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

• leggythegoose
In summary, the block is moving with a force of friction resisting it, which reduces the energy gained by the spring movement.
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

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...

berkeman said:
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?

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?

leggythegoose said:
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.

MatinSAR and leggythegoose
leggythegoose said:
(2.6)(9.81)(0.45)(-0.01)
Oh, apologies! I missed that you did include the work due to friction.

MatinSAR
TSny said:
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.

MatinSAR, jbriggs444 and leggythegoose

## 1. What is the equation for finding the speed of a block using energy and spring equations?

The equation for finding the speed of a block using energy and spring equations is v = √(2kx/m), where v is the speed of the block, k is the spring constant, x is the displacement of the block from its equilibrium position, and m is the mass of the block.

## 2. How do I determine the values for k, x, and m in the equation?

k and x can be determined experimentally by measuring the spring constant and the displacement of the block from its equilibrium position. m can be determined by measuring the mass of the block.

## 3. Can this equation be used for any type of spring?

Yes, this equation can be used for any type of spring as long as the spring constant and displacement are measured accurately.

## 4. What are the units for the values in the equation?

The units for v will be in meters per second (m/s), k will be in Newtons per meter (N/m), x will be in meters (m), and m will be in kilograms (kg).

## 5. Is there a specific method for solving this equation?

There is no specific method for solving this equation, but it is recommended to rearrange the equation to solve for the desired variable, and then plug in the known values to calculate the speed of the block.

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