Sunday, 6 March 2016

newtonian mechanics - Why does the amount of energy transferred depend on distance rather than time?



The change in energy of an object can be determined by the work equation, where work is the change in energy:


$$ W = F \cdot d $$


I conceptualize the transfer of energy as simply a series of small "packets" of energy being transferred at every Planck length. These small "packets" of energy add up to the total energy transferred (i.e. work). I'm not sure if this conceptualization is correct, so correct me if I am wrong.


However, it makes me wonder why the amount of energy transferred is dependent on distance and not time.


$$ m_1 = 10~kg \\ m_2 = 20~kg \\ W_1 = (10~N)\cdot(5~m) = 50~J \\ W_2 = (10~N)\cdot(5~m) = 50~J \\ W_1 = W_2 \\ t_1 \neq t_2 $$


If I apply a constant force on an object, why isn't the energy transferred at a constant rate with respect to time? The energy transfer rate varies dependent on how long it takes to cover the set distance.


In other words: why is the energy transferred consistent per unit of distance, and not per unit of time?



Answer




There is a name for the quantity $F\cdot t$, it's called the impulse. The impulse tells you how much momentum is transferred to the system in a given time interval if you apply a constant force, much as how the work tells you how much energy is transferred over a given distance interval if you apply a constant force.


So, what's up?


Newton's 2nd law can be expressed as


\begin{equation} F = \frac{dp}{dt} \end{equation} where $p=mv$ as usual.


A constant force thus means that momentum is introduced into the system at a constant rate. That is what the force is measuring, the rate at which you are introducing momentum into the system.


Now if we only have kinetic energy, then $p = \sqrt{2 m E} = mv$. So... \begin{equation} F = \frac{d}{dt}(\sqrt{2 m E}) = \frac{\sqrt{2m}}{2\sqrt{E}} \frac{dE}{dt} = \frac{2m}{p} \frac{dE}{dt}=\frac{2}{v}\frac{dE}{dt}=2\frac{dt}{dx}\frac{dE}{dt}=2\frac{dE}{dx} \end{equation}


As you can see, the extra factor of velocity between energy and momentum is crucial. That factor of $dx/dt$ converts the time derivative $dp/dt$ to a space derivative $dE/dx$.


Off the top of my head I can't think of a simple physical reason for the difference. It's just a matter of which variables are more convenient, it turns out to be more meaningful in many problems to study how the momentum changes with time instead of the energy, and so physics is set up to talk about forces measuring the rate of change of momentum, instead of the rate of change of energy. One reason is that momentum is a vector whereas energy is a scalar, so the momentum has more information and so is more useful to track in general.


Incidentally, rate at which you pump energy into the system is called the power. It is related to the force by $P=F\cdot v$ (at least over time intervals short enough that the velocity doesn't change by much).


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...