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Two masses connected by a massless spring, on a frictionless surface , and a force of $60$N is applied to the 15kg mass such that it accelerates at 2 $\frac{m}{s^2}$. What is the acceleration of the $10kg$ mass?

I came across this question. I first thought that that the $10$kg was constrained to move at the same acceleration. But when I work it out, I get $a_2$ = 3 $\frac{m}{s^2}$. And it is the correct answer according to the book.

What I am unable to understand is, isn’t the $10$kg mass constrained to move at the same acceleration as the $15$kg mass? I thought we could replace the massless spring by (or treat it as) a massless string and results would be the same. Am I making a fundamental mistake?

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  • $\begingroup$ Did you make the diagram, or is the diagram from the original source? Either the word "string" is a typing mistake and it should read "spring", or the example has an unusual looking string! Either way, the force, mass, and acceleration of the 15 kg block determines the force in the spring/string, and that force determines the acceleration of the 10 kg block. Since the accelerations are different, the spring/string is stretching. $\endgroup$
    – JohnHoltz
    Commented Apr 8, 2020 at 12:36
  • $\begingroup$ @JohnHoltz It was a typing mistake, sorry. I have corrected it $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 13:55
  • $\begingroup$ @JohnHoltz The 10kg block is accelerating at a rate faster than that of the 15kg block. Doesn't it mean the spring is compressing $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 13:58
  • $\begingroup$ I thought we could replace the massless spring by a massless string and results would be the same. - No you can't replace it with string.In your case 10kg mass will execute SHM as the force is applied on 15 kg mass.In case of string there won't be any SHM. $\endgroup$
    – Madhubala
    Commented Apr 8, 2020 at 14:33
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    $\begingroup$ @πtimese Technically, the spring is compressing if v2 > v1 (where v is the velocity). Of course, if a2>a3 for enough time, then eventually v2 will be > v1. It sounds like the problem was made-up without considering the physical reality of the arrangement. $\endgroup$
    – JohnHoltz
    Commented Apr 8, 2020 at 15:01

3 Answers 3

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The problem is poorly stated. If a 60 Nt force is applied to a 15 kg mass, the acceleration will be 4 m/s/s. The 10 kg mass will start slowly and accelerate as the spring is stretched. The two masses will then oscillate relative to each other. At some later instant when the force from the spring is 30 Nt, the 15 kg will be accelerating at 2 m/s/s and the 10 kg at 3 m/s/s.

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  • $\begingroup$ Thanks, it really clears up the whole thing. Perhaps the question is referring to the instant when $F_{sp}$ is 30N and 15kg mass is accelerating at 3 m/s/s. I agree with you, the question should have been stated clearly $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 16:34
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The thing is, an 'inelastic' massless string ensures constrained motion because it has a definite length.

If x is the distance moved by the first block, then the second block is also constrained to move so that the net extension of the string is 0.

But, a spring can become compressed or stretched. So, if block 1 covers a distance x, three cases arise:

i) spring becomes compressed: the second block moves a distance more than x, and hence has greater acceleration

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ii) spring becomes stretched: the second block moves a distance less than x, and hence has less acceleration

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iii) spring remains in original shape: the blocks have equal acceleration

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  • $\begingroup$ In the question, if the first block accelerates at 2 $\frac{m}{s^2}$ and the block at the back accelerates at 3 $\frac{m}{s^2}$, how can the spring stretch? The block at the back is accelerating faster, so the spring should compress, which is what does not make sense to me $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 9:09
  • $\begingroup$ If the spring compresses, isn't it counter intuitive? Since the first block is pulled to the right and the block at the back accelerates as a result, the spring should stretch. Why does it compress? $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 9:14
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    $\begingroup$ You are imagining that there is a time lag between the acceleration of the first body and that of the second. But in case of an ideal massless spring, there will be no time lag between the two events. $\endgroup$
    – Elendil
    Commented Apr 8, 2020 at 9:22
  • $\begingroup$ You can understand this easily if you just rotate the situation. Imagine two masses connected by a spring falling vertically from some height. Now, you can see that there is no problem in compression of the spring (which suddenly seems more intuitive than elongation of spring ) $\endgroup$
    – Elendil
    Commented Apr 8, 2020 at 9:25
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    $\begingroup$ I'm so sorry if I had confused you. I did not mean gravity or freefall, its just the same situation, a force being applied but you are observing it vertically i.e. without any ground on which it slides so that you realize no time lag is necessary. (I shouldn't have said falling) $\endgroup$
    – Elendil
    Commented Apr 8, 2020 at 14:51
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a string is rigid so cannot be extended or compressed, its both end would move with same acceleration

in spring, it can extend, if spring extends then it would apply equal forces on both bodies i.e $kx$ ,towards left for $15 kg$ and towards right for $10 kg$.

the only force moving $10 kg$ is $kx$

now if you apply this concept to string you will get different acceleration on both sides which is not possible, so acceleration in case of strings is constrained

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  • $\begingroup$ Thanks. I did make a fundamental mistake there. I was wrong to assume that I could replace a massless spring with a massless string in this case. I just revisited what I studied, and from what I understood, I can replace a massless spring with a massless string only when the spring is elongated/compressed, and won't be elongated/compressed any further in the system. Is it correct? $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 14:10
  • $\begingroup$ yes but only if the relative acceleration of both of end of spring pushes spring length it to such extent that it becomes like a straight string $\endgroup$
    – maverick
    Commented Apr 8, 2020 at 14:19
  • $\begingroup$ in saying that "I can replace a massless spring with a massless string", I meant to say that I can treat that massless spring as a massless string $\endgroup$
    – 4d_
    Commented Apr 8, 2020 at 14:19
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    $\begingroup$ why you want to replace it like that a spring would store potential energy but a string cant so it is not appropriate to make it like you are thinking, a spring fully stretched to max length can be treated as string but to only find acceleration of bodies. but it is better to think string as string and spring as spring , otherwise work energy theorem would get dismantled if you think like that $\endgroup$
    – maverick
    Commented Apr 8, 2020 at 14:26

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