# Spring Potential Energy involving two springs

1. Sep 30, 2007

### kdm06

1. The problem statement, all variables and given/known data

A 10kg mass, attached by means of two springs to the ceiling, is held against the floor and is then released. How fast will it be traveling when it hits the ceiling? The spring constant of each spring is 80 N/m, and each spring has an unstretched length of 1 m. Assume that springs become loose and floppy once they're at their rest length.

distance between ceiling and floor: h_o=4m
horizontal distance from mass to each spring: 2m

Springs are at an angle from mass.

2. Relevant equations
PE(sp)=1/2kx^2
PE(g)=mgh
KE=1/2mv^2

3. The attempt at a solution

I used pythagorean theorem to find stretched length of spring:
s=sqrt(4^2+2^2)=sqrt(20)

I found theta to find h after the mass is released:
tan(theta)=4/2
theta=63.43494

I found h using trig:
h=unstretched length of string*sin(theta)=.8944

I used this equation to solve for v with my datum at the floor:

PE(sp)=KE+PE(g)
1/2ks^2=1/2mv^2+mgh
1/2(80)(sqrt(20))^2=1/2(10)v^2+(10)(9.81)(.89)
v=11.93m/s

2. Sep 30, 2007

### AbedeuS

try and give my analysis of the question even though I havent done spring stuff in ages.

We have a few things to work with here, the distance between the Floor and ceiling is 4m, so we have that to work with, and the horizontal distance is 2m, so lets work with that for now to find out how much the spring is actually streached by.

By imagineing a right angled triangle, the hypotamuse is created by the spring, the horizontal and vertical components attached at a right angle, using, as you said, pythagarus theorem, we can find out how far the spring is streached.

[Centre]$$2^{2}+4^{2} = 20$$[/centre]

Square roote 20 to get 4.472Meters, 1 meter is used up by the "Forceless" rest state of the spring, that is to say, if something was "Streached" to 1m, it wouldnt experience a force exherted by the spring, as thats when the spring is fully compressed anyway, so that makes for an effective displacement of 3.472 Meters for the spring from its equilibrium position.

$$E=\frac{1}{2}kx^{2}$$

This is the potential energy that the springs tension supplies theres two springs, so you multiply this potential energy by two.

Now lets assume that all potential energy from the spring transfers as kinetic energy to the Kilogramme weight, using:

$$E=\frac{1}{2}mv^{2}$$

you can find the maximum velocity of the weight if all kinetic energy is transferred to the weight and nothing (such as gravity) was decellerating this kinetic energy transfer. Maybe this helps?

Also this maths doesnt take into account the fact that when the weight reaches the top of the room, the springs would still technically be tensed by 2m, and therefore the kinetic energy given to the weight would actually be 3.472-1 meters = 2.472meters, but i think your teacher implies that all energy is given to the object as the springs become "floppy" at the end...

3. Sep 30, 2007

### kdm06

Thank you for your help. But why would the string still be tensed by 2m after it reaches the top of the room?