Block on Spring with Friction

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  • #1
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A relaxed spring with spring constant k = 50 N/m is stretched a distance di = 63 cm and held there. A block of mass M = 6 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/10.

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


I know i need to use the Work-Energy theorem.

I'm not really sure what i need to do?

The kinetic energy is 0, but i don't know how to find the work done by friction.

Homework Statement





Homework Equations





The Attempt at a Solution

 
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  • #2
Hootenanny
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http://img48.imageshack.us/img48/7973/showmepley5.gif [Broken]

A relaxed spring with spring constant k = 50 N/m is stretched a distance di = 63 cm and held there. A block of mass M = 6 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/10.

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


I know i need to use the Work-Energy theorem.

I'm not really sure what i need to do?

The kinetic energy is 0, but i don't know how to find the work done by friction.
Welcome to Physics Forums,

You're spot on with using the work-energy theorem. What does that theorem state and how can we apply it to this problem?
 
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  • #3
Doc Al
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The kinetic energy is 0, but i don't know how to find the work done by friction.
What's the definition of work?
 
  • #4
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Thanks,
The work-energy theorem is: change in Kinetic Energy=the work done by friction.
I believe the change in kinetic energy is 0, but its the work done by friction i can't seem to find.

Work is a force over a distance, correct?
So would the work done by friction mean its
(Coefficient of friction-M)(Fnormal)(distance)=Work done by friction?
 
  • #5
Doc Al
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The work-energy theorem is: change in Kinetic Energy=the work done by friction.
Make that the change in total mechanical energy (kinetic energy plus spring potential energy).
Work is a force over a distance, correct?
So would the work done by friction mean its
(Coefficient of friction-M)(Fnormal)(distance)=Work done by friction?
Exactly.
 
  • #6
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So the kinetic energy = 0
Work done by friction = M(58.86N)(.567m)
The potential energy = -Force(x2-x1)???

I'm not too familiar with potential energy yet. Is this the right formula i need to use?
If so what is the force i need? Is it F = -50N/m(x)?
If so what distance will i use, the distance it was initially stretched or the distance it comes to rest at?
 
  • #7
Doc Al
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So the kinetic energy = 0
Work done by friction = M(58.86N)(.567m)
Good.
The potential energy = -Force(x2-x1)???
No. (Look up the formula for the energy stored in a stretched spring. Or derive it yourself.)

Since the force is not constant, you can't just use FΔx.
 
  • #8
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I used Elastic Potential Energy Formula: (1/2)kx^2

So i set it up like this:
Work by Friction_______Potential Energy Final____Potential Energy Initial__Kinetic
M(58.86N)(0.567m) = (1/2)(50N/m)(.063)^2 - (1/2)(50N/m)(.63)^2 + 0

I solved for M = -.29, but its saying this is the wrong answer and i thought i had it right.
 
  • #9
Doc Al
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I used Elastic Potential Energy Formula: (1/2)kx^2
Good.

So i set it up like this:
Work by Friction_______Potential Energy Final____Potential Energy Initial__Kinetic
M(58.86N)(0.567m) = (1/2)(50N/m)(.063)^2 - (1/2)(50N/m)(.63)^2 + 0
The work done by friction is negative, since it opposes the motion.
 
  • #10
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I see, i had the right answer it just needed to be positive and have a more exact answer. It was 0.294! Thanks alot for your help. I'm beginning to understand this now. I wish i had you as my teacher, i don't seem to learn too much from the one i have right now. Thanks again.
 

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