Coefficient of Friction and Energy

In summary, a 30kg sled is initially sliding on a frictionless sheet of ice at a velocity of 4m/s. After encountering a rough patch of ice and traveling 3m, the sled's velocity decreases to 2m/s. Using the work-energy theorem and the equation F = µmg, the coefficient of friction between the rough ice and sled can be determined to be approximately 0.204. The sum of forces is equal to the sled's mass multiplied by its acceleration, and the acceleration can be found by using the equation v^2 = v0^2 + 2a(x-x0), where v0 and x0 represent the initial velocity and position, and x represents the distance traveled. The final
  • #1
Blakeasd
11
0

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

FF = uk * FN

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 - (uk * (30kg * 9.8)) = 30kg * (2/3)
uk = 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.
 
Last edited:
Physics news on Phys.org
  • #2
Hi Blakeasd! :smile:
Blakeasd said:
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. :wink:

(and what is the relevance of acceleration? :confused:)
 
  • #3
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!
 
  • #4
Hi Blakeasd! :smile:

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

That change is 180J :wink:

(and you can get µ from W = Fd and F = µmg)
 
  • #5
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?
 
  • #6
Blakeasd said:
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 :wink:

Blakeasd said:
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 :smile:
 
  • #7
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!
 
  • #8
Blakeasd said:
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.
Blakeasd said:
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! :smile:
 
  • #9
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!
 
  • #10
Blakeasd said:
… 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 :wink:
 
  • Like
Likes 1 person

FAQ: Coefficient of Friction and Energy

What is the coefficient of friction?

The coefficient of friction is a dimensionless quantity that represents the amount of resistance or friction between two surfaces in contact with each other. It is a ratio of the force required to move one surface over the other to the normal force between the two surfaces.

How is the coefficient of friction measured?

The coefficient of friction can be measured using a variety of methods such as inclined plane, sliding friction, and rolling resistance. Inclined plane method involves measuring the angle at which an object starts to slide down a ramp. Sliding friction method involves measuring the force required to slide one surface over the other. Rolling resistance method involves measuring the force needed to roll an object over a surface.

What factors affect the coefficient of friction?

The coefficient of friction is affected by a number of factors including the types of surfaces in contact, the nature of the surfaces (rough or smooth), the amount of force applied, the temperature and humidity of the environment, and the presence of any lubricants or contaminants on the surfaces.

What is the relationship between coefficient of friction and energy?

The coefficient of friction determines the amount of energy that is required to move an object over a surface. Higher coefficients of friction mean that more energy is needed to overcome the resistance and move the object. On the other hand, lower coefficients of friction mean that less energy is needed to move the object.

How is the coefficient of friction used in real-life applications?

The coefficient of friction is an important concept in many real-life applications such as designing machines and vehicles, determining the performance of tires and brakes, and understanding the behavior of materials in contact with each other. It is also used in sports and games to optimize performance and reduce the risk of injury.

Similar threads

Back
Top