This is a question that was on my Physics quiz last week, kids in my class solved it in two difference ways (to be explained in a sec), which result in two different answers that are off by a factor of two. No one (including pretty much every physics teacher at my school) has found a flaw in either method... but there can't be two correct answers, right? He ended up giving credit for both answers because no one could come to a consensus about which was correct, but uhh... we still want to know which is.

 

 

 

 

 

 

 

anyway...

 

 

 

 

 

 

 

question:

 

 

 

A .850 kg block is attached to a very light spring (k = 180 N/m) as shown in diagram (Diagram depicts a horizontal spring attached to a wall on one end and a block on the other, the block is also resting on a horizontal surface). When the spring block system is compressed 5 cm, and then released, it stretches out 3.2cm beyond the equilibrium position before stopping and turning back. What is the u (coefficient of friction) of the block and table.

 

 

 

 

 

 

 

 

 

 

Solution 1: Potential Energies

 

 

 

 

 

 

 

Initial potential energy of compressed spring (position 1):

 

 

 

PE=(1/2)kx^2 = (1/2)(180N/m)(.05m)^2 = .225 J

 

 

 

 

 

 

 

Instantaneous potential energy of spring when stretched to 3.2cm (position 2):

 

 

 

PE=(1/2)kx^2 = (1/2)(180N/m)(.032m)^2 = .09216 J

 

 

 

 

 

 

 

Energy lost between positions 1 and 2 is work done by friction over the 8.2cm:

 

 

 

Wf = .225J - .09216J = .13284 J

 

 

 

 

 

 

 

Calculate force of friction:

 

 

 

 

 

 

 

Wf = Ff* d

 

 

 

.13284J = Ff * .082m

 

 

 

 

 

 

 

Ff = 1.62 N

 

 

 

 

 

 

 

Calculate coefficient of friction:

 

 

 

 

 

 

 

Ff = Fn * u

 

 

 

 

 

 

 

1.62 N = (.850 kg)(9.8 m/s^2) * u

 

 

 

 

 

 

 

u = .194

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 2: Forces

 

 

 

 

 

 

 

Force required to compress spring to 5cm:

 

 

 

 

 

 

 

Fp = k*x = (180 N/m)(.05m) = 9 N

 

 

 

 

 

 

 

Force required to stretch spring to 3.2cm past equilibrium:

 

 

 

 

 

 

 

Fp = k*x = (180 N/m)(.032m) = 5.76 N

 

 

 

 

 

 

 

Difference = force applied by friction ?

 

 

 

 

 

 

 

Ff = 9 N - 5.76 N = 3.24 N

 

 

 

 

 

 

 

Calculate coefficient of friction:

 

 

 

 

 

 

 

Ff = Fn * u

 

 

 

 

 

 

 

3.24 N = (.850 kg)(9.8 m/s^2) * u

 

 

 

 

 

 

 

u = .389

 

 

 

 

 

 

 

I do not know which of these two methods is correct (or that either is correct for that matter), but i can conclude that the coefficient of friction is indeed a constant and does not have two values, so at least one of these methods is incorrect.

 

 

 

 

 

 

 

Can anyone provide any reason why one or both methods is wrong, and if indeed both are wrong, provide a proper method?

You lost me.

My younger brother is doing a physics degree right now, he might look at this topic in a bit.

OK I've spent a little while looking stuff up online and in various books, and i think i might have the solution but i would still like someone to confirm this and explain it to me.

 

 

 

 

 

 

 

This is what I've come up with.

 

 

 

the F= kx equation, Hooke's law as it is called, applies to the force required to hold the spring compressed or stretched. That force however, is not applied over the entire distance it travels towards and beyond equilibrium... which is what the calculations are assuming? The actual force decreases as the spring approaches equilibrium, ultimately reaching zero. So the actual "average" force applied over that distance is indeed half (?) that value that my above explained calculations, which would explain why the calculation is off by a factor of two? Therefore the energy method is correct and the force method is incorrect because of the above stated reason?

 

 

 

 

 

 

 

Confirmation and/or further explanation would be appreciated.

:uhh: Oh geez, I'm going to have to learn this stuff to become a naval architect...

