- #1

leggythegoose

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- Homework Statement
- A 2.60 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m. The spring has force constant 855 N/m. The coefficient of kinetic friction between the floor and the block is 0.450. 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 0.0100 m from its initial position? (At this point the spring is compressed 0.0200 m.)

- Relevant Equations
- K=1/2mv^2

U= 1/2kx^2

So here's what I did but it isn't right:

W = (Kf + Uf) - (Ki + Ui)

(2.6)(9.81)(0.45)(-0.01)=(1/2mvf^2 + 1/2kxf^2) - (1/2mvi^2 + 1/2kxi^2)

-0.1 = (1/2(2.6)(vf^2) + 1/2(855)(0.02^2)) - (1/2(855)(0.03^2))

1.3Vf^2 = 0.114

Vf^2 = 0.09

Vf = 0.3 m/s

W = (Kf + Uf) - (Ki + Ui)

(2.6)(9.81)(0.45)(-0.01)=(1/2mvf^2 + 1/2kxf^2) - (1/2mvi^2 + 1/2kxi^2)

-0.1 = (1/2(2.6)(vf^2) + 1/2(855)(0.02^2)) - (1/2(855)(0.03^2))

1.3Vf^2 = 0.114

Vf^2 = 0.09

Vf = 0.3 m/s