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

[hbk69]

Homework Statement

A 60,000 kg train is being pulled up a 1% gradient by an engine exerting 3 kN. The frictional force opposing the motion of the train is 4 kN. 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?

The answer is 275m

Homework Equations

Fpush/pull=mgsinθ (component along the ramp)+Ff (friction force)

I know that the equation above is used to solve the problem but i do not understand it or know how to use it

The Attempt at a Solution

s=ΔE/F

Vf=1/2mv^2=2430000
Vi=1/2mv^2=4320000

ΔE=-1890000

then i did not know how to work out the force to put it in the s=ΔE/F formula to work out the horizontal distance.

But i know my attempt is completely wrong, my major problem with friction problems is identifying which forces are acting on the object and the types of forces and then finding the right formula to work out the answer.

Any help very much appreciated, thank you.

 

[tiny-tim]

hi hbk69!

you need to find the total work done, including by gravity (per distance)

1% gradient means tan = .01, but it's so small i think you're entitled to assume sin = .01 also

 

[Chestermiller]

Let d be the distance that the train travels before its speed is reduced to 9 m/s. What is the net force acting on the train in the direction down the grade during the time it is traveling that distance. In terms of d, how much work do these forces do on the train while it is traveling this distance? What is the change in kinetic energy of the train? How is the change in kinetic energy related to the work done by the external forces?

 

[hbk69]

hi hbk69!

you need to find the total work done, including by gravity (per distance)

1% gradient means tan = .01, but it's so small i think you're entitled to assume sin = .01 also

Hi! thanks for the response. Workdone=F*s, and the s is the horizontal distance which i am looking for? but how can i work out F?

 

[hbk69]

Let d be the distance that the train travels before its speed is reduced to 9 m/s. What is the net force acting on the train in the direction down the grade during the time it is traveling that distance. In terms of d, how much work do these forces do on the train while it is traveling this distance? What is the change in kinetic energy of the train? How is the change in kinetic energy related to the work done by the external forces?

thanks for the reply, how can i determine the net force down the gradient? which i think would help me work out the work done?

And this was the change in kinetic energy i calculated:

Vf=1/2mv^2=2430000
Vi=1/2mv^2=4320000

ΔE=-1890000

i am not sure how it relates to the work done or whether i worked it out correctly, as i don't know how to determine the external forces/net force

 

[tiny-tim]

hi hbk69!

A 60,000 kg train is being pulled up a 1% gradient by an engine exerting 3 kN. The frictional force opposing the motion of the train is 4 kN.

Workdone=F*s, and the s is the horizontal distance which i am looking for?

not really

work done = F.s

s is the displacement …

it's a vector, and (in this case) it's along the slope,

ie it's one unit up for every 100 units horizontally​

F is the total force (from the engine, the friction, and the weight)

but how can i work out F?

F is always the total force (the net force).

along the slope ("horizontally"), it's 3000 N forwards, and 4000 N backwards

vertically, it's mg

 

[Chestermiller]

thanks for the reply, how can i determine the net force down the gradient? which i think would help me work out the work done?

And this was the change in kinetic energy i calculated:

Vf=1/2mv^2=2430000
Vi=1/2mv^2=4320000

ΔE=-1890000

i am not sure how it relates to the work done or whether i worked it out correctly, as i don't know how to determine the external forces/net force

Do a free body diagram on the engine and draw vectors for all the forces acting on the train, in the directions both parallel and perpendicular to the grade. Tiny Tim pretty much laid it out for you, but forgot to mention the component of the weight 0.01 mg acting parallel to the grade.

Recall that the work done on the train by the external forces is equal to its change in kinetic energy. This gives you an equation for calculating s.

Chet

 

[hbk69]

hi hbk69!



not really

work done = F.s

s is the displacement …

it's a vector, and (in this case) it's along the slope,

ie it's one unit up for every 100 units horizontally​

F is the total force (from the engine, the friction, and the weight)


F is always the total force (the net force).

along the slope ("horizontally"), it's 3000 N forwards, and 4000 N backwards

vertically, it's mg

How can the horizontal distance be found mathematically also how could i find the net force mathematically?

If 3000 is forward and 4000 is backwards, would the net force be 4000*mg(vertical component) - 3000*mg

 

[tiny-tim]

How can the horizontal distance be found mathematically also how could i find the net force mathematically?

If 3000 is forward and 4000 is backwards, would the net force be 4000*mg(vertical component) - 3000*mg

nope, you haven't got this at all

let's go over it slowly …

(let's assume for the moment that the train goes a displacement of s = 1 metre forwards along the slope)

there's three forces, and they're all vectors

it's convenient to deal with the weight separately from the other two

the weight of course is F_g = 60,000g downwards

what is the work done (F_g·s) ?​

the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

what is the work done by those forces?​

 

[Chestermiller]

nope, you haven't got this at all

let's go over it slowly …

(let's assume for the moment that the train goes a displacement of s = 1 metre forwards along the slope)

there's three forces, and they're all vectors

it's convenient to deal with the weight separately from the other two

the weight of course is F_g = 60,000g downwards

what is the work done (F_g·s) ?​

the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

what is the work done by those forces?​

Tiny Tim,

It seems you again forgot to include the component of the train weight parallel to the grade:
mg(0.01)=6000 N backwards.

