Question
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction μs=0.100\mu_s = 0.100. (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is μk=0.0850\mu_k = 0.0850, what total distance does it travel before stopping? Assume it starts at the maximum amplitude.
Question by OpenStax is licensed under CC BY 4.0
Final Answer
  1. 4.91 mm4.91 \textrm{ mm}
  2. 9.46 mm9.46 \textrm{ mm}

Solution video

OpenStax College Physics, Chapter 16, Problem 45 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. On a horizontal surface we have this mass, m, connected to a spring and it's bouncing back and forth. But it's dampened simple harmonic motion because there is some friction with the horizontal surface. The spring constant we're told is 150 newtons per meter and the mass is 0.75 kilograms. The coefficient of friction when the mass is moving is 0.0850 and when it is at rest the coefficient of static friction is 0.1. We're asked to figure out what is the maximum amount the spring could be stretched and have the mass at rest and have it not bounce back. Well that is going to be the spring constant times the amplitude, is going to equal the gravity which is the normal force that the mass experiences because force of friction is normal force times the coefficient of friction, be it static or kinetic depending on whether the mass is not moving or moving. The normal force is going to be the weight m g. So that's where this m g comes from. It's being multiplied by mu s which is the static coefficient of friction because we're told that the mass should not be moving. Then we have to solve for this amplitude A one. I gave it a number or subscript one because we're going to use this value in our subsequent calculation in part B. Divide both sides by k to solve for A one and it's going to be m g mu s over k. So that's 0.75 kilograms times 9.81 newtons per meter times 0.1 divided by 150 newtons per meter. This works out to 4.91 millimeters. So if you slowly stretch this mass 4.91 millimeters from the equilibrium position or the un-stretched position of the spring, and you let it go, it will not spring back because the static friction force will equal the spring force up until this point. Okay. Now in part B, we're told that -- let's suppose the initial amplitude is going to be two times what we figured out in part A, that's two times A one in which case it is two times m g mu s over k which I wrote here. So the mass is moved to this position two times A one and then you let it go, in which case it will spring back because the spring force will be greater than the static friction force at that point. Since it's going to be moving the kinetic friction is what's going to be dampening the oscillation and it's going to be dissipating some of the energy up until the point where the spring force equals the kinetic friction force. That is the kinetic friction force where you have mu k and not mu s. Okay. Here is where we talk about the final position of the mass. It's going to be at a point where it's going to stop when the spring force which is the spring constant multiplied by the final amplitude, equals the kinetic friction force. So that's the coefficient of kinetic friction times the normal force which equals m g. We'll divide both sides by k to solve for A f. So the final amplitude when this thing stops oscillating at a certain distance from the un-stretched length of spring, and we'll call that amplitude final, A f, is mu k m g over k. The total initial elastic potential energy of the mass, or the system I should say, is one half k times the initial amplitude squared. We're going to dissipate some of that energy due to friction and this is friction force multiplied by distance. This is the total distance that the mass travels as it's sliding back and forth, back and forth, back and forth. Once this energy is dissipated we'll be left over with the elastic potential energy when the mass comes to rest at this final amplitude. Okay. Then we substitute for A i and A f. So A initial, amplitude initial is two m g mu s over k. So we substitute that in, and amplitude final is mu k m g over k. We square that and we get that from here. Then it's a bunch of algebra to solve for d. This two squared becomes four but divided by two makes two in the end, times m squared g squared mu s squared, all over k squared times k makes k to the power of one in the denominator, minus mu k m g d equals mu k squared m squared g squared over two and then over k to the power of one because it's k squared in the denominator times k makes k to the one down there. Then divide everything by m g or multiply everything by one over m g, the way I really like to say it. That makes two m to the power of one, g to the power of one, coefficient of static friction squared over k, minus mu k d equals mu k squared m g each to the power of one, over two k. Then we add mu k d to both sides because we want to isolate d on one side of the equation. We also subtract mu k squared m g over two k from both sides. Then switch the sides around and we end up with mu k d equals two m g mu s squared over k minus mu k squared m g over two k. So the m g over k is a common factor between these two terms so we factor that out to get this line here. Then we divide both sides by the coefficient of kinetic friction. The final formula we're going to use is that d equals the total distance that the mass slides back and forth such that the kinetic friction force will dissipate an amount of energy enough to reduce the initial elastic potential to this final elastic potential energy. This is going to be m g over mu k times the spring constant, times two times the coefficient of static friction squared minus coefficient of kinetic friction squared over two. There we go! So d is 0.75 kilograms times 9.81 meters per second squared, over 0.085 times 150 newtons per meter, times two times 0.1 squared, minus 0.085 squared over two. This works out to 9.46 millimeters. Now this formula is not something that we can use intuition to verify very easily because it's a very strange-looking formula. What we do expect though is that this number should be greater than the answer for part A and it doesn't tell us if we're for sure correct or not, but at least the fact that it's bigger than this number indicates that it could be correct and I think that it is. There!

