1. ## conservation of energy

an experimental rocket sled on a level , frictionless track has a mass of 1.4*10^4 Kg. It propels itself by expelling gases from its rocket engines at a rate of 10 Kg/s, at an exhaust speed of 2.5*10^4 m/s relative to the rocket. For how many seconds must the engines burn if the sled is to acquire a speed of 50m/s starting from rest? You may ignore the small decrease in mass of the sled and the small speed of the rocket compared to the exhaust gas.
ps: i am learning conservation of energy.. this question confused me coz the reading.. somebody help. thanks_

2. ives,

If the subject is conservation of energy, then you are likely supposed to approach the problem by taking the gas and the rocket as an isolated system. Notice that when both are stopped mechanical energy is zero.

By the law of conservation, the energy is the same in an isolated system if no external forces are applied. Since all forces involved are interactions between of the rocket and the gas, all forces are internal and energy must conserve.

So, what is the kinetic energy of the gas after it is propelled out? What is the relation of this energy with the kinetic energy of the rocket? Tip: you can convention energy to be negative in a specific situation.

3. .. i was confused with the wording, could u list out the equations and list out procedures? that would be great appreciated~~
Originally Posted by Rafael Almeida
ives,

If the subject is conservation of energy, then you are likely supposed to approach the problem by taking the gas and the rocket as an isolated system. Notice that when both are stopped mechanical energy is zero.

By the law of conservation, the energy is the same in an isolated system if no external forces are applied. Since all forces involved are interactions between of the rocket and the gas, all forces are internal and energy must conserve.

So, what is the kinetic energy of the gas after it is propelled out? What is the relation of this energy with the kinetic energy of the rocket? Tip: you can convention energy to be negative in a specific situation.

4. OK let me walk you through a little closer:

1) Isolate the system
Consider the whole gas body and the rocket as an isolated system. Notice that no matter where the gas and the rocket are they are still part of the same system in your model.

Also, notice that all forces in the experiment are internal, as they are interactions of the rocket and the gas. Forces such as weight or friction can be disconsidered because their resultant is null, since they are either being countered by opposite forces (such as weight-normal) or disconsidered.

2) Conservation of energy
If no external forces are applied, then the energy in a given system remains contant. In our particular case:
$\displaystyle E_{system} = E_{gas} + E_{rocket} = constant$
Notice that in the starting scenarion, when both are stopped, $\displaystyle E_{system} = 0$, since you can't have kinetic energy if you have no speed (i.e. body is stopped implies that it has no kinetic energy).

3) After the burn starts
When the gas is burned, it transforms chemical energy into kinetic via expansion. Obviously the details of how this transformation happens are totally irrelevant in the context, so don't worry about them(*).

The kinetic energy that the gas aquires can be measured by the data you have via the kinetic energy expression:
$\displaystyle E_{gas} = \frac{m_{gas}v_{gas}^2}{2}$
Let me suggest that you do not change the letters for numbers yet. (**)

But you are given the rate at which the gas burns, so you know the mass of gas burned (and because of this, in movement) after a time t:
$\displaystyle E_{gas}(t) = \frac{10t(v_{gas}^2)}{2} = 5t(v_{gas}^2)$
4) "Signal" of energy
Energy is a scalar. Because of such, you can't say that it has a direction or a sense. But it is somewhat easy to see, and somewhat overkill to get into the details of, that you can safetly consider that a body moving to the right has positive kinetic energy and one moving to the left has negative kinetic energy. This makes our application of the conservation law easier.

Another approach would be to consider the rocket and the gas in separate systems, and notice that all forces of the gas act upon the rocket and vice versa.

So we can say that, since the energy within the system is constant, both bodies have the same energy, with reverse signs:
$\displaystyle E_{rocket} = -E_{gas}$
5) Conclusion
Since you want the speed you want the rocket to move by, then you know the energy you want it to have:
$\displaystyle E_{rocket} = \frac{m_{rocket}v_{rocket}^{2}}{2}$
Combining the facts we derived previously. I'll jump a step here and say that energies are equal, forgetting about the minus sign. This is because it's a simple situation, and you can perfectly visualize that if the gas is going one sense then the rocket is going the opposite one.
$\displaystyle E_{rocket} = E_{gas}$
$\displaystyle \frac{m_{rocket}v_{rocket}^{2}}{\rlap{/}2} = \frac{10t(v_{gas}^2)}{\rlap{/}2}$
$\displaystyle t = \frac{m_{rocket}v_{rocket}}{10v_{gas}^2} s$
Now it's all joy: just sub in the info you have and you have the number of seconds desired.

(*) A teacher of mine said that lots of students tend to get confused in situations like this one, worrying about unnecessary details. So, focus in the data you are given, as you will 'measure' stuff from them. This is also like it's done in the real world: sometimes you choose to measure what's easier to. In this particular case, for example, you can determine the internal energy of the gas (and from this obtain lots of information about it, had you had more data) indirectly by measuring its kinetic energy after it's burned and knowing how they relate.
(**) Also a tip from the same teacher. Get used to doing this, because in Physics problems numbers often have lots of algarisms, making it boring to work with them all the time. Also, leaving it as letters it's easier to spot simplifications. This tip also works for most computational exercises: it may be a bad idea to efetuate a multiplication, or even an addition, during an intermediate step of the solution because they may simplify later.

Hope I've helped,

5. very detail, thanks!