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Question :-

A block is attached to the free end of a spring of spring constant $50\ \mathrm{N/m}$. Initially the spring was at rest. A $3\ \mathrm N$ force was applied to the block until it came to rest again. Find the maximum displacement of the block, take initial displacement as $0$.

My first try :-

We know the spring force is $-kx$.

So at rest only horizontal forces acting on the block would be spring force and applied force of 3 N.

$$\therefore -kx + 3N = 0$$

$$\implies x = \frac 3{50} = 0.06\ \mathrm m$$ Which is incorrect.

My second try :-

Let work done by spring force and applied force be $W_s$ and$ W_a$ respectively.

$$ \begin{aligned} \Delta E_k &= W_s + W_a\\ &\implies0 = \displaystyle -\frac12 kx^2 + Fx\\ &\implies\displaystyle \frac12 kx = F\\ &\implies\displaystyle kx = 6\\ &\implies\displaystyle x = \frac{6}{50} = 0.12\ \mathrm m \end{aligned} $$

Which is correct.

I am still not getting why my first try failed. What was my error ?

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  • $\begingroup$ @Downvoter Sorry can you please tell what I haven't included in the question that you want to me to include ?. I object you down vote because 1) :- I am not giving my homework to anyone to do. 2) :- I have included all my tries and the question verbatim. 3) :- How is my question worse than physics.stackexchange.com/questions/299723/… (he did even write a single equation for his claim). $\endgroup$ – A---B Dec 26 '16 at 19:12
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    $\begingroup$ don't be demotivated by random downvotes. Your question is well suited in my point of view. $\endgroup$ – user_na Dec 26 '16 at 19:14
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Your first try is considering a static system. So the result is true if you very very slowly increase the force until you reach 3N. (With the friction dampening all oscillatory motion of the mass)

So the real question is what is different in the problem you posted.

Try thinking it from the beginning, you switch on the force and initially it is much greater than the counter force of the spring. So the mass accelerates, building up kinetic energy. The part the work which is transformed into kinetic energy decreases until you reach 0.06m as you calculated in your first try. From now on the force is decelerating the mass until it comes to a rest. As you calculated this is 0.12m from its original position. This scenario is dynamic, the mass will now oscillate around the center-point.

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  • $\begingroup$ +1; Oh right, your point seems valid. But in my first try I considered the time when the system is at rest after being pulled by the external force. Then the box will be at rest, So why isn't at this moment I can consider the system to be static and equate forces ? $\endgroup$ – A---B Dec 26 '16 at 19:21
  • $\begingroup$ Everything you say is correct. But look at the question again, it asks for the maximum displacement. Remember that you are in the dynamic case and the question is not asking for the average one (which you calculated in a). And the max displacement is at the turning point of the oscillatory motion, which is what you calculated in b. $\endgroup$ – user_na Dec 26 '16 at 19:27
  • $\begingroup$ Thanks for the posting such a precise answer, I want to upvote your answer again :))). Newtonian mechanics is by far the most difficult part of physics. I find Electromagnetism far more easier than this :). $\endgroup$ – A---B Dec 26 '16 at 19:33
  • $\begingroup$ Don't bother to have a look at quantum field theory then ;-) $\endgroup$ – user_na Dec 26 '16 at 19:37
  • $\begingroup$ What I learnt in physical chemistry class on wave mechanical model of hydrogen was pretty easy. I think main problem is that I find newtonian mechanics vape and mundane whereas I find quantum mechanics quite interesting atleast from physical chemistry class. ^_^ $\endgroup$ – A---B Dec 26 '16 at 19:41
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You say "A force was applied to the block until it came to rest again" The block doesn't come to rest at x = 0.06 m, only acceleration becomes zero); it overshoots because of its non-zero velocity and comes to its instantaneous rest (instantaneous speed zero) at x=0.12 m. So for your question, (b) is the correct answer. Of course there are two answers in your part (b), x=0 and x=0.12 - the block is at rest instantaneously at both places as it oscillates. But the maximum displacement is at x = 0.12 m.

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  • $\begingroup$ Welcome to PSE. Thanks for the answer. And happy new year.$$ \displaystyle \huge \ddot \smile$$ $\endgroup$ – A---B Jan 1 '17 at 12:35

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