5 come from? Shouldn't a = -.625 (the value for acceleration we figured out earlier)?

And, look at your equation: m cancels!

 

[Peterson]

First, where did 5 come from? Shouldn't a = -.625 (the value for acceleration we figured out earlier)?

And, look at your equation: m cancels!

I used the v0 value of 5 instead of the a value of .625.

(mu)mg = ma
(mu)m(9.8) = m(.625)
(mu) = (9.8)/(.625)
mu = 15.68N

is it Newtons?

 

[Peterson]

I did that wrong as well...

(mu)mg = ma
(mu)m(9.8) = m(.625)
(mu) = (.625)/(9.8)
mu = .0637N

 

[gabee]

There you go, you've got it! It's not Newton's; actually, \mu is a dimensionless value, so it has no units. Can you see why? In your equation, you have mu = .625 m/s^2 / 9.8 m/s^2. The m/s^2 cancel, leaving you with just a pure number, with no units.

 

[Peterson]

It's not Newton's; actually, \mu is a dimensionless value, so it has no units. Can you see why? In your equation, you have mu = 9.8 m/s^2 / .625 m/s^2. The m/s^2 cancel, leaving you with just a pure number, with no units.

that makes sense.

So final answer to the problem is:
\mu = .0637