- #1

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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