The horizontal distance the train will move before its speed reduces?

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  • #26
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i honestly don't see the difficulty … surely the question will tell you whether to add or subtract? :confused:

(almost always you add: in this question, the 3000 N is forwards, and the 4000 N is backwards: when you add them, you get 1000 N backwards)

can you give an example of a question where you've been confused?


you split them into components by multiplying by cosθ or sinθ


you should really use (scosθ, ssinθ),

but cosθ is so close to 1 that you may as well call it 1


that's correct … the work done by gravity is minus 600 J per metre of displacement


no

(why are you using 60,000 ? :confused:)

the 4000 and the 3000 are in exactly the same direction as the displacement s, so you don't need to split anything into components, do you? :wink:

Hi chet,

I had posted a threat not so long ago with similar friction problems but never understood the help given, i wanted information to help my understanding to solve friction problems generally in a variety of situations. This was the thread, take a look:

https://www.physicsforums.com/showthread.php?t=718970

I managed to solve the questions by getting help from here and there but my understanding of the physics is still weak, understanding this physics and then knowing what formula to use automatically having read the situation correctly which i find most challenging.

In regards to using scosθ, ssinθ, is the s the distance/displacement? and are scosθ and ssinθ forces as a whole acting on the train? if so how can one know what direction they act on the train? i assume the sin is the horizontal along the ramp? and the cos component vertical acting downwards which would be mg while the sin component would be the net force?

Since you say 4,000 and 3,000 in same direction then it would be (1,000) (1, 0.01) is that correct?
 
  • #27
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The net force on the train the the direction parallel to the incline is F = (3000-4000-6000)=-7000N. If d is the distance in meters over which the force is applied, the loss in kinetic energy of the train is 7000d. So,
[tex]7000d=\left(\frac{60000}{2}\right)12^2-\left(\frac{60000}{2}\right)9^2[/tex]
So, solve for d.
Chet

Got to understand the first bit before getting to the final solution, but in this case could you show me where you derived the formula/variables involved to work out the answer please as i do not know what is going on and how you came to that solution.

Is the left hand side of the equation 7000d the work? but what is the right hand side of the equation? why are you dividing 6000 the Fg by 2 and mutiplying it by the square of the velocity before subtracting the output by 6000/2*g (not sure where that came from either)
 
  • #28
haruspex
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Is the left hand side of the equation 7000d the work?
Yes.
but what is the right hand side of the equation?
The decrease in kinetic energy, ##\frac 12 mv_0^2 - \frac 12 m v_1^2##.
why are you dividing 6000 the Fg by 2 and mutiplying it by the square of the velocity before subtracting the output by 6000/2*g (not sure where that came from either)
It's 60,000, not 6000, and its the mass of the train in kg, not its weight (Fg). Do you understand the difference between mass and weight?
 
  • #29
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Hi chet,

I had posted a threat not so long ago with similar friction problems but never understood the help given, i wanted information to help my understanding to solve friction problems generally in a variety of situations. This was the thread, take a look:

https://www.physicsforums.com/showthread.php?t=718970

I managed to solve the questions by getting help from here and there but my understanding of the physics is still weak, understanding this physics and then knowing what formula to use automatically having read the situation correctly which i find most challenging.

In regards to using scosθ, ssinθ, is the s the distance/displacement? and are scosθ and ssinθ forces as a whole acting on the train? if so how can one know what direction they act on the train? i assume the sin is the horizontal along the ramp? and the cos component vertical acting downwards which would be mg while the sin component would be the net force?

Since you say 4,000 and 3,000 in same direction then it would be (1,000) (1, 0.01) is that correct?
Tiny Tim and I discussed two different, but equivalent, methods of getting the same answer. Tim separated out the change in potential energy of the train from the work done by the engine and frictional force. In this approach, it is not necessary to include the component of the gravitational force acting along the grade. The potential energy change takes care of that.

In the approach that I discussed, I resolved the gravitational force into components perpendicular and parallel to the incline, and included the parallel component of weight explicitly in calculating the work done against external forces as the train moves along the incline. The work done against external forces is equal to the decrease in kinetic energy of the engine.

The method I gave and the method Tiny Tim gave are entirely equivalent, and lead to the exact same energy balance equation and the same distance traveled by the train.

Chet
 
  • #30
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Yes.
The decrease in kinetic energy, ##\frac 12 mv_0^2 - \frac 12 m v_1^2##.

It's 60,000, not 6000, and its the mass of the train in kg, not its weight (Fg). Do you understand the difference between mass and weight?

Yes that was a mistake regarding the mass/weight,Mass is the same everywhere while weigh depends on effects of gravity.

