Recent content by gary32

Would the reason for this be to achieve the target value faster where an overshoot doesn't matter all too much? If not then what would the reason be? Thanks everyone so far :)

The circuit isn't very clear but if there's 2 power sources then Kirchhoff's voltage law and the use of simultaneous equations would be helpful

I would say what they don't have in common, they appear to do the opposite, Look how an inductor (solenoid) affects current/voltage and how a capacitor affects current/voltage, should be relatively straight forward. That's one thing you could look at anyway. Is that the full question?

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?

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

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.

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

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...

You've lost me now :confused: Please give me a clue or example,

I calculated tractive resistance force as F = 80000 x 0.07 = 5600N That is just a force, not work done? I know work done is Force x Distance, but I don't have an actual distance other than the one I calculate in the end (and tractive resistance needs to be implemented before you work out the...

Initial energy + Work energy = Final total energy? What is Newtons per kg physically? to me, Newtons is a force, as in weight, as in, mass x gravity and then kg is a mass so to me i see it as mass x gravity / mass which is just gravity so its confusing. How can I formulate N/Kg into work...

I think I've got it, KE = 0.5 x Mass x Velocity2 KE = 15054400J So that is the amount of energy dissipated when it comes to a stop so height = energy / m x g height = 15054400 / 80000 x 9.81 = 19.18m using trig, hyp (distance) = opp (height) / sin(8) distance = 19.18/sin(8) = 137.8m...

Potential energy = mass x gravity x distance? So, Potential energy = 80000 x 9.81 x ? Energy where train stops is Potential energy = 80000 x 9.81 x ? Sin8 probably goes into either of them formulas some how and then transpose them both to find distances?

Please elaborate, I was supposed to be working the problem out using Conservation of energy (which now looking at it, I have not done silly me I can't remember if I mentioned. What would be the difference in calculations and what method have I used?) After calculating my answer using...

Mass is measured in Kilograms (SI unit) Weight is measured in Newtons (SI unit) 1kilogram is 2.2lbs, Lbs (pounds) is another unit of measuring weight