I think I understand, but where do you find force?

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In summary, the conversation was about finding the coefficient of kinetic friction to stop a train engine with a velocity of 30 m/s in 30 seconds. The equation Ffric = μN = μmg was suggested to be used, and the acceleration and friction force were calculated to be 1 m/s and 10,000 N, respectively. The concept of air resistance was also briefly discussed.
  • #1
Laura0901

Homework Statement


Find the coefficient of kinetic friction that is used to stop a train engine along its track of the trains velocity is 30 m/s when the brakes are applied, and it takes 30 sec for the 10,000 kg train to stop

Homework Equations


Ffric=MN=Mmg
(I think that's the equation to use.

The Attempt at a Solution


Ffric=M(10000kg)(9.8m/s)
Ffric=M(98,000)
What do you use for the force? 30m/s?
 

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  • #2
Laura0901 said:

Homework Statement


Find the coefficient of kinetic friction that is used to stop a train engine along its track of the trains velocity is 30 m/s when the brakes are applied, and it takes 30 sec for the 10,000 kg train to stop

Homework Equations


Ffric=MN=Mmg
(I think that's the equation to use.

The Attempt at a Solution


Ffric=M(10000kg)(9.8m/s)
Ffric=M(98,000)
What do you use for the force? 30m/s?
Hello Laura0901. Welcome to PF !

What is the acceleration of the train ?
 
  • #3
It is not said in the problem.
 
  • #4
Laura0901 said:
It is not said in the problem.
Of course you must figure it out from the given information.
 
  • #5
And how would you do that? Using the kinematic equations?
 
  • #6
Why would we need to find the acceleration? I need the friction force.
 
  • #7
Laura0901 said:
Why would we need to find the acceleration? I need the friction force.
And, just what force is it that is slowing the train?
 
  • #8
Either gravity or velocity..I think.
 
  • #9
Laura0901 said:
Either gravity or velocity..I think.
The train slows to a stop. That's an acceleration. That requires a force. (Ask Mr. Newton.)

Gravity is a force, but it acts in a vertical, the train's motion and its acceleration are in the horizontal direction. Gravity is not the force which slows the train.

Velocity is not a force.
 
  • #10
1) Figure out the force in the x-direction that SammyS is talking about, the force that causes the train to stop.
2) Use kinematics to find a
3)Use Fx=max to find the magnitude of the force mentioned in step one
Then you should have enough data to solve Ffric=M(98,000)
 
  • #11
Thanks for trying to help me understand
 
  • #12
There aren't any forces in the x direction. I don't think length is a force. I don't know how to find friction.
 
  • #13
Laura0901 said:
There aren't any forces in the x direction. .
Then the train's going to keep going in the x direction.

By the way: What direction is the friction force ?
 
  • #14
The Positive direction
 
  • #15
Is it 2N?
 
  • #16
Laura0901 said:
The Positive direction

Positive in what sense?
Positive x-direction?
or
Positive as in upward?
 
  • #17
The x direction, it is horizontal
 
  • #18
Okay... I hope you are asking if fk is 2N because M is not in Newtons. I don't think fk=2N either though...
 
  • #19
stephen8686 said:
Okay... I hope you are asking if fk is 2N because M is not in Newtons. I don't think fk=2N either though...
Right. It would take very long time to stop.
 
  • #20
Ughh I'm so bad at everything sorry guys.
 
  • #21
I don't understand how you'd even get a number just because it's a horizontal force?
 
  • #23
I think I understand friction now thanks for the video.
 
  • #24
So is air resistance part of it?
 
  • #25
Laura0901 said:
So is air resistance part of it?
Probably not much. The friction force is stopping a train.

Calculate the acceleration as has been suggested previously.
 
  • #26
I got one by using v=at
 
  • #27
I'm so bad at this :/
 
  • #28
Air resistance is a totally different thing. You'll learn that later in physics. Until then, assume there is no air resistance. Kinetic friction is between two objects sliding on each other.
 
  • #29
I agree a=1 (or -1 depending on your positive x direction).
 
  • #30
Thank you for your time. If I get this problem wrong on the quiz, I'll get a 75%. I'll take that..
 
  • #31
Laura0901 said:
I got one by using v=at
Well what did you get?
 
  • #32
I used F=ma so: F=(10,000kg)(1)= 10,000
 
  • #33
Which I think is definitely wrong! I don't see how to use acceleration in the equation F=Mmg..
 
  • #34
Laura0901 said:
I used F=ma so: F=(10,000kg)(1)= 10,000
Why is that wrong? (Except of course you left the units off.) 10,000 N

Now use you "friction" equation: Ff = μN = μmg.
 
  • #35
10,000 N is the force?
 

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