Chet

 

[hbk69]

nope, you haven't got this at all

let's go over it slowly …

(let's assume for the moment that the train goes a displacement of s = 1 metre forwards along the slope)

there's three forces, and they're all vectors

it's convenient to deal with the weight separately from the other two

the weight of course is F_g = 60,000g downwards

what is the work done (F_g·s) ?​

the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

what is the work done by those forces?​

Sorry, i have a very very weak understanding of physics

So in this case would i add the Fnet to the Fg downwards? when s=1m, W= (1000+60,000) *1 to get the answer?

 

[tiny-tim]

So in this case would i add the Fnet to the Fg downwards? when s=1m, W= (1000+60,000) *1 to get the answer?

no, you can't add vectors like that

you can't add 60,000 in one direction to 1000 in a different direction to get 61,000, can you?

you have to add them as vectors, ie either use a vector triangle, or use components (eg x and y), and add the components, separately

in this case, adding the vectors (as vectors) is a bit awkward (since they're not exactly perpendicular), so i recommended that instead of adding the vectors (as vectors), and then "dotting" with s, you "dot" each vector with s first, and then add the results (which of course are scalars)

ok, now answer my previous questions (assuming for the moment that s = 1 m) …

the weight of course is F_g = 60,000g downwards

what is the work done (F_g·s) ?​

the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

what is the work done by those forces?​

Tiny Tim,

It seems you again forgot to include the component of the train weight parallel to the grade:
mg(0.01)=6000 N backwards.

Chet

no, this says it all …

the weight of course is F_g = 60,000g downwards

 

[hbk69]

no, you can't add vectors like that

you can't add 60,000 in one direction to 1000 in a different direction to get 61,000, can you?

you have to add them as vectors, ie either use a vector triangle, or use components (eg x and y), and add the components, separately

in this case, adding the vectors (as vectors) is a bit awkward (since they're not exactly perpendicular), so i recommended that instead of adding the vectors (as vectors), and then "dotting" with s, you "dot" each vector with s first, and then add the results (which of course are scalars)

ok, now answer my previous questions (assuming for the moment that s = 1 m) …

the weight of course is F_g = 60,000g downwards

what is the work done (F_g·s) ?​

the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

what is the work done by those forces?​

no, this says it all …

lol sorry still confused, not sure how i can add the vectors in that case because as they are not perpendicular i can't use pythogras theorem either where i square both of them add the result and take the square root to get Fnet, and what do you mean by "dot each vector with s" don't know how to to it.

and answering that question, you said s=1m, Fg=60,000 so the Work=Fg*s= 60,000J? it's wrong again isn't it.

 

[tiny-tim]

and answering that question, you said s=1m, Fg=60,000 so the Work=Fg*s= 60,000J? it's wrong again isn't it.

no, is said that the displacement vector s is 1 m along the slope

since you don't know how to do a dot product, yes it's wrong

… what do you mean by "dot each vector with s" don't know how to to it.

for dot product, see eg http://en.wikipedia.org/wiki/Dot_product

i don't see how you can understand work done if you don't understand the dot product

do you have a ta or a tutor or a study-buddy who can help you with this?

if not, i think you ought to ask your professor to go over it with you

lol sorry still confused, not sure how i can add the vectors in that case because as they are not perpendicular i can't use pythogras theorem either where i square both of them add the result and take the square root to get Fnet, and

you have to add the x and y components of the vectors (separately)

have you not done addition of vectors?

 

[hbk69]

I have done it in maths but don't know how to apply it to physics problems takes a bit longer before i understand the maths side of things, don't have a tutor re-taking a physics module without attendance. Thanks for trying to help me

 

[tiny-tim]

ok, let's see …

can you add the vectors (4,6,0) and (1,1,3) ?

 

[hbk69]

ok, let's see …

can you add the vectors (4,6,0) and (1,1,3) ?

(5,7,3)

 

[tiny-tim]

that's right

now when you want to add two vectors, you split each vector into its x y and z components, and then you add the components (exactly like that example)

that's called the component method for adding vectors (you can also use the vector triangle method)

ok, now let's do dot product

(dot product is the same as inner product, but most people call it dot product because you write it with a dot, unlike the cross product!)

can you do the dot product of the vectors (4,5,6) and (2,3,4) ?

 

[hbk69]

that's right

now when you want to add two vectors, you split each vector into its x y and z components, and then you add the components (exactly like that example)

that's called the component method for adding vectors (you can also use the vector triangle method)

ok, now let's do dot product

(dot product is the same as inner product, but most people call it dot product because you write it with a dot, unlike the cross product!)

can you do the dot product of the vectors (4,5,6) and (2,3,4) ?

but sometimes you may have to substract them and involve angles etc and that is where it gets really complicated for me with friction problems involving inclined planes and finding the right formulas to solve the problem when it come to friction

oohhh is that what the dot product its like one of those where you know someone by face but not their name, i think i can do that but applying it to physics is the problem

(4)(2),(5)(3),(6)(4)

(8,15,24)

 

[haruspex]

oohhh is that what the dot product its like one of those where you know someone by face but not their name, i think i can do that but applying it to physics is the problem

(4)(2),(5)(3),(6)(4)

(8,15,24)

Almost... but the result is not a vector. You add up the products to get a scalar: 8+15+24.

