Locomotivation

A locomotive of mass m m starts moving on the x x -axis from the origin such that its velocity (v) is directly proportional to the square root of its displacement; v = k x v=k\sqrt{x} . Find the total work done by all of the forces acting on the locomotive in the first t t seconds of its motion.

The answer will be a function f ( m , k , t ) f(m,k,t) and the constant of proportionality which relates the velocity and the square root of the displacement. Find the sum of the exponents on m , k , m,k, and t t in the function f f .

PS: In fact, there is a much shorter way to solve this problem, one which requires only two steps. If you can figure it out, put it down as a solution. This alternative solution does not require calculus or any difficult mathematics. It simply requires you to see the problem which I have given in a little more detail

Credits: Irodov's General Problems in Physics.


The answer is 7.

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2 solutions

Kishore S. Shenoy
Aug 19, 2015

v = k x \displaystyle v = k\sqrt{x}

v 2 = k 2 x \displaystyle \Rightarrow v^2 = k^2x

And Eq. of motion v 2 = v 0 2 + 2 a x \displaystyle v^2 = v_0^2 + 2ax

a = k 2 2 , v 0 = 0 \displaystyle \Rightarrow a = \frac{k^2}{2},~~~v_0 = 0

s = 1 2 a t 2 = k 2 t 2 4 \displaystyle s = \frac{1}{2}at^2 = \frac{k^2 t^2}{4}

W d o n e = m k 2 k 2 t 2 8 = m k 4 t 2 8 \displaystyle \Rightarrow W_{done} = \frac{mk^2\cdot k^2t^2}{8} = \frac{mk^4t^2}{8}

7 \displaystyle \boxed{\therefore~ 7}

I hope this is the easiest method! (Well I used Calculus!)

Haha, of course, this is the computational approach. The other approach is dimensional analysis. The simple reason for this is because I've already given information about the variables which constitute the function and therefore, the fastest way to solve it, without having to know almost any physics at all, is through dimensional analysis :)

Abhijeet Vats - 5 years, 9 months ago

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Hehe, you gave the all the basic dimensions!! I didn't take it in my mind. BTW, what is the question number of this question in Irodov?

Kishore S. Shenoy - 5 years, 9 months ago

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Oh I forgot that one. Let me check later and I'll tell you then.

Abhijeet Vats - 5 years, 9 months ago

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@Abhijeet Vats Thanks! I think its in the Kinematics chapter. Seems to me that I've done it...

Kishore S. Shenoy - 5 years, 9 months ago

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@Kishore S. Shenoy Your answer is definitely correct. You've got the exponents correct and the important variables in the equation.

The point of asking, though, for a shorter solution is to point out that, within an exam setting, you have to be very smart in the way you approach these kinds of problems and sometimes, you have to have some common sense, rather than know a lot of physics.

Of course, someone who truly loves physics will aim to solve this problem conceptually but if runs out of time, he will lose out to the guy who used the simpler shortcut.

Abhijeet Vats - 5 years, 9 months ago

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@Abhijeet Vats Yup that's true! I'll try to use the easiest method from now. Thanks!

Kishore S. Shenoy - 5 years, 9 months ago

Hey Abhijeet, ur question isn't clear. Please tell to sum up all the exponents of the function or else I was thinking only the exponents of m and t. Ur problem statement made me input 3 1st. Next i tried 7 nd it matches!

Md Zuhair - 3 years, 11 months ago

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Report it.

Kishore S. Shenoy - 3 years, 9 months ago
Abhijeet Vats
Aug 6, 2015

I'm not going to give a very complicated solution which will involve math. I'm going to relay the concepts required to solve the problem, so others can read up on those concepts and make full use of them in solving the problem.

The first thing that must be done is to define the relationship between the velocity and displacement. Now, it is clear that this problem requires differentiation and integration and so, the first thing that should be done is to differentiate the velocity to obtain an expression for the acceleration. Before this, it should be quite obvious that v is equal to some constant multiplied by the square root of the velocity, which is equal to the first derivative of x with respect to time. Now, we have obtained a differential equation which will make it easy for us to obtain quantities which will help us later on in the problem

What you will find is that the acceleration is a constant, This is really neat, isn't it? Now, the work done is simply the integral F. dx. But wait, a minute, we do not have an expression for dx, right?

Actually, we do. dx is simple equal to v dt, where v has already been given as a function of x. The aim now, is to find v as a function of time, so that we can do the integration F.v dt. Finding v as a function of time is simply a matter of solving a simple differential equation or finding a solution online. Because the force is a constant by itself, what we end up with, eventually, is the constant c multiplied by the integral of t.dt., which is equal to the work done by the force. Solving the integral, we find that m has an exponent of 1, the constant k, which relates the velocity and position, has an exponent of 4 and the time has an exponent of 2. Adding them gives us the answer 7.

I try, very much, not to include a lot of mathematical notation in these types of solutions, simply because I do not want to confuse those who are simply starting out on doing physics. I also want to emphasize that this was my thought process as I was solving the problem. What matters here are the concepts. The equations simply relay the concepts in an easy to understand way but it is a skill to relay the concepts without the equations. Anybody who has solved the problem is welcome to put in numerical solution to the problem :)

explanation was quite easy and helpful... but u misssed one thing... for finding relationship between x and t we have to equate the expression of velocity with dx/dt.

Rajdeep Paul - 5 years, 10 months ago

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Oh yes, give me a few minutes to make the relevant changes.

Abhijeet Vats - 5 years, 10 months ago

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i think the final answer of workdone will be m(k^4)(t^2)/8? reply me if i did it right

Rajdeep Paul - 5 years, 10 months ago

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@Rajdeep Paul Yes, that is correct :)

Abhijeet Vats - 5 years, 10 months ago

You are saying to find the exponents on just m and t! The answer should be 3 then!

tom engelsman - 3 years, 9 months ago

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