# Spring Potential Energy

CLAIMED BY Taraji Long Fall 2019 11/15/2019 This topic covers Spring Potential Energy.

## The Main Idea

Spring potential energy, also known as elastic potential energy,is the stored energy in a spring, that can potentially be converted into kinetic energy. The energy stored in the spring is due to the deformation of the spring, often from stretching and compressing. The force excreted to stretch or compress a spring is known as Hooke's law, **F _{s} = -k_{s}x**, where

**F**is force,

_{s}**x**is the displacement, and

**-k**is the spring constant. The spring constant being unique for every spring depends on factors such as material and thickness of coiled wire. If a spring is not stretched or compressed, then it is at equilibrium. At equilibrium a spring has no potential energy, assuming there is no force being applied to the spring.

_{s}

### A Mathematical Model

The formula for Force of a Spring:

**F _{s} = -k_{s}x**

The formula for Spring Potential Energy:

**U _{s} = ^{1}⁄_{2}k_{s}x^{2}**

where:

**k _{s}** = spring constant

**x ^{2}** = stretch measured from the equilibrium point;

### A Computational Model

**An oscillating spring on the floor:**

GlowScript 2.9 VPython display(width=600,height=600,center=vector(6,0,0),background=color.black) mbox=2 L0 = vector(9,0,0) ks = 1 deltat = .01 t = 0 wall=box(pos=vector(0,1,0),size=vector(0.2,3,2),color=color.cyan) floor=box(pos=vector(7.2,-0.6,0),size=vector(14,0.2,4),color=color.cyan) box=box(pos=vector(12,0,0),size=vector(1,1,1),color=color.red) pivot=vector(0,0,0) spring=helix(pos=pivot,axis=box.pos-pivot,radius=0.4,constant=1,thickness=0.1,coils=20,color=color.orange) box.p = vector(0,0,0) while (t<50): rate(100) s = wall.pos - box.pos Fspring=(L0-box.pos)*(ks) box.p= box.p +Fspring*deltat box.pos = box.pos + box.p*deltat spring.axis = box.pos - spring.pos t = t+ deltat

Link to Simulation: https://trinket.io/glowscript/3146b836dc

**A hanging oscillating spring:**

from __future__ import division from visual import * from visual.graph import * scene.width=600 scene.height = 760 g = 9.8 mball = .2 Lo = 0.3 ks = 12 deltat = 1e-3 t = 0 ceiling = box(pos=(0,0,0), size = (0.5, 0.01, 0.2)) ball = sphere(pos=(0,-0.3,0), radius=0.025, color=color.yellow) spring = helix(pos=ceiling.pos, color=color.green, thickness=.005, coils=10, radius=0.01) spring.axis = ball.pos - ceiling.pos vball = vector(0.02,0,0) ball.p = mball*vball scene.autoscale = 0 scene.center = vector(0,-Lo,0) while t < 10: rate(1000) L_vector = (mag(ball.pos) - Lo)* ball.pos.norm() Fspring = -ks * L_vector Fgrav = vector(0,-mball * g,0) Fnet = Fspring + Fgrav ball.p = ball.p + Fnet * deltat ball.pos = ball.pos + (ball.p/mball) * deltat spring.axis = ball.pos-ceiling.pos t = t + deltat

**Code to visualize a spring's motion given x and y coordinate input:**

Web VPython 3.2 scene.background = color.white ball = sphere(radius=0.03, color=color.blue) trail = curve(color=ball.color) origin = sphere(pos=vector(0,0,0), color=color.yellow, radius=0.015) spring = helix(color=color.cyan, thickness=0.006, coils=40, radius=0.015) spring.pos = origin.pos xplot = graph(title="x-position vs time", xtitle="time (s)", ytitle="x-position (m)") xposcurve = gcurve(color=color.blue, width=4, label="model") xpos2curve = gcurve(color=color.red, width=4, label="experiment") yplot = graph(title="y-position vs time", xtitle="time (s)", ytitle="y-position (m)") yposcurve = gcurve(color=color.blue, width=4, label="model") ypos2curve = gcurve(color=color.red, width=4, label="experiment") eplot = graph(title="Change in Energy vs Time", xtitle="Time (s)", ytitle="Change in Energy (J)") dKcurve = gcurve(color=color.blue, width=4, label="deltaK") dUgcurve = gcurve(color=color.red, width=4, label="deltaUgrav") dUscurve = gcurve(color=color.green, width=4, label="deltaUspring") dEcurve = gcurve(color=color.orange, width=4, label="deltaE") ball2 = sphere(radius=0.025, color=color.red) ball.m = 0.402 ball.pos = vector(0.55,-0.0039,0) ball.vel = vector(0,0,0) X = [] Y = [] obs = read_local_file(scene.title_anchor).text; for line in obs.split('\n'):

if line !=:line = line.split(',') X.append(float(line[0])) Y.append(float(line[1]))

idx = 0 #variable used to select data from list. cnt = 0 #variable to keep track of predictions made between each measurement t = 0 deltat = (5.5/len(X))/20 #choose this small AND an integer multiple of the time interval between frames of experiment video g = 9.8 k_s = 8.87 L0 = 0.123 L = spring.pos-ball.pos Lhat = L/mag(L) s = mag(L)-L0 K = (1/2)*ball.m*mag(ball.vel)**2 # kinetic energy Ug = ball.m * g * ball.pos.y # gravitational potential energy Us = 1/2*k_s*s**2 # spring potential energy E = K + Ug + Us # total energy while t < 5.5:

