Distance travelled whilst decelerating

  • Thread starter gary32
  • Start date
In summary: I seemed to have made a mistake in the calculations, I am struggling as I am not sure what formula to use to calculate the distance, I can't find a formula that seems to contain the parameters I have.I also understand that there are multiple forces acting on the train, including gravity which I didn't take into account before. In summary, the train of 80 tonnes moves at a velocity of 70 km/hr up a gradient of 8 degrees. When the power is turned off, the train coasts up the gradient and comes to a rest. The tractive resistance to motion is 70 Newtons per tonne. To calculate the distance moved by the train up the gradient before coming to rest, the formula F=
  • #36
gary32 said:
I calculated tractive resistance force as F = 80000 x 0.07 = 5600N
That is just a force, not work done?

Correct.

I know work done is Force x Distance

Correct, too.

but I don't have an actual distance other than the one I calculate in the end

So make it some unknown variable, and solve for it.
 
Physics news on Phys.org
  • #37
voko said:
So make it some unknown variable, and solve for it.

You've lost me now :confused:

Please give me a clue or example,
 
  • #38
If force is F, and displacement is s, what is the work of the force?

If the displacement is s along the gradient angle a, what is the elevation in the end? And what is the corresponding potential energy?
 
  • #39
voko said:
If force is F, and displacement is s, what is the work of the force?

If the displacement is s along the gradient angle a, what is the elevation in the end? And what is the corresponding potential energy?

Work of the force is F x displacement?

If displacement is s along the gradient a,
The end elevation is Sin8 x s
Corresponding potential energy is 15054400J?

I still am lost, honestly, I am probably as dumb as it gets when it comes to mechanics, you need to dumb things down to my level :frown:
 
  • #40
gary32 said:
Work of the force is F x displacement?

Use the symbol ##s## for displacement.

If displacement is s along the gradient a,
The end elevation is Sin8 x s
Corresponding potential energy is 15054400J?

This cannot be a number, because the end elevation is ## s \cdot \sin a ##, where ##s## is unknown. So you have to write that down as a formula.
 
  • #41
voko said:
Use the symbol ##s## for displacement.



This cannot be a number, because the end elevation is ## s \cdot \sin a ##, where ##s## is unknown. So you have to write that down as a formula.

So what now? Also, what was the principle I used to initially work the problem out?
 
  • #42
gary32 said:
So what now?

Write the work-energy equation in terms of known and unknown variables, and solve it.

Also, what was the principle I used to initially work the problem out?

Not sure whether there is a particular name to it. FBD analysis and first principles of motion, perhaps.
 
  • #43
voko said:
Write the work-energy equation in terms of known and unknown variables, and solve it.

Not sure whether there is a particular name to it. FBD analysis and first principles of motion, perhaps.

There are 2 unknown variables (from what I can see) so I can't see a way for solving it,
I will have to not submit this work and not get the mark as I am away until the day it needs submitting.

Thanks for everyones help, it has been appreciated.
 
  • #44
Which two unknowns do you think you have here?
 
  • #45
voko said:
Which two unknowns do you think you have here?

Work done is unknown because that's what I am trying to find and s appears to be unknown, i know you've said s sin but that means nothing to me. Of course this is solvable but I don't have a clue how
 
  • #46
Work depends on s. Just write that down as a formula. Then plug that formula into the energy-work equation. That will eliminate the work as an unknown.
 
  • #47
voko said:
Work depends on s. Just write that down as a formula. Then plug that formula into the energy-work equation. That will eliminate the work as an unknown.

Work = F x s
Energy-work equation = F d = (1/2 mv2) - (1/2 mv2)

Energy-work equation = F d = (1/2 x 80000 x 02) - (1/2 x 80000 x 19.42)

= -15054400J

d = W/F
d = -15054400J/what force?
 
  • #48
gary32 said:
Work = F x s
Energy-work equation = F d = (1/2 mv2) - (1/2 mv2)

This is not what you had in #33.
 

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • General Math
Replies
6
Views
699
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
6
Views
973
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
Back
Top