How Does a Spring's Compression Relate to Work and Energy in Physics Problems?

In summary, the problem involves a spring with a constant of 18 N/cm and a mass of 0.5 kg being compressed 7 cm and then released. It then skids across a frictional surface with a length of 1.6 m and a coefficient of friction of 0.17 before compressing a second spring with a constant of 5 N/cm. The goal is to find the distance the second spring will compress in order to bring the mass to a stop. Using the equations for potential and kinetic energy, as well as work done by non-conservative forces, it is possible to solve for the distance by setting the initial potential energy equal to the final potential energy and subtracting the work done by non-conservative forces
  • #1
Mathieu
8
0

Homework Statement


A spring of constant 18 N/cm is compressed a distance 7cm by a 0.5 kg mass, then released. It skids over a frictional surface of length 1.6m with coefficient of friction 0.17, then compresses the second spring of constant 5 N/cm. the acceleration of gravity is 9.8m/s^2


Homework Equations


PE=1/2kx^2 --- potential energy = 1/2 kinetic length ^2
KE=1/2mv^2 --- kinetic energy = 1/2 mass velocity ^2
W=Fd ---work = force distance



The Attempt at a Solution


i converted all the cm to m.. then plugged them into EP=1/2kx^2.. so
PE=1/2(.18)(0.07)^2
PE=0.000441J
but this already didn't look right, however, I still used it in W=Fd.
W = 0.000441*1.6
W = 0.000705600
now this REALLY isn't looking right, and I really have NO idea what I'm doing, except getting random numbers. As much as I love physics (really, I use to want to be a physicist) i have never really understood what I'm doing as far as these harder problems go. If anyone could help me, that would be amazing. thanks!


Mathieu
 
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  • #2
Mathieu said:

Homework Statement


A spring of constant 18 N/cm is compressed a distance 7cm by a 0.5 kg mass, then released. It skids over a frictional surface of length 1.6m with coefficient of friction 0.17, then compresses the second spring of constant 5 N/cm. the acceleration of gravity is 9.8m/s^2

Homework Equations


PE=1/2kx^2 --- potential energy = 1/2 kinetic length ^2
KE=1/2mv^2 --- kinetic energy = 1/2 mass velocity ^2
W=Fd ---work = force distance

The Attempt at a Solution


i converted all the cm to m.. then plugged them into EP=1/2kx^2.. so
PE=1/2(.18)(0.07)^2
PE=0.000441J
but this already didn't look right, however, I still used it in W=Fd.
W = 0.000441*1.6
W = 0.000705600
now this REALLY isn't looking right, and I really have NO idea what I'm doing, except getting random numbers. As much as I love physics (really, I use to want to be a physicist) i have never really understood what I'm doing as far as these harder problems go. If anyone could help me, that would be amazing. thanks!Mathieu

Are you solving for energy lost by friction from the initial position (first compressed spring) to the final position (second compressed spring)? Is the mass connected to the two springs, or is the mass launched off the first spring and into the second spring? In you're attempt at a solution, you seem to substitute PE for F in the work equation. When dealing with energy, I think you use:
W = [tex]\Delta[/tex]KE = -[tex]\Delta[/tex]PE = [tex]\int[/tex][tex]\vec{F}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex], because [tex]\vec{F_{spring}}[/tex]= -k[tex]\vec{x}[/tex]. If the mass is launched, then the initial spring energy is converted into an initial kinetic energy of the mass (-[tex]\Delta[/tex][tex]SE_{i}[/tex] = [tex]\Delta[/tex][tex]KE_{i}[/tex]). Some of that initial kinetic energy is lost as frictional energy, and the mass collides with the second spring with a final kinetic energy ([tex]KE_{i}[/tex] - FE = [tex]KE_{f}[/tex]). The final kinetic energy is converted into spring energy, and the mass stops (-[tex]\Delta[/tex][tex]KE_{f}[/tex]= [tex]\Delta[/tex][tex]SE_{f}[/tex]). I'm not sure what you're solving for, but I think this is how you would solve for energy lost. Feel free to correct me.
 