K_i = K Ug_i = Ug Us_i = Us E_i = E Fgrav = vector(0,-ball.m*g,0) Fspring = -k_s*s*Lhat Fnet = Fspring + Fgrav ball.vel = ball.vel+(Fnet/ball.m)*deltat ball.pos = ball.pos+ball.vel*deltat L = ball.pos-spring.pos Lhat = L/mag(L) s = mag(L)-L0 spring.axis = L trail.append(pos=ball.pos) K = (1/2)*ball.m*mag(ball.vel)**2 deltaK = K-K_i Ug = ball.m*g*ball.pos.y deltaUg = Ug-Ug_i Us = 1/2*k_s*s**2 deltaUs = Us-Us_i E = K + Ug + Us deltaE = deltaK+deltaUg+deltaUs dKcurve.plot(t,deltaK) # blue dUgcurve.plot(t,deltaUg) # red dUscurve.plot(t,deltaUs) # green dEcurve.plot(t,deltaE) # orange xposcurve.plot(t,ball.pos.x) yposcurve.plot(t,ball.pos.y) # Update time t = t + deltat rate(1000)

## Examples

Be sure to show all steps in your solution and include diagrams whenever possible

### Simple

**Question**

If a spring's spring constant is 200 N/m and it is stretched 1.5 meters from rest, what is the potential spring energy?

**Solution**

**k _{s}**= 200 N/m

**s**= 1.5 m

**U _{s}**=(0.5)k

_{s}s

^{2 }

**U _{s}**= (0.5)(200 N/m)(1.5 m)

^{2 }

**U _{s}**= 225 J

### Middling

**Question**

A horizontal spring with stiffness 0.6 N/m has a relaxed length of 10 cm. A mass of 25 g is attached and you stretch the spring to a length of 20 cm. The mass is released and moves with little friction. What is the speed of the mass at the moment when the spring returns to its relaxed length of 10cm?

**Solution**

**k _{s}**= 0.6 N/m

**s**= 0.1 m

**U _{s}**=(.5)k

_{s}s

^{2}= (.5)(0.6 N/m)(0.1 m)

^{2 }

**U _{s}** = 0.003 J

Potential Energy is Converted into Kinetic Energy (K):

**U _{s}** = K

**U _{s}** =(0.5)mv

^{2 }

0.003 J = (0.5)(0.025 kg)v^{2
}

**v ^{2}** =

^{(0.003 J)}⁄

_{((0.5)(0.025 kg))}

**v ^{2}** = 0.24 J/kg*s

**v** = 0.49 m/s

### Difficult

**Question**

A package of mass 9 kg sits on an airless asteroid with mass 8.0x10^{20} kg and radius 8.7x10^{5} m. Your goal is to launch the package so that it will never come back and when it is very far away it will have a speed of 226 m/s. You have a spring whose stiffness is 2.8x10^{5} N/m. How much must you compress the spring?

**Solution**

The initial condition for escape from the asteroid is:

K_{i}+U_{i}=^{1}⁄_{2}mv_{esc}^{2} + (-G*^{Mm}⁄_{R)}=0

Potential energy of the spring equals the total energy in the system.

** ^{1}⁄_{2}k_{s}s^{2}**=

^{1}⁄

_{2}mv

_{esc}

^{2}+(-G*

^{Mm}⁄

_{R})

**s ^{2}**=

^{m}⁄

_{ks}(

^{2GM}⁄

_{R}+v

^{2})=s

^{2}=

^{9}⁄

_{2.8x105}(

^{2G8.0x1020}⁄

_{8.7x105}+226

^{2})

**s ^{2}**=5.58 m

**s**=2.36 m

### Graphing

**Energy graph**

The energy graph of a oscillating spring continuously switches between kinetic and potential energy because springs are continuously returning to equilibrium. The following is an example of a computational model of a hanging spring with an initial force in the positive x direction. As seen below, the energies the spring contains will always equal out to zero despite each fluctuating individually.

## Connectedness

Because springs are all around us, from Slinkies to parts in automobiles, spring potential energy is useful in everyday life. One example of this is a trampoline. Without potential spring energy to allow for bounce, a trampoline would simply be a boring stretch of fabric. Spring potential is also used to absorb shock in vehicles. This allows for a smoother ride while traveling over bumps in the road. An extreme example of spring energy we commonly come across is the mechanism for garage doors. These doors contain large springs that allow them to open and close. These springs can lift an average of 400 pounds, giving them a huge amount of potential energy.

## History

Elastic Potential Energy stemmed from the ideas of Robert Hooke, a 17th century British physicist who studied the relationship between forces applied to springs and elasticity. Hooke’s Law, which is a principle that states that the that the force needed to extend or compress a spring by a distance is proportional to that distance.

## See also

Spring potential energy is related to Hooke's Law and Potential Energy.

### Further reading

http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

https://openstax.org/books/university-physics-volume-1/pages/8-1-potential-energy-of-a-system

### External links

## References

- http://www.universetoday.com/55027/hookes-law/
- http://www.scienceclarified.com/everyday/Real-Life-Physics-Vol-2/Oscillation-Real-life-applications.html
- http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html
- Chabay and Bruce A. Sherwood. Matter & Interactions. 4th ed.