 

[hbk69]

Almost... but the result is not a vector. You add up the products to get a scalar: 8+15+24.

Forgot to add the results

 

[tiny-tim]

now when you want to add two vectors, you split each vector into its x y and z components, and then you add the components (exactly like that example)

that's called the component method for adding vectors (you can also use the vector triangle method)

but sometimes you may have to substract them …

if you need to subtract two vectors, you just subtract the components, eg (4,6,0) - (1,1,3) = (3,5,-3)

(essentially, you're adding minus the second vector to the first vector)

… and involve angles etc and that is where it gets really complicated for me with friction problems involving inclined planes and finding the right formulas to solve the problem when it come to friction

if there are angles (as here), you have to start by splitting the vector into its components

in exam questions, it's nearly always a 2D situation (as here) …

you split a vector along the slope, of magnitude s,

into (scosθ, ssinθ) where θ is the angle to the x direction (the horizontal)

that is always the formula … you should learn it!

(in this case, because θ is very small, you can take cosθ = 1 and sinθ = .01)

oohhh is that what the dot product its like one of those where you know someone by face but not their name, i think i can do that but applying it to physics is the problem

(4)(2),(5)(3),(6)(4)

(8,15,24)

Forgot to add the results

yes it's 8 + 15 + 24 = 47

ok, apply that to the dot product F_g·s

F_g is (0, -60,000g) and s is (1cosθ, 1sinθ) = (1, 0.01)

(and what about the dot product of the other two forces with s ?)

 

[hbk69]

if you need to subtract two vectors, you just subtract the components, eg (4,6,0) - (1,1,3) = (3,5,-3)

(essentially, you're adding minus the second vector to the first vector)if there are angles (as here), you have to start by splitting the vector into its components

in exam questions, it's nearly always a 2D situation (as here) …

you split a vector along the slope, of magnitude s,

into (scosθ, ssinθ) where θ is the angle to the x direction (the horizontal)

that is always the formula … you should learn it!

(in this case, because θ is very small, you can take cosθ = 1 and sinθ = .01)
yes it's 8 + 15 + 24 = 47

ok, apply that to the dot product F_g·s

F_g is (0, -60,000g) and s is (1cosθ, 1sinθ) = (1, 0.01)

(and what about the dot product of the other two forces with s ?)

How would i know if i substract or add the vectors in a specific problem?What is the easiest way to identify the components? because then that way i could split them into their x and y components before adding/substracting them as i don't know how to split themwhich formula do you mean, the scosθ (x) , ssinθ (y) ? and when you say cosθ=1 did that come from the 1% gradient? because for sin the angle is 0.01? but we don't actually take the cos or sin of 0.01? but why?

And attempting the question:

Fg*s

(0)(1) + (-60,000)(0.01) = -600 J, which is work done to pull the train?

and for the two other forces, i am unsure how i resolve them into their components but i will give it a try:

(4000, -60,000) (1 , 0.01)

(3000, -60,000) (1, 0.01)

is that correct? i feel its wrong.

Thanks

 

[tiny-tim]

How would i know if i substract or add the vectors in a specific problem?

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

(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?

What is the easiest way to identify the components? because then that way i could split them into their x and y components before adding/substracting them as i don't know how to split them

you split them into components by multiplying by cosθ or sinθ

which formula do you mean, the scosθ (x) , ssinθ (y) ? and when you say cosθ=1 did that come from the 1% gradient? because for sin the angle is 0.01? but we don't actually take the cos or sin of 0.01? but why?

you should really use (scosθ, ssinθ),

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

And attempting the question:

Fg*s

(0)(1) + (-60,000)(0.01) = -600 J, which is work done to pull the train

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

and for the two other forces, i am unsure how i resolve them into their components but i will give it a try:

(4000, -60,000) (1 , 0.01)

(3000, -60,000) (1, 0.01)

is that correct?

no

(why are you using 60,000 ? )

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?

 

[Chestermiller]

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,
$$7000d=\left(\frac{60000}{2}\right)12^2-\left(\frac{60000}{2}\right)9^2$$
So, solve for d.
Chet

 

[hbk69]

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

(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 sinθyou should really use (scosθ, ssinθ),

but cosθ is so close to 1 that you may as well call it 1that's correct … the work done by gravity is minus 600 J per metre of displacementno

(why are you using 60,000 ? )

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?

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?

 

[hbk69]

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,
$$7000d=\left(\frac{60000}{2}\right)12^2-\left(\frac{60000}{2}\right)9^2$$
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)

 

[haruspex]

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?

 

[Chestermiller]

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

 

[hbk69]

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

 

[hbk69]

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

 

[haruspex]

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.

 

[hbk69]

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.

 

[haruspex]

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.

 

[Chestermiller]

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