Spring Potential Energy

Work in progress by scarswell3 This topic covers Spring Potential Energy.

Spring Potential Energy

Elastic Potential Energy is the energy stored in elastic materials due to their deformation. Often this refers to the stretching or compressing of a spring.

A Mathematical Model

The formula for Ideal Spring Energy is:

U_s=¹⁄₂k_ss²

where:

k_s= spring constant

s²= stretch measured from the equilibrium point;

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Spring Potential Energy

A Computational Model

An oscillating spring can be modeled by the following:

 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

Examples

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

Simple

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?

k_s= 200 N/m

s= 1.5 m

U_s=(0.5)k_ss²

U_s= (0.5)(200 N/m)(1.5 m)²

U_s= 225 J

Middling

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?

k_s= 0.6 N/m

s= 0.1 m

U_s=(.5)k_{ss^{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)mv²

0.003 J=(0.5)(0.025 kg)v²

v²=^{(0.003 J)}⁄_{((0.5)(0.025 kg)}

v²=0.24 J/kg*s

v=0.49 m/s

Difficult

A package of mass 9 kg sits on an airless asteroid with mass 8.0x10²⁰ kg and radius 8.7x10⁵ 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 2.8x10⁵ N/m. How much must you compress the spring?

The initial condition for escape from the asteroid is:

K_i+U_i=¹⁄₂mv_esc² + (-G*^Mm⁄_R)=0

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

¹⁄₂k_ss²=¹⁄₂mv_esc² +(-G*^Mm⁄_R)

s²= ^m⁄_{k_s}(^2GM⁄_R +v²)=s² = ⁹⁄_2.8x10⁵(^{2G8.0x10²⁰}⁄_8.7x10⁵+226²)

s²=5.58 m

s=2.36 m

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.

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.

Spring Potential Energy

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