# Suppose That You Have a Mass of 457 Kg

Suppose That You Have a Mass of 457 Kg

Question

Suppose y’all have a 0.750-kg object on a horizontal surface connected to a spring that has a forcefulness constant of 150 N/m. In that location is simple friction betwixt the object and surface with a static coefficient of friction $\mu_s = 0.100$. (a) How far tin can the leap 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 full distance does it travel before stopping? Assume it starts at the maximum amplitude.

- $4.91 \textrm{ mm}$
- $ix.46 \textrm{ mm}$

Solution Video

## OpenStax College Physics Solution, Chapter sixteen, Problem 45 (Bug & Exercises) (8:07)

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Video Transcript

This is College Physics Answers with Shaun Dychko. On a horizontal surface we have this mass, chiliad, connected to a jump and information technology’s billowy back and forth. But it’south dampened unproblematic harmonic motion considering 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 information technology is at balance the coefficient of static friction is 0.1. Nosotros’re asked to figure out what is the maximum corporeality the spring could be stretched and have the mass at residual and have it non bounciness dorsum. Well that is going to be the bound constant times the amplitude, is going to equal the gravity which is the normal strength that the mass experiences because strength of friction is normal strength times the coefficient of friction, be it static or kinetic depending on whether the mass is not moving or moving. The normal strength is going to be the weight

*m yard*. And so that’s where this

*m k*

comes from. It’south being multiplied by

*mu south*

which is the static coefficient of friction because we’re told that the mass should not be moving. Then nosotros have to solve for this amplitude

*A one*. I gave it a number or subscript i because we’re going to use this value in our subsequent calculation in part B. Carve up both sides by

*chiliad*

to solve for

*A 1*

and information technology’s going to be

*chiliad one thousand mu s*

over

*k*. So that’southward 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. Then if you slowly stretch this mass 4.91 millimeters from the equilibrium position or the un-stretched position of the spring, and you let information technology go, it will not jump dorsum because the static friction forcefulness will equal the spring force up until this point. Okay. Now in part B, nosotros’re told that — let’s suppose the initial amplitude is going to be two times what we figured out in part A, that’s ii times

*A one*

in which case information technology is two times

*m g mu s*

over

*chiliad*

which I wrote here. So the mass is moved to this position two times

*A ane*

and then you let information technology go, in which case it will jump back because the leap strength will exist greater than the static friction force at that point. Since it’s going to exist moving the kinetic friction is what’s going to be dampening the oscillation and information technology’southward going to exist dissipating some of the free energy up until the point where the bound forcefulness equals the kinetic friction force. That is the kinetic friction strength where you take

*mu k*

and not

*mu s*. Okay. Here is where we talk well-nigh the final position of the mass. It’southward going to be at a point where information technology’southward going to stop when the spring force which is the spring constant multiplied by the final amplitude, equals the kinetic friction strength. So that’south the coefficient of kinetic friction times the normal force which equals

*one thousand k*. We’ll split both sides by

*grand*

to solve for

*A f*. So the final amplitude when this thing stops oscillating at a certain distance from the un-stretched length of jump, and we’ll call that amplitude final,

*A f*, is

*mu k m chiliad*

over

*k*. The total initial elastic potential energy of the mass, or the system I should say, is one one-half

*k*

times the initial aamplitude squared. Nosotros’re going to dissipate some of that energy due to friction and this is friction strength multiplied by distance. This is the total distance that the mass travels as information technology’southward sliding back and forth, back and forth, dorsum and along. One time this energy is dissipated we’ll be left over with the elastic potential energy when the mass comes to rest at this concluding aamplitude. Okay. Then we substitute for

*A i*

and

*A f*. So

*A initial*, amplitude initial is two

*thou yard mu due south*

over

*thousand*. And so we substitute that in, and aamplitude final is

*mu grand one thousand g*

over

*k*. We square that and we get that from here. So it’s a bunch of algebra to solve for

*d*. This two squared becomes 4 just divided past two makes two in the cease, times

*m*

squared

*chiliad*

squared

*mu s*

squared, all over

*thousand*

squared times

*g*

makes

*k*

to the ability of i in the denominator, minus

*mu k grand g d*

equals

*mu k*

squared

*m*

squared

*g*

squared over ii and so over

*k*

to the power of one because it’southward

*k*

squared in the denominator times

*k*

makes

*k*

to the one down there. And then carve up everything past

*m k*

or multiply everything by one over

*one thousand g*, the way I really like to say it. That makes two

*m*

to the ability of one,

*g*

to the ability of one, coefficient of static friction squared over

*k*, minus

*mu one thousand d*

equals

*mu chiliad*

squared

*thou g*

each to the power of one, over 2

*thou*. And then we add

*mu g d*

to both sides because we want to isolate

*d*

on 1 side of the equation. Nosotros too subtract

*mu k*

squared

*m g*

over two

*one thousand*

from both sides. And then switch the sides around and we end upwards with

*mu k d*

equals two

*m m mu s*

squared over

*k*

minus

*mu k*

squared

*m g*

over two

*k*. And so the

*chiliad g*

over

*thousand*

is a common factor betwixt these two terms so nosotros factor that out to get this line here. Then nosotros dissever both sides by the coefficient of kinetic friction. The terminal formula we’re going to apply is that

*d*

equals the full distance that the mass slides back and along such that the kinetic friction forcefulness will dissipate an amount of free energy enough to reduce the initial elastic potential to this final rubberband potential free energy. This is going to be

*yard g*

over

*mu grand*

times the spring abiding, times ii times the coefficient of static friction squared minus coefficient of kinetic friction squared over ii. There we become! Then

*d*

is 0.75 kilograms times nine.81 meters per second squared, over 0.085 times 150 newtons per meter, times ii times 0.1 squared, minus 0.085 squared over two. This works out to ix.46 millimeters. Now this formula is non something that we can apply intuition to verify very easily considering it’south a very foreign-looking formula. What we practise wait though is that this number should be greater than the respond for part A and it doesn’t tell us if we’re for certain correct or not, but at to the lowest degree the fact that it’due south bigger than this number indicates that it could be right and I think that it is. In that location!

## Solutions for problems in chapter 16

### Suppose That You Have a Mass of 457 Kg

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