 

 

 

 

 

 

 

*Is scared*

*sob* *sob*

 

 

 

 

 

 

 

this hurts my head....

 

 

 

 

 

 

 

what type of quiz was this on.. And what kind of teacher doesnt know the correct answer to a quiz he sets?

*sob* *sob*

 

 

 

 

 

 

 

this hurts my head....

 

 

 

 

 

 

 

what type of quiz was this on.. And what kind of teacher doesnt know the correct answer to a quiz he sets?

 

 

 

 

 

 

 

the kind that makes his own questions rather than taking boring questions out of books that every teacher in the galaxy uses?

Thats some insane math...your gonna have to search har 2 find someone that knows that. :ohnoes:

It's been about 6 months since I've done my last Physics problem, but isn't Hooke's Law F = -kx?

 

 

 

 

 

 

 

Secondly, when doing a problem like this, isn't it assumed that the friction of the spring itself is minimal so it is negelcted? You're looking for the force of friction exerted on the block and to find that you would have to use the potential energy equation because Hooke's Law would give you a force of friction for the spring itself?

 

 

 

 

 

 

 

I have to bust out my Physics book but this is just off the top of my head.

It's been about 6 months since I've done my last Physics problem, but isn't Hooke's Law F = -kx?

 

 

 

 

 

 

 

Secondly, when doing a problem like this, isn't it assumed that the friction of the spring itself is minimal so it is negelcted? You're looking for the force of friction exerted on the block and to find that you would have to use the potential energy equation because Hooke's Law would give you a force of friction for the spring itself?

 

 

 

 

 

 

 

I have to bust out my Physics book but this is just off the top of my head.

 

 

 

 

 

 

 

yes Hooke's law is F = -kx

 

 

 

where F is the restoring force of the spring, which is negative because its acting in the opposite direction to the force compressing or stretching the spring (which is of equal magnitude).

 

 

 

 

 

 

 

I think the idea in the calculations was that the restoring force of the initial compressed spring becomes the stretching force once it passes equilibrium (two forces in the same direction, so the negative sign becomes irrelevant), although as i stated above, that logic seems to be faulty because that force ultimately reaches zero at equilibrium.

 

It's been about 6 months since I've done my last Physics problem, but isn't Hooke's Law F = -kx?

 

 

 

 

 

 

 

Secondly, when doing a problem like this, isn't it assumed that the friction of the spring itself is minimal so it is negelcted? You're looking for the force of friction exerted on the block and to find that you would have to use the potential energy equation because Hooke's Law would give you a force of friction for the spring itself?

 

 

 

 

 

 

 

I have to bust out my Physics book but this is just off the top of my head.

 

 

 

 

 

 

 

yes Hooke's law is F = -kx

 

 

 

where F is the restoring force of the spring, which is negative because its acting in the opposite direction to the force compressing or stretching the spring (which is of equal magnitude).

 

 

 

 

 

 

 

I think the idea in the calculations was that the restoring force of the initial compressed spring becomes the stretching force once it passes equilibrium (two forces in the same direction, so the negative sign becomes irrelevant), although as i stated above, that logic seems to be faulty because that force ultimately reaches zero at equilibrium.

 

 

 

 

 

 

 

I think your explanation that you found seems reasonable. The potential energy equation would have to be used because it is calculating the force of the block for the entire length of time. I recall using the potential energy equation for solving these pretty often, never Hooke's law unless I had to find an unknown variable.

Ha, this is funny. I literally just had an exam on Springs, Friction, and Gravity. I'm in college mind you, but we never really got much into compression for some reason. Anyway here's what I see that's amiss, the first calculation in each:

 

 

 

 

 

 

 

-Initial potential energy of 5cm compressed spring:

 

 

 

PE=(1/2)kx^2 = (1/2)(180N/m)(.05m)^2 = .225 J

 

 

 

 

 

 

 

-Force required to compress spring to 5cm:

 

 

 

Fp = k*x = (180 N/m)(.05m) = 9 N

 

 

 

 

 

 

 

Something has to be wrong there. The rest of the calculations check out. But just what exactly... I can't put my finger on it :?