  • #3
I'm sorry..i didn't post the actual question...i'm so dumb haha its:

How far will the second spring compress in order to bring the mass to a stop? answer in units of cm.

and the mass isn't connected, the first spring launches it across the surface to the second spring
 
  • #4
Hi Mathieu, welcome to PF.
Check your conversions. 1 cm = 0.01 m.
Then 18 N/cm = ... N/m?
 
  • #5
0.18 N/m...correct?
 
  • #6
The spring used in the problem requires 18 N force to stretch 1 cm. How much force is needed to stretch the same spring through 1 m?
 
  • #7
ki = 1800 N/m
kf = 500 N/m
xi = 0.07 m
m = 0.5 kg
r = 1.6 m
[tex]\mu_{k}[/tex] = 0.17
g = 9.81 m/s2

-[tex]\Delta[/tex][tex]SE_{i}[/tex] = [tex]\Delta[/tex][tex]KE_{i}[/tex]
-0.5kixi = 0.5mvi^2

[tex]KE_{i}[/tex] - FE = [tex]KE_{f}[/tex]
0.5mvi^2 - [tex]\mu_{k}[/tex]mgr = 0.5mvf^2

-[tex]\Delta[/tex][tex]KE_{f}[/tex]= [tex]\Delta[/tex][tex]SE_{f}[/tex]
-0.5mvf^2 = 0.5kfxf

Substituting the KEs for PEs into the middle equation, we get
-0.5kixi - [tex]\mu_{k}[/tex]mgr = -0.5kfxf
xf = (kixi + 2[tex]\mu_{k}[/tex]mgr)/kf
xf is how far the second spring will compress in order to bring the mass to a stop, and is in m, so convert to cm.
 
  • #8
Hey everyone, I'd thought I'd let everyone know that I did figure this out, our teacher conveniently forgot to teach us the equations for work done by non-conservative forces, so after learning that, it just randomly occurred to me how to solve it.

The short version is this, take PE of the first spring, make that me initial energy in the Work done by non-conservative forces equation (Eo-Wnc=E) and take PE of the second spring and make that the final energy, so basically PE1 - Wnc = Pe2

I explain in more detail here:

PE=1/2kv^2, and Eo-Wnc=E
(Wnc is work of nonconservative forces, Eo is starting energy and E is...final energy)

Because of the law of conservation of energy, I thought I could do this:

PE = E

Or rather, because there are two springs:

PE1 = Eo
E = PE2

So what I did was this:

PE1=1/2k1x2
PE1=1/2(1800)(0.07)2
PE = 4.41J

PE = Eo

Eo-Wnc = E

PE - Wnc = E

PE - (mgμk*d) = E (Because W=Fd and F is = to Nμk (normal force mu k)

4.41 - (0.5*9.8*0.17*1.67) = PE2

3.01889 = PE2

PE = 1/2 kx2

3.01889 = 1/2(500)x2

2(3.01889) = 500x2

0.123088 = x2

X = 0.109888852937866

now because I converted everything to meters and the answer has to be in centimeters, the answer would be 10.9888852937866cm.

And I'm proud to say I figured it all out by myself ^^ hahaha i feel smart now :)THANK YOU EVERY FOR YOUR HELP!

I really appreciate it!Mathieu
 
Last edited:

Related to How Does a Spring's Compression Relate to Work and Energy in Physics Problems?

1. How do springs store and release energy?

Springs work by storing potential energy when they are stretched or compressed. This energy is stored in the bonds between the atoms in the spring, giving it potential to do work. When the spring is released, this potential energy is converted into kinetic energy, causing the spring to move back to its original state.

2. What factors affect the amount of energy stored in a spring?

The amount of energy stored in a spring is affected by its stiffness, or spring constant, and the distance it is stretched or compressed. A stiffer spring or a greater distance of compression or stretch will result in more potential energy being stored.

3. How is the potential energy of a spring calculated?

The potential energy of a spring can be calculated using the equation U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the distance the spring is stretched or compressed.

4. Can springs be used to do work?

Yes, springs can be used to do work. When a spring is stretched or compressed, it has the potential to do work. This potential energy can be harnessed and used to power various devices, such as clocks, toys, and even vehicles.

5. How are springs used in everyday life?

Springs are used in a variety of everyday objects, such as mattresses, trampolines, car suspensions, and pens. They are also used in many machines and devices to control motion and absorb energy, such as in shock absorbers and door hinges.

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