Questions about locomotive start up questions Come on! I can add ten points to be able to explain c

Your answer is wrong.

The car is accelerated at rated power, and the traction force f satisfies f = p v, which is not a constant value.

It cannot be combined with f.

It is not good to think of him as an average, the first formula f is the average value of t, and the second formula f is the average value of s. These two are different.

Averages are a vague concept, and he averages different objects with different results.

There is an object m, velocity v1, another object m, velocity v2;

We usually say that their average velocity v=(v1+v2) 2 is used to calculate the conservation of momentum, and he is correct: 2mv=mv1+mv2 holds.

Used to calculate the conservation of kinetic energy, it is wrong: it is not true.

The reason for this is that the average velocity v of the establishment of the conservation of momentum is the average of v once.

And the average velocity v to make the conservation of kinetic energy establish is the average of v (e.g. the average velocity v = v1 + v2 is correct).

I hope it helps you, if you have any questions, please ask or hi to chat about o( o haha

The f-conjunction in f-junction t=mvt-mv0 is the average value of the variable force versus time, and the f-junction's = 1 2 * mvt 2-1 2 * mv0 2 in f'is the average value of the variable force to the displacement, and the f-combination is not equal to the f-combination'！

This question is actually a difficult problem in high school physics, but it is not the point.

The displacement of variable acceleration motion is easy to find, p=f*v, v can be found, and the acceleration can also be found, 2as v square, the whole displacement s can be thrown away s.

As for time, it actually tends to infinity.

Because p=f*v, and f-g=ma, the larger the v, the smaller f is, resulting in a smaller a. The slower the growth of V, the time of this process is infinite, and I don't know if it's clear.

The first process.

a = (f-g)/m

vt1 = p/f

t = vt1/a

s1 = 1/2 * a * t)^2

The second process. s2 = s - s1

vi2 = vt1

s2 is equal to v + at in the integral of t from 0 to t2 where a=(f2-g) m , f2=p v solves the system of equations, and t2 is the time of process two.

When the power of the car has reached the maximum, and the resistance f is equal to the traction force, the speed reaches the maximum, so it is obtained by p=fv: f=p vm=4000n when the car does a uniform acceleration movement, the traction force is always greater than the resistance, so the maximum speed of the uniform acceleration process is less than the maximum speed that the car can reach.

The magnitude of the traction force in the process of uniform acceleration is calculated by Newton's second law: f-f=ma solution f=8000n

The maximum speed that can be achieved by the uniform acceleration process: VM'=p f=10m s, so the time is: t=vm'/a=5s

3s unvelocity, v=at=6m s

Resistance instantaneous power P resistance = FV = 24000W

The displacement of the car in this process: l=

From the kinetic energy theorem: w-fl='2

It can be calculated that the work done by traction force w = 200000w

Since the first question you have made, you have already found the size of the car resistance, the whole process only traction and resistance to the car to do work, traction to do work: pt = 8800 * 10 3w * 550s resistance to do negative work, the size is: f*s

The kinetic energy of the train changes as follows: du=1 2m*v 2=kg*(100ms) 2

According to du=pt-fs, the distance s can be solved

Can you be more specific? The question is too abstract.

When p is constant, it is obtained by p=fv, and with the increase of speed, the traction force f decreases; And the net force acting on the object f=ma, so the acceleration a also decreases. When the acceleration a is constant, it can be seen that the net force f acting on the object is unchanged, and when the influence of drag is not considered, the power of the locomotive also increases with the increase of speed.

No, the f here refers to the forward power received by the car, and the ground has a backward friction force on the car, so the resultant force should be traction minus power.

When starting at rated power, p=fv, the speed of the car is very low at start, so a lot of traction can be obtained, and then as the speed increases, the traction gradually decreases.

No, the resultant force is the sum of the traction force and the resistance experienced by the locomotive.

.The first question relates to force analysis.

Drive Force * Speed = Power.

Since at the beginning, the speed was only 10 meters per second, which is a driving force of 6,000 N, so it can drive forward (you need to be subjected to force analysis, although the horizontal direction is only two forces, but the basic work of physics is force analysis, accurate analysis: horizontal, driving force and resistance vertical: gravity and support force).

6000 N is greater than 2500 Nm, so, the car accelerates, but due to the constant power, the speed increases, the driving force decreases (but still greater than the resistance), when the speed increases to this point, the horizontal direction is balanced by two forces, so the speed no longer increases (so, at this point it is the maximum speed): driving force = resistance.

Maximum speed = constant power Resistance = 60000 watts 2500 Nn = 24 meters per second.

2.The second question is to use the kinetic energy theorem, the kinetic energy theorem: the difference in kinetic energy (later kinetic energy - initial kinetic energy) = the total work done during this period (including positive and negative work).

m * v 2 - m * v0 2 = p*t - resistance * distance.

m = 5000 kg.

v=24 m/s v0=10 m/s.

p = 60000 watts t = 72 seconds.

Resistance = 2500 Bull.

I took the data in and did the math, and the distance was 1252 meters.

Good luck with your studies!

Hope it helps.

And adopt me as.

Thank you!

In fact, there are nothing more than two cases, uniform acceleration start, uniform power start, both grasp the constant quantitative equation on the line, uniform power start, engine power p=fv, is the same car, continuous acceleration, therefore, the traction force f of the engine will continue to decrease, when it is small to balance with the friction, the car can no longer accelerate, the second case, uniform acceleration start, f-f is unchanged, so the car keeps accelerating, the engine power continues to increase, when the maximum power of the engine is reached, it can not accelerate, So, first lz just figure out what is constant, and then.

Second, when can I stop accelerating, just make sure that the landlord exam is okay.

One is to accelerate with a certain acceleration, i.e., f is certain, v keeps increasing until fv=p, and then the acceleration slowly decreases until f=f (friction) and finally fv=p anyway

The second is to drive at maximum power, always maintaining fv=p, but during the period f keeps decreasing, v keeps increasing, and finally fv=p

Just revolve around f=am and fv=p.

That analyzes constant acceleration start and constant power start.