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As the title says, if one end of a spring of mass $m$ is fixed to say, a wall, and the other one is pulled at a constant velocity $v$ by some external agent, we have to find the kinetic energy of the whole spring as a body.

Now, the book where i saw this question just gave the answer with a hint to consider that this is akin to a mass of $m/3$ being pulled while attached to a massless spring.

My problem is, isnt this in multivariable calculus? Or have I gone wrong. My education level is high school currently. So thats why I am facing the problem. Please suggest a solution.

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closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind, JamalS, tom Mar 15 '15 at 13:20

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  • $\begingroup$ Does the external agent keep pulling with velocity $v$ no matter how much force the springs acts on it ie does the external agent pulls with a constant or varying force? $\endgroup$ – Gonenc Mar 14 '15 at 11:40
  • $\begingroup$ @gonenc yes, that must be assumed $\endgroup$ – Saurabh Raje Mar 14 '15 at 13:24
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I don't know if you have already done integration in high school but at least here is a solution without using multivariable calc.

The kinetic energy of the spring with length $L$ is the following $$E_{kin}=\int_m \frac{u^2}{2} \, \text{d}m$$

where $u$ is the speed of a little mass element d$m$ and the integration is over the mass of the spring. Assuming uniform mass density $\mu = \frac{m}{L}$ such that $m(x)=\mu x \implies \text{d}m = \mu \, \text{d}x$

$$\implies E_{kin}=\int_0^L \frac{u^2 \cdot \mu}{2} \, \text{d}x $$

The velocity of a mass element is directly proportional to the velocity of the external agent, assuming that it persistently is pulling with velocity $v$, namely $u=\frac{v\cdot x }{L}$. Notice that this gives the correct boundary conditions at $x=0 \implies u=0$ and $x=L \implies u=v$. You can fairly easy convince yourself that this relation holds. $$\implies E_{kin} = \frac{\mu}{2} \int_0^L \left( \frac{v\cdot x }{L} \right)^2 \text{d}x = \frac{\mu v^2}{2L^2} \int_0^L x^2 \text{d}x = \left. \frac{\mu v^2 x^3}{2\cdot 3 L^2} \right|_0^L = \frac{\mu v^2 L}{2\cdot 3} = \frac{m}{3} \cdot \frac{v^2}{2}$$

where I used the $\mu = \frac{m}{L}$ in the last step. Notice that this is equal to the kinetic energy of an object with mass $\frac{m}{3}$.

I don't know any other way of getting this equation without using integration and I doubt that one exists because my intuition always says that 3 is a sneaky number, which almost always arises from an integral.

Edit after comments: For those who cannot read the names of points they are from left to right $A$ $D$ $B$ and $B'$

(For those who cannot read the names of points they are from left to right $A$ $D$ $B$ and $B'$)

I have seen that the formula for $u$ created some confusion so let me elaborate on the formula with an example. Assume that the spring is of the length $|AB|=L$ and the point that we are interested in is the middle of the spring ie the point $D$. If I stretch the string by an amount $AB$ in $t$ seconds ie I double the length ($AB'$) of the spring then the middle of the stretched spring is the point $B$. Although the end point of the spring has travelled a distance $AB$ the midpoint of the string has travelled a distance of $|DB|=L/2$ as can be seen from the (poor) drawing that I made. Thus the speed of the little mass element on the point $D$ is $\frac{v}{2}$ ,where $v=\frac{L}{t}$.

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  • $\begingroup$ I agree with the last line. And yes, i know basic integration. The only prob in the answer was the assumption for u(x). Is there any way u can prove it? $\endgroup$ – Saurabh Raje Mar 14 '15 at 13:27
  • $\begingroup$ by prove, i mean arrive at this result and not assume it? $\endgroup$ – Saurabh Raje Mar 14 '15 at 13:28
  • $\begingroup$ I cannot say that I can prove it but I can try to convince you that it is correct. Think of this way: If you pull the string 2m then the middle of the string moves 1m and 1/4 of the string moves 1/2m and so on. If you do that in 1s then u is equal to 1m/s and 1/2 m/s respectively. $\endgroup$ – Gonenc Mar 14 '15 at 13:30
  • $\begingroup$ i did not get that. How can u conclude that if u pull spring 2m, then its mid point comes out by 1 m? $\endgroup$ – Saurabh Raje Mar 14 '15 at 13:35
  • $\begingroup$ @SaurabhRaje I have edited my answer you can check the answer above. $\endgroup$ – Gonenc Mar 14 '15 at 13:56

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