[douglasm6 (OP)]

Consider motion of an object in a straight line.  The following equation applies:

Work done on object  (W) = Force (F) applied in the direction of travel * displacement (x)

Consider the object of mass 1000 kg initially at rest.  A constant force is applied.  For simplicity, assume that frictionless motion.

Initial velocity, v0 = 0. 
Acceleration, a,  is constant (constant force) = 1 m/s2

F = 1000 N



Consider first period of 10 seconds.

t1 = 10 seconds

x1 = 50 m


Work done in first 10 seconds, W1 = 1000 * 50 = 50 000 J.

v1 = 10 m/s

Kinetic energy = 0.5 * 1000 * 10^2 = 50 000 J.

So, all work done has been converted to kinetic energy.

================================

Consider second interval of 10 seconds.

t2 = 20 seconds

t2 - t1 = 10 s

Distance travelled during 2nd interval, x2 = 150 m

Work done in 2nd interval, W2 = 1000 * 150 = 150 000 J.

v2 = 20 m/s

Kinetic energy = 0.5 * 1000 * 20^2 = 200 000 J.

This equals W1 + W2.

=====================================

All very simple so far!

Now, this constant force was the constant thrust of a rocket engine.  Thus the W1 and W2 was done by the rocket.

Since thrust was constant, the same amount of fuel was consumed in the 1st 10 seconds as the 2nd interval of 10 seconds.  (A)

If the rocket's efficiency of converting fuel energy into work is assumed to be constant at 0.5.

Fuel energy in 1st interval = W1 / 0.5 = 50 000 / 0.5 = 100 000 J
 
Fuel energy in 2nd interval = W2 / 0.5 = 150 000 / 0.5 = 300 000 J

This contradicts (A) and common sense, which states that the rate of fuel burnt is constant!!!

What is your explanation of this paradox? []

Doug

[lyner]

There is no paradox really.
The (constant)  energy from combustion is shared between the KE of the propellant and the rocket.
In a given time, there is an equal increase in Momentum, not KE, of the rocket.
(Impulse = force times time = change in momentum)
If you start off at zero velocity the craft will end up with a velocity of
force  X time / mass
after a given interval.
The velocity of the craft will be twice this at the end of the second interval and the KE quadruples.
During the second interval, the propellant gets less KE.

A rocket engine is hugely inefficient at takeoff.

[douglasm6 (OP)]

Thanks for your reply!

If the source of the force is an electric motor (to get rid of those exhaust gases!), everything else being the same in the example given above, how does one explain that as the velocity eventually increases to a point where the tractive force, 1000 N, multiplied by the velocity is a very large number, exceeding the power rating of the motor.

(Originally, what gave rise to my question, was that I was doing some calculations about my train journey, and I could not get my calculation to work.  The train engines draw 100 A at 25 kV, power factor 0.8 to give a shaft power of 2000 kW.  I measured the acceleration using a stopwatch, found out the mass of the train and hence could calculate the tractive force.  The nett force was worked out by calculating the dynamic friction and air resistance, the coefficients of which are published for trains.  What got me thinking was that there is obviously an upper limit on the engine power, but that this was soon exceeded by the tractive force multiplied by the velocity  – dimensionally the same as the power – in my calculation.)

I am obviously missing some basics here!

Doug

[lyner]

An electric motor can't provide the same force at 'any speed'.
Power is force times speed so, if speed increases, the force must decrease. This, despite your measurements, has to cause your model to fail.
I suggest that there were practicalities which produced this apparent anomaly. The tractive force 'should' go down and down.
Of course, there is no theoretical maximum to the speed achieved (changing gear etc.) but the time taken gets longer and longer.

[douglasm6 (OP)]

Thanks! The cobwebs have been cleared.  Gosh what a blockhead I am!

Doug

[lyner]

Don't beat yourself up about it.
Physics is such a nit picking subject.