Coefficient of Friction and Energy

[Blakeasd]

Homework Statement

A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled's velocity is 2m/s. Determine the coefficient of friction between the rough ice and sled.

Homework Equations

KE = (1/2)mv²

F_F = u_k * F_N

W = f * d

v² = vi² + 2a(x - xi)

The Attempt at a Solution

I begin by finding the Kinetic energy when the sled is on the frictionless sheet of ice:

KE = (1/2)mv²

KE = (1/2) * (30) * 4²

KE = 240J

Then I find the Kinetic Energy when the sled is on the rough patch:

KE = (1/2)mv²

KE = (1/2) * (30) * 2²

KE = 60J

I now note the amount of energy released due to friction:

240J - 60J = 180J

I continue to evaluate the force acting on the sled when on the rough patch:

W = fd
60J = f * 3m
f = 20N

Then I solve for acceleration in the rough patch:

v² = vi² + 2a(x - xi)

4 = 0 + 2*a*3
a= (2/3)

Now I sum the forces:

20N - (u_k * (30kg * 9.8)) = 30kg * (2/3)
u_k = 0

This is obviously not correct as it states friction is present. I don't know where I'm making a mistake. Presumably my logic is incorrect, not the math itself. I think finding the kinetic energy for the frictionless surface was useless with the approach I'm taking.

Could anyone please help me understand what's incorrect?

Thanks.

 

[tiny-tim]

Hi Blakeasd!

I continue to evaluate the force acting on the sled when on the rough patch:

W = fd
60J = f * 3m
f = 20N

No, W = 180J.

(and what is the relevance of acceleration? )

 

[Blakeasd]

Thank you for your response.

When I sum up the forces, I set them equal to ma. I need acceleration so I can solve for the coefficient of friction.

Can you please explain how you found work to be 180J .

Thanks!

 

[tiny-tim]

Hi Blakeasd!

The work energy theorem says that the change in mechanical energy equals the work done

That change is 180J

(and you can get µ from W = Fd and F = µmg)

 

[Blakeasd]

Ok, I see why it is 180J, thanks! Here is my revamped work:

W = fd
180J = f * 3m
f = 60N

60N - (u * (30*9.8)) = (2/3) * 30
u = -.14

This is not a choice, though:

a.) 0.07
b.) 0.12
c.) 0.20
d.) 0.27
e.) 0.60

When I use W = Fd and F = umg , I get .204 (an answer choice).

So why doesn't this work when I sum forces? If I were to assume constant velocity, then summing forces would work, but how could I tell it is constant velocity? Why is it appropriate to assume this?

 

[tiny-tim]

Then I solve for acceleration in the rough patch:

v² = vi² + 2a(x - xi)

4 = 0 + 2*a*3
a= (2/3)

No, that should be

16 = 4 + 2*a*3
a= 2 ​

60N - (u * (30*9.8)) = (2/3) * 30
…
So why doesn't this work when I sum forces? If I were to assume constant velocity, then summing forces would work, but how could I tell it is constant velocity? Why is it appropriate to assume this?

corrected for a, that's 60N - (µ * (30*9.8)) = 2 * 30

which is F - µmg = ma

that's wrong because you've used the same force twice

there is only one force (= ma), and it's either F or µmg

 

[Blakeasd]

I don't understand how I'm using the same force twice. I take the force pushing the object forward and subtract the force of friction pushing in the other direction. When summing forces I thought it is necessary to set their sum equal to ma. Am I missing something fundamental?

Thanks!

 

[tiny-tim]

A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled's velocity is 2m/s. Determine the coefficient of friction between the rough ice and sled.

I don't understand how I'm using the same force twice. I take the force pushing the object forward and subtract the force of friction pushing in the other direction.

there is no force pushing the sled!

if it wasn't for the rough patch, the sled would go on for ever at 4m/s

the only force is the friction!

 

[Blakeasd]

Ah... so if I were to sum the forces:

0 - (u*30*9.8) = 60

u = .204

It works!

If you don't mind me asking another question, can you explain you reasoning for: v2 = vi2 + 2a(x - xi)

16 = 4 + 2*a*3

Why don't you use 2m/s for the final velocity?

Thanks!

 

[tiny-tim]

… can you explain you reasoning for: v2 = vi2 + 2a(x - xi)

16 = 4 + 2*a*3

Why don't you use 2m/s for the final velocity?

ohhh, that should have been

4 = 16 + (-2)*a*3
a= -2 …​

the acceleration from the friction is actually -2