Kinetic Energy Problem???

[Mat_D]

You are sitting at the side of a railway track, armed with a laser pointer velocity measuring gun.  To accurately measure the velocity of any moving object, you just point the laser at it and pull the trigger.

A train comes past and you spot a friend of yours sitting in a seat.  You point the laser at him and measure his velocity.  It tells you it is 10m/s.  You conclude that the train is travelling at 10m/s. (1)

Now, the train track is a large loop and soon the train comes past again at the same velocity, but this time your friend is running down the train in the direction of motion.  You measure his velocity and you get a reading of 12m/s. You conclude that the train is travelling at 10m/s and he is running at 2m/s down the train. (2)

You then work out his Kinetic Energy in both cases (1) and (2)

KE = � * Mass * Velocity Squared.

(You know your friend weighs 100 kg)

(1) When he is sitting;

KE = � * 100 * (10 * 10) = 5000 Joules

(2) When he is running;

KE = � * 100 * (12 * 12) = 7200 Joules

Therefore by running he has increased his kinetic energy by 2200 Joules. (7200 � 5000)

Now, imagine the same thing happens but this time the train is travelling twice as fast, at 20m/s.

(3) When he is sitting;

KE = � * 100 * (20 * 20) = 20000 Joules

(4) When he is running at 2m/s (combined velocity is 22m/s);

KE = � * 100 * (22 * 22) = 24200 Joules

This time, by running, he has increased his kinetic energy by 4200 Joules. (24200 � 20000)

This would suggest that the faster the train is moving, the harder it is to move, run or walk down the train! (as you have to put more energy into moving and therefore do more work)  This simply cannot be right, as people on a supersonic jet would be unable to move! 

What is going on?

[Loosehead]

Your calculations are correct but your interpretation of them is wrong. The faster the train moves the more energy he has to start with. Your friends effort is the same but his resultatant velocity is greater as he is helped along by the train. Do your calculations relative to the train and you'll always get the same answer. Consider also that the train speed you give is only relative to the ground you are standing on. The ground itself is moveing around the sun at 60000mph!

[Mat_D]

Surely it is only the man that is adding energy to the system (by running)  The train is not adding extra energy, the earth rotating certainly isn't.  The only object adding energy is the man himself!  The mans resultant velocity is greater of course, but we are dealing with delta E, the change in energy of the system. Therefore the starting energy is irrelevant.  If the man is the only object adding energy, why does he have to add more when the train is going faster.

Of course, this is a scientific puzzle that I know the answer to.  I'm not sure if you have given the correct answer yet..........

[gen2]

How about this as a suggestion:

Every action has an equal and opposite reaction. As the person on the train accelerates from 0m/s to 2m/s they have minutely slowed the train down. The train will have to use up more energy to regain its original velocity.  The extra joules have come from this source, not the person's own energy reserves.

[Mat_D]

Good answer but not correct I'm afraid.  The velocity of the train is constant in this thought experiment.  It is nor possible to slow a train down if you are inside it as you cant introduce any opposing force!

[Space]

I totally disagree with you when you say "only the man is adding energy to the system".
What makes you say that???

The man is adding enough energy to go from 0 to 2 m/s, which is 200 J whatever the speed of the train is. The rest of the energy comes from the train.

When the man accelerates forward, it pushes the train in the opposite direction. I disagree when you say you can't introduce an opposing force when inside a train. You can, when you accelerate (or decelerate). Replace the train by a skate board and experiment, you'll see !

Think about what is happening at the contact between the man's shoes and the floor of the train. There is a force that makes the man accelerate forwards, and an equal an opposite reaction that tries to make the train decelerate.

The train engine has to provide some extra force ( or torque ) to compensate for this reaction whithout slowing down.
That means the engine gives extra power. That extra power is in proportion with the train speed and the force ( power = force * velocity ) , which is why in the first case it gives
2200 - 200 = 2000 J
and in the second case, with twice the speed, it gives
4200 - 200 = 4000 J
which is twice the energy of the first case.

