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Please don't report. It's not a homework question. Yesterday on my physics test there was this question. there is a block of mass $m$ connected to a spring as shown in the figure. the spring constant is $k$ and the friction coefficient between the block and the floor is $\mu$. they have asked what is the minimum horizontal force $F$ applied as shown in the figure so that the block starts to move. I answered $\mu mg$ considering the whole spring mass system as a single system of mass $m$. but a friend said afterwards that it would be $\frac12 \mu mg$ because the force is being applied on the spring and if the force elongates it by $x$ length, $$kx=\mu mg$$ and, $$Fx=\frac12kx^2$$ so solving, $$F=\frac12µmg$$ . Is he right ? please explain in detail why he is right or wrong. And please point out the problem in my thinking if I am wrong.enter image description here

what he is saying is that he is equating the increase in potential energy to the work done by the force on the spring. I don't understand his point.

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  • $\begingroup$ This spring questions drive me crazy, I hope you get an answer because I would like to know how to solve it too. Apart of that I think your approach is not correct as you didnt take into account the spring force. $\endgroup$
    – Suriya
    Commented Jan 24, 2016 at 12:21
  • $\begingroup$ "homework" is used as a generalized category, so , sorry but exam questions are treated the same way. Anyway, since apparently the spring is considered to be massless, and more important, the question asks when the block will move, not the spring, and not how far the block will move, you should be able to see the way to the correct answer. $\endgroup$
    – Carl Witthoft
    Commented Jan 24, 2016 at 13:15
  • $\begingroup$ If the spring is attached to the mass, how can you stretch the spring at all? Any force you put on the spring will pull the connected mass with it. Or is that the point you're making? $\endgroup$
    – user854
    Commented Jan 24, 2016 at 15:06
  • $\begingroup$ Hint #1: you pull the spring. It's the spring that pulls the mass. $\endgroup$
    – garyp
    Commented Jan 24, 2016 at 15:30
  • $\begingroup$ I suggest you edit the question and use math formatting for readability. Enclose any expressions in $...$ and use special keywords like \frac{a}{b} for $\frac{a}{b}$, x^2 for $x^2$ and x_1 for $x_1$. $\endgroup$
    – John Alexiou
    Commented Jan 24, 2016 at 16:27

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What would the force be if there was no spring? Each side of the spring feels the same force - so if you put a black box around the spring and only saw the string "going in" and a string "coming out" of the box, with the same tension on each, the force needed to move the box would be the same.

This means your approach is correct.

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  • $\begingroup$ But if the force applied stretches the spring, it now pulls both the mass and whoever creates the force, so wouldn't you need more than just the friction to pull the mass? . I never understand this variable force problems :P $\endgroup$
    – Suriya
    Commented Jan 24, 2016 at 23:08
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    $\begingroup$ @Pablo - the force would have done more work, yes - but until the block moves you are only doing work against the spring; once the block moves, you do no further work on the spring (since both ends of the spring will move by the same amount, the spring doesn't stretch any more). If there was a different force between the two ends of the spring, there would be a net force on the spring and it would accelerate; but it is massless, so it would accelerate infinitely fast until the forces were once again the same. $\endgroup$
    – Floris
    Commented Jan 24, 2016 at 23:11
  • $\begingroup$ Thanks a lot Floris, are these kind of physics explained in classical dynamics text like goldstein? $\endgroup$
    – Suriya
    Commented Jan 24, 2016 at 23:18
  • $\begingroup$ I am not familiar with that text; but typically this is explained in most high school physics courses. Although mileage varies with physics teachers... from the very good to the terminally bewildering. $\endgroup$
    – Floris
    Commented Jan 24, 2016 at 23:21
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    $\begingroup$ @Pablo good thing you have the rest of the world to ask for help... :-) $\endgroup$
    – Floris
    Commented Jan 25, 2016 at 16:15
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F(x) = kx, according to Hooke's Law. This means that your friend is incorrect, and you got the question correct.

Note: the potential energy of the spring is (kx^2)/2, so your friend confused potential energy with force. A bit of dimensional analysis would greatly decrease the chance of mixing the units in this way.

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  • $\begingroup$ sir thank you for your answer. but what he is saying is that he is equating the increase in potential energy to the work done. i don't understand who's potential energy. and sir isn't the dimensional analysis correct ? $\endgroup$
    – Subhranil Sinha
    Commented Jan 25, 2016 at 6:21
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If it's an ideal spring, your approach is right. Try to understand what your friend did: for mass $M$ to just start slipping, $kx = \mu mg$. This implies that the other, free end of the spring has to be displaced through a distance of $x = \frac{\mu mg}{k}$.

Here lies the crux of the problem: if the spring were not ideal, it would have gained some velocity during its elongation. Inertia would have kept it going beyond the equilibrium position and the maximum elongation then would be found out using the equation $F_{ext}x = \frac{1}{2}kx²$ (by the Work-Energy Principle), at which point $F_{spring} = 2F_{ext}$. Essentially, the force requirement would be halved ($=\frac{1}{2}\mu mg$) , courtesy of inertia.

But if the spring were ideal, any unbalanced force would cause it to have an infinite acceleration ($F_{net}=m \times a$; $m \to 0 \implies |a| \to \infty$), such that it would always remain at the equilibrium position. Therefore the maximum deformation would be $F_{ext}/k$, implying minimum $F_{ext} = \mu mg$.

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