 

 

 

 

 

 

 

Lmao I 2x that..I'm so confused. :shock:

Pker_dude_jr, is that problem from the Physics textbook by Douglas Giancoli? I just did a problem exactly like that last week :lol:

Pker_dude_jr, is that problem from the Physics textbook by Douglas Giancoli? I just did a problem exactly like that last week :lol:

 

 

 

 

 

 

 

that would be funny, cuz we do use the Giancoli book.. but my teacher claims he made the problem up.. if it came from a book he wouldn't be disputing the answer... Nevertheless ill flip through the text and see if i can find a similar problem

Thought about it for a bit, and I think the reasoning behind the forces was slightly off. I compared this to a similar situation where I move a block with a first force a certain distance, stop, then move it again further with a second force, stopping again then. After I stopped the first time, friction was done during that first movement. When I started moving again, friction acted only on that second force, not the first one or both combined.

 

 

 

 

 

 

 

The two spring forces acting on the block (force1 from first position to 0, and force2 from 0 to second position [0 being the equilibrium point]) both have fricton acting upon each separately. Since this is the case, you'd require two friction forces in the equation instead of just one. You woud only use one friction force in the equation if both spring forces acted at the same time on the block (which would be the case if two springs were pushing on the single block simultaneously). This would make the solution obtained by using forces equal the solution from energy equations:

 

 

 

 

 

 

 

Fnet = Fs1 + Fs2 - Ff1 - Ff2 = 0

 

 

 

 

 

 

 

Ff1 = Ff2 = Ff, since mass, g, and u are all constant

 

 

 

 

 

 

 

Fs1 + Fs2 = 2Ff

 

 

 

 

 

 

 

3.24 N = (2)(.850 kg)(9.8 m/s^2) * u

 

 

 

 

 

 

 

Solving this last equation gives u = 0.194 since now you are deviding by 2 also. You have to count the force of friction twice since both spring forces to not act during the same displacement interval.

 

 

 

 

 

 

 

In the energy equations you were allowed to use only the first and last positions because changes in energy within the block did not dipend on the path taken, only on the final and initial states (IE: you could have moved the block back and worth a million times, put it in the same final spot, and have the energy equations give the same answer).

 

 

 

 

 

 

 

This is the only reasoning I could come up with, but couldn't think of anything else.

OK I've spent a little while looking stuff up online and in various books, and i think i might have the solution but i would still like someone to confirm this and explain it to me.

 

 

 

 

 

 

 

This is what I've come up with.

 

 

 

the F= kx equation, Hooke's law as it is called, applies to the force required to hold the spring compressed or stretched. That force however, is not applied over the entire distance it travels towards and beyond equilibrium... which is what the calculations are assuming? The actual force decreases as the spring approaches equilibrium, ultimately reaching zero. So the actual "average" force applied over that distance is indeed half (?) that value that my above explained calculations, which would explain why the calculation is off by a factor of two? Therefore the energy method is correct and the force method is incorrect because of the above stated reason?

 

 

 

 

 

 

 

Confirmation and/or further explanation would be appreciated.

 

 

 

 

 

 

 

I'd say that's your answer right there.

 

 

 

 

 

 

 

(The force from the coil on the block is first +, then 0 at the equilibrium, then - once it has past the equilibrium. The friction remains a - force throughout the entire movement, but it's actual size depends on the velocity of the block, which depends on the time and acceleration, and the acceleration depends on the distance to the equilibrium...)

:-k ............. #-o :wall: ,

 

 

 

sorry im bad at this, im the type that studies the most important stuff 2 days before the exam or the night before, cheers!

i would say that the energy method is most likely correct, because a spring problem is so easily solved with energy that it's a much higher probability of having the correct answer.

 

 

 

 

 

 

 

ive had enough of physics that i dont really care do work it out either way myself, but the only time i can ever remember using forces to work a spring problem, it involved rotation, a pulley with mass, and a few torques.

the best way to solve this problem is by using energy. You are quite correct in noting that the force isn't constant throughtout the entire thing. Actually, the force -kx is certainly constant, but note the negative sign. Friction applies, thus the net force is the changing force (hence the deceleration and then the bouncing back). However, it is the rebound part that is hard to calculate (I should think so; you college dudes correct me if I'm wrong), so you would probably use potential energy formulas. (not sure which one you're using there...)