Suppose a spring with spring constant 1 N m^-1 is held horizontally. If a pull of 1 N is applied to its left end and a pull of 2 N is applied to its right end, how much longer would the spring be as compared to its unstretched length?
I came up with this question because the spring related qurstions I've seen all have one end fixed some where. I am totally stuck on this one and I really have no idea where to start.
All I can figure out is that the spring will accelerate due to a net force. It seems inappropriate to be taking net forces, since pulling both ends with a larger force makes the spring more elongated.
P.S. this is not homework
Answer
Solution 1---Guess-work: If the forces applied on the two ends are equal, say both $1.5\ N$, the spring will get stretched $1.5\ m$. A natural guess is that the stretch is determined by the average of the two forces at the ends, which in this case are both equal to $1.5\ N$. Therefore, for the case you mentioned ($1\ N$ applied to one end, and $2\ N$ to the other end), the answer is again obtained from the average: the spring is stretched $1.5\ m$.
Solution 2---Precise analysis: If the spring is massless, the question is ill-posed; the nonzero net force yields infinite acceleration.
If the spring has mass $1\ kg$ (you can extend the following analysis to arbitrary mass), then since the net force on it is $2\ N-1\ N =1\ N$, the spring will have $1\ m/s^2$ acceleration to the right.
Now let us, as observers, accelerate along with the spring so at all times we are in rest with respect to each other. So we (the observers) have $1\ m/s^2$ acceleration to the right.
Since we are accelerating to the right, we are no longer inertial observers; so if we want to use Newton's laws we should think everything we see is subject to a "gravitational field" with strength $1\ m/s^2$ to the left (this is the push-back force you experience when a car you sit inside accelerates). Therefore we see a spring with $1\ N$ force applied to its left end, $2\ N$ applied to its right end, and a gravitational pull of $1\ N$ on it to the left; the total force is zero in our (non-inertial) accelerated frame, as it should be, since we see the spring at rest.
Now, applying Newton's second law (we are still in the non-inertial frame remember) to the tiniest bit of spring at the left end, we learn that the tension of the spring is $1\ N$ at the left end; similarly the tension of the spring is $2\ N$ at the other end. The tension in the middle of the spring linearly interpolates between these two values. This linear gradient in spring's tension is due to the left-wards gravitational field; a similar (gravitational) effect results in linear vertical pressure gradient in liquids at rest.
Now you see why this problem is much more difficult than those with equal forces applied to both ends of the spring. Here the tension varies along the spring, leading to further complications. However it's not too difficult to overcome these complications ...
The tiniest bit of the spring at the left end is subject to $1\ N$ at its both ends; if every bit of the spring was like that, the spring would stretch $1 m$. The tiniest bit of the spring at the right end is subject to $2\ N$ at its both ends; if every bit of the spring was like that, the spring would stretch $2 m$. But the parts of the spring in the middle are subject to a tension which linearly interpolates these two values, therefore on average the tiny pieces of the spring stretch in a way that results in the whole spring stretching $1.5\ m$.
This confirms the guess-work in Solutions 1.
You can convince yourself that although the ($1\ kg$) mass we introduced for the spring is necessary to make the problem well-defined, the final answer doesn't depend on the mass of the spring.
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