Spring problem using work energy theorem

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Homework Help Overview

The problem involves a block attached to a spring on a horizontal surface, where the spring is initially compressed. The scenario includes considerations of kinetic friction and the work-energy theorem to determine the speed of the block after it has moved a certain distance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the work-energy theorem, particularly the calculation of work done by friction and its sign. Questions arise regarding the nature of work done by friction and its impact on the energy balance.

Discussion Status

Participants are actively engaging with the problem, correcting each other's understanding of the work done by friction. There is recognition of errors in calculations, and some guidance is provided regarding the signs of work done by friction.

Contextual Notes

There are indications of confusion regarding the application of the work-energy theorem and the correct setup of the energy equations, particularly in relation to the initial and final states of the system.

Garrit
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Homework Statement


A 2.90 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0360 m . The spring has force constant 860 N/m . The coefficient of kinetic friction between the floor and the block is 0.35 . The block and spring are released from rest and the block slides along the floor.

What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0160 m)

Homework Equations


I'm using the equation Work( of the nonconservative forces) = Eb - Ea where Eb equals the final kinetic energy plus the final potential energy and Ea equals the initial kinetic energy plus the initial potential energy. I'm calculating kinetic energy using (1/2)mv^2 and for potential energy (1/2)kx^2.

The Attempt at a Solution


My attempt at it:

The work done by friction should equal (mass)(gravity)(coefficient of friction)(displacement) which would equal 2.9(9.8)(0.35)(0.02)

Now for the other side of the equation:

(1/2)(m)vf^2 + (1/2)k(x2)^2 - ( (1/2)(m)vi^2 + 1/2k(x1)^2 )

So this is what I get:

2.9(9.8)(0.35)(0.02) = (1/2)(2.9)(vf)^2 + (1/2)(860)(0.016) - (1/2)(2.9)(0) - (1/2)(860)(0.036)

I punched this into wolfram and got 2. something. I know the answer is 0.414, so I must be doing something wrong. I'm just using the equation I learned in class. Any suggestions?
 
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Is the work done by friction positive or negative?
 
Garrit said:
The work done by friction should equal (mass)(gravity)(coefficient of friction)(displacement) which would equal 2.9(9.8)(0.35)(0.02)
So the friction is adding energy to the block?
 
Ah so the work done by friction should be negative, right? That still doesn't give me 0.414 though. What else is wrong?
 
Garrit said:
1/2k(x1)^2 )

Garrit said:
1/2)(860)(0.036)
Compare.
 
ah man. Even writing out the equations, I still couldn't punch in the numbers right! Thanks so much guys. I made the corrections and got the answer I was looking for!
 

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