As you are using the ΔKE are yout not supposed to involve the final speed - the initial speed rather then Vo-V1? because in the question they say The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s. Why does it have to be Vo (12)^2 - V1=(9.8)^2", where 12m/s is the initial speed and the g is the acceleration due to gravity
 
  • #31
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Tiny Tim and I discussed two different, but equivalent, methods of getting the same answer. Tim separated out the change in potential energy of the train from the work done by the engine and frictional force. In this approach, it is not necessary to include the component of the gravitational force acting along the grade. The potential energy change takes care of that.

In the approach that I discussed, I resolved the gravitational force into components perpendicular and parallel to the incline, and included the parallel component of weight explicitly in calculating the work done against external forces as the train moves along the incline. The work done against external forces is equal to the decrease in kinetic energy of the engine.

The method I gave and the method Tiny Tim gave are entirely equivalent, and lead to the exact same energy balance equation and the same distance traveled by the train.

Chet

PE? i thought it was ΔKE.

The energy method seems easier to grasp, although your method could probably be applied to most friction scenario? which means will need to understand it better
 
  • #32
haruspex
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As you are using the ΔKE are you not supposed to involve the final speed - the initial speed rather then Vo-V1? because in the question they say The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s./QUOTE]
Final KE - initial KE gives you the gain in KE, which would of course be negative, so you'd write Final KE - initial KE = -7000d. Or you can write 7000d = loss of KE = initial - final.
 
  • #33
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As you are using the ΔKE are you not supposed to involve the final speed - the initial speed rather then Vo-V1? because in the question they say The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s./QUOTE]
Final KE - initial KE gives you the gain in KE, which would of course be negative, so you'd write Final KE - initial KE = -7000d. Or you can write 7000d = loss of KE = initial - final.

thanks understood, it was the math.

But i don't understand the way in which you got -7000, F = (3000-4000-6000) which formula are you using when you try and get the net force and why do you substract everything what is thetheory behind it?, also what is the 6000 value? where did it come from.
 
  • #34
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But i don't understand the way in which you got -7000, F = (3000-4000-6000) which formula are you using when you try and get the net force and why do you substract everything what is thetheory behind it?, also what is the 6000 value? where did it come from.
You have 3000N attempting to accelerate the train, 4000N of resistance tending to slow it down, and the component of gravity parallel to the track is also slowing it down. The gravity component is Mg*1%, where M is the mass of the train. Taking g = 10N/kg gives the 6000 result.
 
  • #35
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You have 3000N attempting to accelerate the train, 4000N of resistance tending to slow it down, and the component of gravity parallel to the track is also slowing it down. The gravity component is Mg*1%, where M is the mass of the train. Taking g = 10N/kg gives the 6000 result.

Very well said. I don't see how it could possibly be explained any better.

Chet
 
  • #36
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You have 3000N attempting to accelerate the train, 4000N of resistance tending to slow it down, and the component of gravity parallel to the track is also slowing it down. The gravity component is Mg*1%, where M is the mass of the train. Taking g = 10N/kg gives the 6000 result.

thanks, i understand the math of it. But with the physics in order to get the net force you substract all the forces acting in various directions from the force which is accelerating the train?
 
  • #37
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thanks, i understand the math of it. But with the physics in order to get the net force you substract all the forces acting in various directions from the force which is accelerating the train?
Not 'in various directions'. You can only add and subtract magnitudes of forces that are acting parallel. These are all parallel to the slope of the track. When you stand still, gravity applies a force to you pushing you down, and the floor applies an equal and opposite force pushing you up. Subtracting the one from the other gives you 0, so no net force, so no acceleration. It's the same principle here, except that you do end up with a nonzero net force.
(You can add and subtract forces that are not parallel, but that's done using vector addition, which is different.)
 
  • #38
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Not 'in various directions'. You can only add and subtract magnitudes of forces that are acting parallel. These are all parallel to the slope of the track. When you stand still, gravity applies a force to you pushing you down, and the floor applies an equal and opposite force pushing you up. Subtracting the one from the other gives you 0, so no net force, so no acceleration. It's the same principle here, except that you do end up with a nonzero net force.
(You can add and subtract forces that are not parallel, but that's done using vector addition, which is different.)

How do we know if a force is a magnitude? in this case did you calculate the magnitude? if so how
 
  • #39
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How do we know if a force is a magnitude?
A force is not a magnitude. There are two aspects to a force, its magnitude (how big it is) and its direction. Every force has a magnitude.
If two forces are in the same direction you can add the forces simply by adding their magnitudes. If they are in opposite directions you can subtract the magnitudes. If they not parallel then its more complicated.
in this case did you calculate the magnitude? if so how
We've already been through that. The three magnitudes were 3000, 4000 and 6000. The 3000 was in one direction, the 4000 and 6000 in the completely opposite direction. So we combine them by adding the two in the same direction and subtracting the one in the other direction.
 
  • #40
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thanks a million to everyone who has helped me
 

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