Stretched spring -find coeff of friction

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Homework Help Overview

The problem involves a spring with a known spring constant that is stretched and then releases a block across a rough surface. The objective is to determine the coefficient of kinetic friction between the block and the floor based on the spring's behavior and the forces acting on the block.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss breaking down the problem into segments, considering forces, accelerations, and the use of differential equations to analyze motion.
  • Some participants mention using the work-energy theorem to relate initial and final energies to the work done by friction.
  • There are attempts to equate forces from the spring and friction, leading to confusion about static versus kinetic friction.
  • Questions arise regarding the conversion of forces and energies into consistent units for analysis.

Discussion Status

The discussion is active, with various approaches being explored. Some participants have provided guidance on using energy principles, while others are questioning their assumptions about static and kinetic friction. There is no explicit consensus, but multiple interpretations and methods are being considered.

Contextual Notes

Participants are navigating the complexities of applying different friction coefficients and ensuring consistent units across equations. There is an acknowledgment of the need to clarify the distinction between static and kinetic friction in the context of the problem.

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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.
 
Last edited:
  • #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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