Stretched spring -find coeff of friction

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The discussion revolves around calculating the coefficient of kinetic friction (µk) for a block attached to a spring. The spring constant is given as k = 70 N/m, and the block has a mass of 10 kg. Participants explore the relationship between the spring's potential energy and the work done by friction as the block moves across a rough surface. They emphasize using the work-energy theorem to relate initial and final mechanical energies to find µk, noting the importance of distinguishing between static and kinetic friction coefficients. The conversation highlights the need to equate forces and energies correctly to derive the desired coefficient.
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stretched spring --find coeff of friction

Homework Statement


A relaxed spring with spring constant k = 70 N/m is stretched a distance di = 66 cm and held there. A block of mass M = 10 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/7


Homework Equations


7What is the coefficient of kinetic friction µk between the block and the floor?


The Attempt at a Solution


This problem is a problem for me.
i know about spring constants and friction but can't visualize where to start?

I know:
M= 10kg
df = .094m
Fs = .09 * 70 = 6.3N
N = 9.8*10= 98N

dont know
µk (µk = Fk/N)

Also i know (obviously) that block stops because µk force must be greater than the spring constant force.

thanks for any help
 
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Fun problem. I don't know the answer right off the bat, but the way I'd approach a problem like this is to initially break it into understandable segments. Like, you know the initial force on the block from the spring and its mass, so what would be the possible accelerations for the block initially based on the mu values? And if you were going to set up a differential equation for the instantaneous velocity of the block based on the things that you are given, what would that equation look like? And given that instantaneous differential equation, how would you integrate or solve for the motion versus time, based on the mu value? And given that and the answer that you are given for the final V=0 position, how can you then solve for the mu?
 
I'd use the work-energy theorem... The final energy - initial energy = work done by non-conservative forces (ie work done by friction).
 
this is what i did but its wrong sumhow.

Fsp = -K delta x
Fk = mu N

so i equated equations

-K delta x = mu N

mu = -K delta s / N

mu = -70(.094) / (10*9.8)

mu = .067

=wrong?
 
mujadeo said:
this is what i did but its wrong sumhow.

Fsp = -K delta x
Fk = mu N

so i equated equations

-K delta x = mu N

mu = -K delta s / N

mu = -70(.094) / (10*9.8)

mu = .067

=wrong?

That gives the coefficient of static friction, not kinetic.
 
coeff of static and kinetic use same formula don't they?
Fk = mu*N
Fs = mu*N
 
oh so whole picture is wrong
i am looking at it like : equate forces, then when they become equal (and opposit) that is mu-max for static.\
should be..
 
mujadeo said:
coeff of static and kinetic use same formula don't they?
Fk = mu*N
Fs = mu*N

There are 2 different mu's... mustatic and mukinetic
 
so is kinetic coeff, --get that from KE=.5mv^2 somehow?
 
  • #10
mujadeo said:
oh so whole picture is wrong
i am looking at it like : equate forces, then when they become equal (and opposit) that is mu-max for static.\
should be..

Yes. To get mu-kinetic, you need to use work-energy...
 
  • #11
mujadeo said:
so is kinetic coeff, --get that from KE=.5mv^2 somehow?

You don't need that... what's the initial mechical energy of the system (includes all kinetic and potential energies)... what's the final mechanical energy of the system... It starts at rest and ends at rest so the kinetic energies are 0. There is kinetic energy in between, but we don't need to worry about that... we just need to worry about the initial and final states...

work done by friction = final mechanical energy - initial mechanical energy

From the left hand side you can get mu-kinetic.
 
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  • #12
OK so i am equating PE of spring with force of spring??

I know its these 3 equations :

U=1/2kx^2
Fsp = -K delta x
Fk = muN

But Fsp and Fk are in in Newtons and U is in Joules.
I want everything in joules to use conserv of energy??

How can i get both equations in J?
 
  • #13
mujadeo said:
OK so i am equating PE of spring with force of spring??

I know its these 3 equations :

U=1/2kx^2
Fsp = -K delta x
Fk = muN

But Fsp and Fk are in in Newtons and U is in Joules.
I want everything in joules to use conserv of energy??

How can i get both equations in J?

Just use the first equation for now... what's the initial energy, and what's the final energy?
 
  • #14
mujadeo said:
OK so i am equating PE of spring with force of spring??

I know its these 3 equations :

U=1/2kx^2
Fsp = -K delta x
Fk = muN

But Fsp and Fk are in in Newtons and U is in Joules.
I want everything in joules to use conserv of energy??

How can i get both equations in J?

Work done = Force * distance... which is N*m which is the same as J.
 
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