When you say "I know the answer", do you mean "I know the real answer made by someone really competent" ?
What you write definately doesnt come from a physics teacher, so how can you say things such as "good answer but not correct" ?
No offense, but it seems to me that you are the one who makes mistakes, not the one who gives marks to students...
gen2's answer is correct if you consider that the train's engine doesnt react immediately ( which is more realistic that my answer )

[londonbarry]

With the train and him moving at 10 m/s or either 20 m/s he need only to add enough energy to move him just the 2 m/s,

i.e.
(1) When he is sitting;

KE = � * 100 * (10 * 10) + 0 = 5000 Joules (plus zero as he is putting no energy in)

(2) When he is running;

KE = � * 100 * (10 * 10) + � * 100 * (2 * 2) = 5200 Joules

... and likewise at 20 m/s

(3) When he is sitting;

KE = � * 100 * (20 * 20) = 20000 Joules

(4) When he is running at 2m/s (combined velocity is 22m/s);

KE = � * 100 * (20 * 20) + � * 100 * (2 * 2)  = 20200 Joules

[Mat_D]

Wow, Space.  Are you having a bad day?

This was mean't to be a fun, spot the error, puzzle, not a Physics exam!

This thought experiment is carried out like most simple physics problems - in a frictionless vacuum, to make them simple!  Therefore there is no need to go too deep on this and worry about friction, slowing the train down etc...That was what I was getting at...

Maybe I should have posted this on the puzzles page so that I didn't offend the sensibilities of serious scientist's like yourself.

[Space]

londonbarry, what do these sums (2) and (4) mean?
They are not the kinetic energy of the person,
Mat_D calculated them correctly in his question,
you have 2000 J and 4000 J missing in (2) and (4) respectively.

[Mat_D]

Londonbarry.

You have missed the point that it is the man at the side of the tracks making the measurements and doing the calculations!  To him, what he is seeing is real, and his calulations are indeed correct.

[Space]

sorry Mat_D, but I thought that solving a puzzle is more "fun" if you look for a valid answer.

The train can be in a vacuum, that doesnt change the problem at all.

[Mat_D]

Ok.

I suppose the question is why, although the man has done his school boy science calculations entirely correctly, do they give him a false answer? i.e. it appears to the observer, that the man, who is the only body putting a change of energy in to the system, has to put more energy in when the train is going faster.

Why don't the KE equations work in this case?

[gen2]

Quote Mat_D
<<This thought experiment is carried out like most simple physics problems - in a frictionless vacuum, to make them simple!  Therefore there is no need to go too deep on this and worry about friction, slowing the train down etc...That was what I was getting at...>>

Without friction, the wheels would not get a grip on the track and the man's feet would not get a grip on the floor so nothing goes anywhere. Speed = 0m/s.

In a vacuum, both guys would suffocate so no measurements either.

End of experiment.

[mfewell]

It's a bit like a fly doing 10mph colliding with an oncoming train doing 100mph. Since the fly ends up doing 100mph plastered over the windscreen, it has gone from 10mph in one direction to 100mph in the opposite. Therefore at one point it was stationary. Since it was in contact with the train, the train must have been stationary at that moment too. A good one for parties for people who've never heard it before and had a few to drink.

[Space]

"since it was in contact with the train, the train must have been stationary at that moment too" ...  that is complete nonsense... not even funny ...

Try this problem:
http://www.geocities.com/activityworkshop/puzzlesgames/monkey/index.html

[Mat_D]

Still waiting for a clear, concise, lucid answer!

 

Anyone?

[Space]

2 people gave you the answer already. Let me try to make it concise and clear:
You shouldn't assume that only the man is providing energy.

[petsim]

I think that to calculate this and compare the results you must start from a common point. Therefore your original perception I'm afraid is wrong. You can only calculate the enrgy, by taking your friend's mass and speed relative to the train which in both cases is 100kg and 2m/s. Therefore you come up with the same answer each time. The speed of the train is irrelevant.

[Mat_D]

Space - yet again you have failed to understand the problem.  From the point of the observer who is calculating the KE, it IS only the man who is adding or changing the energy by running - the train stays at a constant velocity.

 

Petsim -THANKYOU!  It's all about FRAMES OF REFERENCE.  The man is outside the frame of reference of the passanger and therefore the Kinetic Energy equation fails.

[Space]

I understand your problem perfectly, thanks.
Your problem is that you made an assumption that you can't justify. And that assumption happens to be false.
None of what you wrote has justified your assumption "it is only the man who is adding energy".
The train velocity being constant doesnt prove it neither. Was that meant to be a proof?

[Space]

petsim,
you can calculate energy in any inertial frame of reference you want, as long as you keep it the same in all your calculations. In the question, the reference is the ground, which is valid, all energy is calculated according to this reference and the equations are valid.