Comments

Hello,
When system is not moving in final position , should be energy equal to zero??.a.ka equation is 1/2k(A)^2 = Mmgd.
I think potenial energy is zero in final position because A=0-distance from equalibrium point. Where i get wrong???
Thanks.

Below is copy/pasted from an emailed question from a student that I'm including here in case it helps others:

Hi Shaun,

Problem 45 is a rather devious one in which we find the maximum distance a block can be displaced without slipping on a surface with static friction. In part (b), we displace it to twice the distance and the problem asks for the total distance the block travels before coming to a halt. I found an answer different from your own, then tried follow your work from your transcript. I wasn't able to follow your work in its entirety, but I believe I found a mistake in your reasoning. You say that the maximum amplitude occurs when the kinetic friction force is equal to the spring force. Here are two excerpts from the transcript:
"Since it's going to be moving the kinetic friction is what's going to be dampening the oscillation and it's going to be dissipating some of the energy up until the point where the spring force equals the kinetic friction force."
"It's going to be at a point where it's going to stop when the spring force which is the spring constant multiplied by the final amplitude, equals the kinetic friction force."

If the spring and friction forces were opposite one another, the position where they are equal would only demarcate where the block is not accelerating, not where its velocity is zero. Worse, if the block has overshot the equilibrium position (practically guaranteed from the problem statement), the spring and friction forces will act in the same direction, opposite the motion of the block.

One rather clear indication that something is wrong is that if we allow the coefficient of kinetic friction to approach zero in your solution, the distance traveled approaches infinity, which doesn't agree with a half-oscillation solution and fails to conserve energy. Instead, we would expect the maximum amplitude to be the same as its initial amplitude, 2*mu_s*m*g/k. It is possible that your solution considers an arbitrary number of back-and-forth oscillations, which might manifest as some kind of geometric series, but from what I've skimmed, I don't see that reasoning in there. The coefficient of static friction that was linear in part (a) turns quadratic in part (b) and it's difficult for me to reconcile the discrepancy.

My own solution is X = 2*m*g/k * ( 2*mu_s - mu_k) = 6.4 mm. This was obtained from conservation of energy principles, demanding that the initial potential energy of the spring plus the (negative) external work performed by friction be equal to the final potential energy of the spring. In equation form:
1/2*k*x_0^2 - mu_k*m*g*(x_0+x_1) = 1/2*k*x_1^2
x_0 = 2*mu_s*m*g/k

where x_0 is the initial displacement of the spring and x_1 is the final displacement, assumed to be past the equilibrium position. This turned into a rather messy application of the quadratic formula and the simplicity of my solution led me to believe either there's an easier way to approach this or I'd made a mistake along the way. It's necessary to confirm that the x_1 overshoot is less than the maximum static displacement found in part (a); otherwise, the oscillations will continue until the static friction is enough to overcome the spring force to bring the block to rest.

The fact that as mu_k approaches zero, we retrieve x_1 = x_0 gives me reason to trust my own answer. On the other hand, I'm still struggling to find an easier approach to the problem that avoids direct application of the quadratic formula, plus I'm confused about the physical interpretation of x_1 approaching zero as mu_k approaches mu_s. It may be a "coincidence".

I've included a screenshot of my own work below for your inspection. Any thoughts you can provide would be welcome. Thank you!

Hello, and thank you for this thorough analysis of the problem. Firstly, thank you for showing the shortcoming of my explanation: you're quite right that "If the spring and friction forces were opposite one another, the position where they are equal would only demarcate where the block is not accelerating, not where its velocity is zero." I should have explained that while the spring and friction forces will be equal in magnitude many times during oscillation, it's the final resting position that is of concern here. The final resting position will not be at x=0x = 0, but instead will be at a position where the kinetic friction force equals the restoring force of the spring and there is the special case that the velocity is zero. Velocity is zero twice per oscillation when the mass is at the furthest position from equilibrium. This infinitesimally small moment in time is small enough that it's still kinetic friction in play, not static. Usually the restoring force of the spring will be greater than kinetic friction at this point and the mass will change direction and continue oscillating, but, at long last, there will come a point since the amplitude is gradually decreasing as total mechanical energy of the system is dissipated where the restoring force just equals the force of kinetic friction in magnitude (but opposite direction in this position), and the mass will no longer move. The elastic potential energy of the mass at this final position is the difference between the initial elastic potential energy and the total energy dissipated by kinetic friction during all of the oscillations. I make this point at 4:03 in the video. The total energy dissipated by kinetic friction is proportional to the total distance traveled and we can solve for that distance.

You almost wrote the accounting of energy in:
1/2*k*x_0^2 - mu_k*m*g*(x_0+x_1) = 1/2*k*x_1^2
but the error is in (x_0 + x_1) in the second term. That factor should be the total distance traveled instead of the sum of the final and initial amplitudes.

The textbook actually derives a formula for us in Example 16.7, and it can be used to verify that the solution for part (b) is correct.

Thank you again for your comments, and all the best,
Shaun