# Homework Help: To find out the compression of the spring

1. Jun 15, 2012

### icecool8

1. The problem statement, all variables and given/known data
Hi Guys, I am trying to solve a classic problem involves conservation energy, spring, compression, I got my attempt and answer and I think my step was correct but the answer is different with the solution, can any one please have look for me, any suggestions will be really appreciated.
Here is the question:
2kg block is at rest at the top of a inclined plane, spring constant is 100N/m is at the bottom of the incline, the equilibrium position of the spring is 4m from the block's position, the θ=30, the block is released at some instant in time. find out the compression of the spring if the kinetic coefficient of friction is 0.1

2. Relevant equations
ΔK+ΔU=-fd
Vf=√(Vi+2aΔx)
g=9.8

3. The attempt at a solution
n=mgcosθ, therefore f=μN=0.1*2*9.8*cos30=1.6974
ƩFx=mgsinθ-f=ma
a=(2*9.8*0.5-1.6974)/2=4.0513
Vf=√0+2*4.0513*4=5.69
then ΔK+ΔU=-fd (because the block slided down and contacted to the spring, the Vf become Vi, the Vf become 0)
(Kf+Uf)+(Ki+Ui)=-fd
0+0.5*100*x^2-0.5*2*5.69^2-2*9.8*sin30*x^2=-1.6974*4
x=0.797, but the answer in the solution sheet is 0.933m,
I don't know where I went wrong....

2. Jun 15, 2012

### BruceW

welcome to physicsforums, icecool8! hmm, before I look through the working, I better check a few things: The equilibrium position is 4m from the block's position, does that mean the equilibrium position is 4m down the slope from the block? Also, is the spring attached to the block from the start? Or does the block only come into contact with the spring at the equilibrium position?

3. Jun 15, 2012

### PhanthomJay

Welcome to PF!
It looks like you are using both the kinematic and energy methods and mixing up your approach. You could forget the kinematic method and just use the energy method from start to finish. Note that your gravitational potential energy change is incorrect. It is not mgsintheta(x^2); and the work done by friction in your last equation does not act over a distance of 4 m, it acts over x meters.

4. Jun 15, 2012

### icecool8

Hi BruceW,
yes, the equilibrium position is 4m down the slop from the block, the spring is unstretched(assume it is massless) it is not attached to the block from the start, it is at the bottom of the inclined plane, and block slides down then contact to the spring, the speed of the block is zero when the block compresses the spring.

5. Jun 15, 2012

### icecool8

Hi PhanthomJay, Thanks for that,
:surprised,thats right, once the block contacted the spring, it should only act over x meter ,not the 4m, Thanks for that
My understanding for the nonconservative force such as friction are involved is total initial and final energy=decrease in mechanical energy which is the friction in this case...and is the energy = potential energy +kinetic?

6. Jun 15, 2012

### azizlwl

Find the energy when the block just barely touches the spring, kinetic and potential(with reference to final compression)
This energy will compress the spring.
Remember when compression is carried out, take account of friction too.

7. Jun 15, 2012

### PhanthomJay

You have the correct energy equation when non conservative forces act, W_nc = ΔK+ΔU. Your '2 step' approach using the kinematic equations and Newton's laws for the first part of the motion, and the energy approach for the second part of the motion, is not necessary, but it is nevertheless OK. So after you make the correction for the friction force distance using 'x' instead of '4 m', the other correction you need to make relates to the change in the gravitational potential energy, whch is a function of x, not x^2 as you have written.

8. Jun 16, 2012