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Consider the image above. The block has a mass $m\;kg$, the friction coefficient between the block and the surface is $μ$, the spring is massless with spring constant is $k$.

The question asks to find the minimum force which must be applied to the spring so that the block just slides on the surface.

I am unable to proceed but I can tell the answer $\frac{1}{2}mg\mu $.

Clearly, the minimum horizontal force required to pull the block(without the spring) is $mgμ$. And if the answer is correct, then with a spring, half of the force is required. So, it can be concluded that the spring pulls the block with a force which is just double of the applied force.(I need to know whether this conclusion is correct)

The block too applies an equal and opposite force on the spring. So the force so the spring is $F$ (applied force) towards right and $2F$ (by the block) towards left. Since the spring is massless, therefore,

$F-2F =0×a \implies F=2F$

Which is surely not correct. Even if the spring has a mass then, the it should accelerate towards left which is again contradicting. So, where am I going wrong?

Kartikey
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  • I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. – David Z May 30 '20 at 09:49

1 Answers1

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As far as I see, the 'given' answer of $\frac12mg\mu$ is simply wrong. A mechanical system that gives a mechanical advantage (lever, screw, pully-system, etc.) could reduce the required force, but a simple spring does not do so.

Now, if you had a pulley at the loose end of the spring, w/ a rope/cord going about that pulley with one end tied to a fixed point, and the other being used to apply the force, now that would halve the required force.

If the answer is supposed to be $\frac12mg\mu$, are you certain you got the problem and diagram right? Where did this problem come from?

WRSomsky
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  • This is a question from HC Verma's Concepts of physics. I too thought it to be wrong. But I saw springs working as force multiplier in many case. – Kartikey May 30 '20 at 08:42
  • Do you have some example of a spring 'force multiplier' you can point me to? If you has some mechanism for compressing multiple springs one at a time then applying them all at once, I could see it, but not just putting one in series. – WRSomsky May 30 '20 at 08:45
  • In another question in which a block is connected to a spring and the spring is connected to a string which passes over a pulley. The question asks the minimum force to be applied to the string hanging over the other side of pulley, which just lifts the block. The answer-½mg. In this case too the spring acted like a force multiplier. – Kartikey May 30 '20 at 08:46
  • Do you have the chapter&number for these questions? – WRSomsky May 30 '20 at 08:49
  • I cannot find the first one, the second one is from another book. but I can give you another questions which suggests the same. Chapter work and energy. Question-45.--Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring. – Kartikey May 30 '20 at 09:05
  • the expected answer to me is mg/k but is actually 2mg/k. – Kartikey May 30 '20 at 09:07
  • I got a link to the second question-https://www.excellup.com/testntricks/iitPhysics/friction6.aspx. Skip all and read question 10. I have this question in different book and cannot post the image in comments – Kartikey May 30 '20 at 09:24
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    @Kartikey I strongly believe that the book is wrong about the concept of springs. When a massless spring is stretched by a length $x$, the force is $kx$, uniformly throughout the entire spring. The force on each end is also $kx$ and not $kx/2$. See this question. Also see my comment on DoubtNut's answer. – Vincent Thacker May 30 '20 at 09:24
  • @user7777777 I have often seen the concept used by DountNut. What's wrong in his solution?? – Kartikey May 30 '20 at 09:33
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    @Kartikey There are 2 things you need to understand. Firstly, regardless of what is being used to pull the block, the fact remains that the block must feel a force of $\mu mg$ in order for it to move. If the force on the block is less than this, the laws of static friction dictate that friction exactly cancels out the applied force. Mathematically, if the applied force is $F$, the friction is $-F$ when $F \lt \mu mg$. In other words, the block behaves like a brick wall. This is why the block does not move when the applied force is less than the static friction threshold. – Vincent Thacker May 30 '20 at 11:43
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    The second thing is what I was saying above. For a massless spring, the force at each end is $kx$. It is not possible for the forces at the ends to be unequal. Consider the case in which the spring constant is very large. In this case, the massless spring essentially becomes a massless string. Except for the energy required, this case is no different from the original. – Vincent Thacker May 30 '20 at 11:46
  • @Kartikey Regarding you Work&Energy #45 problem, the equilibrium distance, where the spring force balances the gravitational force is indeed $mg/k$, but but if there is no source of friction -- no energy loss -- when the weight falls to that position, it will have kinetic energy, and will stretch the spring further, up to a maximum extention of $2mg/k$. Conservation of energy. The weight is going to oscillate up and down until the excess energy is dissipated, at the extention finally rests at $mg/k$ – WRSomsky May 30 '20 at 21:43