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 $\mu_s = 0.100$. (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 $\mu_k = 0.0850, what total distance does it travel before stopping? Assume it starts at the maximum amplitude.

Final Answer

- $4.91 \textrm{ mm}$
- $9.46 \textrm{ mm}$

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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!