Where v, a and s are what?george95 said:
No, that was the right assumption to make. But isn't this stated in the problem? Please post the entire question as given to you.george95 said:
That isn't quite right, you left out a factor of 2, and you should specify that this assumes acceleration is constant and there is zero initial velocity. But it is irrelevant here since we are taking speed as constant.george95 said:
That's closer to being relevant, but applies to an object moving up or down a ramp under the influence of gravity and nothing else - no friction, no engine.george95 said:
Nothing to be honest.
Yes, based on the question as stated, that is all you need to consider.george95 said:
Correct. (Spelled joules).george95 said:
I was not going for power just yet. Though that is where I was heading, yes.george95 said:
Provide values for those parameters and calculate how many joules in one second. Then we can make the easy jump to how much power that is.george95 said:
Good. Nicely summarized.george95 said:
If the 1 kilogram mass were moving horizontally on a sheet of ice, the change in altitude is zero meters. The required force is zero Newtons. It takes no work at all to slide a mass across a sheet of ice.george95 said:
Think about starting state and ending state. 50000 Newtons of weight is correct. But what is the change in altitude?george95 said:
Energy ##E## is equal to a force ##F## times a distance ##d##, so the simple answer to your question is: No, you cannot calculate the energy consumption with the given parameters, because you are missing the distance traveled.george95 said:
Why sheet of ice? Because we neglected the friction forces?jbriggs444 said:
If we use the length of the ramp to be 1m, then I guess the change should be:
Maybe better the change of altitude over one meter of track.
Should we increase or decrease the slope?
The one that counts is ##F_S##, i.e the force in the direction of the displacement.george95 said:Thank you for the explanation Jack.
The "F" force from your final formula has two components: "Fn" and "Fs" from this screenshot:
Or am I mistaken?
Yes, just so.george95 said:
Back a couple of posts you'd calculated using the cosine of 10 degrees. That would correspond to an 80 degree slope -- near vertical. Here you have corrected yourself to the sine of 10 degrees.george95 said:
jbriggs444 said:
As I suggested (editted into previous post), figure how much energy it takes to lift a 50000 N train over 0.18 meters of vertical displacement.george95 said:
You show a net force from the tracks that is slanted forward. That will not turn out to be correct. You stipulated that the train's speed is constant. It is not accelerating along the tracks. It is most certainly not accelerating up away from or down into the tracks. Accordingly there can be no net acceleration in any direction.george95 said:
The "net force" is Fn?
You already did it for 1 kg moving vertically through 1 meter.george95 said:
No. ##F_n## would typically denote the "normal" force.george95 said:
According to your video, ##F_S = mg\sin \theta## and that what is important to your train problem. It represents the portion of the gravity that is pulling your train downhill. Therefore, you need to produce an equivalent and opposite force (##F_S##) to counteract it. Multiplying that force with the train velocity, you get the power produced by the train.george95 said:
Thank you jbriggs.jbriggs444 said:
1194 kW equals 1194 kJ/s, so the energy consumption is 1194 kilojoules per second.george95 said:
Then the gravity is pushing the train downhill, so without any other energy input or braking, the train will accelerate (##a = \frac{F}{m} ##).george95 said:
When you go downhill with a bicycle, do you need to use the pedals (i.e. consuming energy)?george95 said:
No.
So if I assume that the train needs to keep a constant speed of 50 km/h -> 13.89 m/sec both uphill, and downhill - when going downhill an equal amount of energy needs to be consumed in order to prevent the train accelerating beyond 50km/h?
Yes.george95 said:
When you go uphill, you have to produce energy - hence, consuming it - and when you go downhill you need to store or dissipate energy. A typical brake system convert the energy into heat and dissipates it, but you can also store it (in a battery for example) and reuse it at a later time.george95 said:
A commonly used formula for estimating train energy consumption is the tractive effort formula, which takes into account the train's weight, speed, and grade of the track. It is expressed as: Tractive Effort (TE) = Train Weight (W) x Speed (V) x Grade (G).
The accuracy of the formula depends on various factors such as the accuracy of the input data and the assumptions made in the formula. In general, it provides a good estimate but may not be completely accurate in all cases.
The simple formula can be used for most types of trains, including diesel, electric, and hybrid trains. However, it may not be suitable for high-speed trains or trains with unique designs that deviate from the standard weight, speed, and grade parameters.
Yes, there are other factors that can affect train energy consumption, such as weather conditions, track conditions, and train maintenance. These factors may not be accounted for in the simple formula and can impact the accuracy of the estimate.
The simple formula can be improved by taking into account additional factors such as track curvature, train acceleration and deceleration, and regenerative braking. It can also be refined by using more specific data for each train journey, rather than relying on general weight, speed, and grade parameters.