Find speed of block using energy and spring equations? Help please

[leggythegoose]

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

 

[berkeman]

Welcome to PF.

The friction makes the motion lossy, so you can't just use energy considerations (KE + PE). You will need to draw the Free Body Diagram (FBD) for the block and account for the force of friction resisting the motion of the block while it is moving...

 

[leggythegoose]

Welcome to PF.

The friction makes the motion lossy, so you can't just use energy considerations (KE + PE). You will need to draw the Free Body Diagram (FBD) for the block and account for the force of friction resisting the motion of the block while it is moving...

Okay thanks! How would I change my equation to include that?

 

[berkeman]

Or I guess put another way, the work done by the retarding friction force over that distance needs to be in your energy equation... The Potential Energy of the spring versus its compressed position is still accurate, but the KE gained by the spring movement is reduced by the work done by friction over that distance. Does that make sense?

 

[TSny]

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

Your work looks correct to me. But you might have too much "round off error". The data is given to 3 significant figures. So, maybe you are expected to express your final answer with 3 significant figures.

 

[berkeman]

(2.6)(9.81)(0.45)(-0.01)

Oh, apologies! I missed that you did include the work due to friction.

 

[kuruman]

Your work looks correct to me. But you might have too much "round off error". The data is given to 3 significant figures. So, maybe you are expected to express your final answer with 3 significant figures.

Yes, my calculation is about 10% off OP's number. That may be too much for a machine-graded answer.
To @leggythegoose :
The best way do such problems is to derive a symbolic expression, of the form $~v_f=\sqrt{2(W_f-\Delta U)/m}~,$ load the given numerical variables and the expression on a spreadsheet and let it rip. The advantage of the spreadsheet is that if the scoring algorithm says that your answer is incorrect, you can easily trace the error to either the input variables or the expression that you put in.