Potential Energy on an ideal spring

[clope023]

[SOLVED] Potential Energy on an ideal spring

Homework Statement

An ideal spring of negligible mass is 12.00cm long when nothing is attached to it. When you hang a 3.15kg weight from it, you measure its length to be 13.30cm.

If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law. (also says there might be 2 answers)

Homework Equations

K1 + U1 + Wother
= K2 + U2

or

Wel = 1/2ky1^2 - 1/2ky2^2
= Uel1 - Uel2

U = mgy

The Attempt at a Solution

12cm = .12m = y1
13.30cm = .133m = y2
10J = Wother

1/2ky1^2 = 10J + mgy1 + 1/2ky2^2

1/2k(.0144m^2) = 10J + (3.15kg)(9.8m/s^2)(.133m) + 1/2k(.0177m^2)

-14.11J = 1/2k(.0177m^2-.0144m^2)

-14.11J = 1/2k(.0033m^2)

-28.22J = k(.0033m^2)

-28.22m = (.003m^2)

is this right? would I just divide -28.22m by .003m^2 and get a #? I'm pretty sure it's not right but I'm not sure of other equations to use, should I use kinematics as well?

edit: or perhaps I should have another potential energy in there for y1? or use the potential energy for mass in y1 and solve for y2?

or perhaps use this equation?

K1 + Ugrav1 + Uel1 + Wother = K2 + Ugrav2 + Uel2

seems like this would work better than what I was using?

 

[Kurdt]

The potential energy of a spring is given by:

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

You can work out the spring constant from the initial info they give you.

 

[clope023]

The potential energy of a spring is given by:

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

You can work out the spring constant from the initial info they give you.

you saying I should do:

10J = 1/2k((.133m)^2-(.12m)^2))

20J = k(.0033m^2)

20J/.003m^2 = k

k = 6060?

or should I use the gravitational potentiona energy from the 3.15kg mass?

(3.15)gy = 1/2k((.133m)^2-(.12m)^2))?

 

[Kurdt]

You need to work out the spring constant by how far the mass stretches the spring, then use the potential energy formula I gave to solve for x. Since the answer will involve a square root there will be two answers.

 

[clope023]

You need to work out the spring constant by how far the mass stretches the spring, then use the potential energy formula I gave to solve for x. Since the answer will involve a square root there will be two answers.

so:

10J = 1/2k(.133m)^2

20J = k(.0177m^2)

20J/.0177m^2 = k

1129.9 = k

and mgy = 1/2kx^2

(3.15(9.8)(.133m)2/k)^1/2 = x?

 

[clope023]

x = -8.5cm, 8.5cm ?

 

[Kurdt]

The spring constant is a force per unit length. You can work it out by dividing the weight of the mass by how much the spring moves from equilibrium. Then solve for x in the potential energy equation.

 

[clope023]

The spring constant is a force per unit length. You can work it out by dividing the weight of the mass by how much the spring moves from equilibrium. Then solve for x in the potential energy equation.

so k = (mg)/y2 = (13.5*9.8)/.133m = 232J?

and is the potential energy in the equation you gave gravitational or elastic?

cause I could solve it like this

mgy = 1/2kx^2

4.11J = 1/2kx^2

8.22J = kx^2

8.22J/232J = x^2

.035 = x^2

x = -.188m, .188m = -19cm, 19cm?\

or is it:

10J = 1/2kx^2

20J = kx^2

20J/232J = x^2

.086 = x^2

x = -.29m, .29m = -29cm, 29cm?

 

[Kurdt]

so k = (mg)/y2 = (13.5*9.8)/.133m = 232J?

No, the spring stretches 1.3cm from equilibrium. The units of the spring constant are not joules. The only purpose of the information about the mass is so you can determine the spring constant.

and is the potential energy in the equation you gave gravitational or elastic?

Its the elastic potential energy. That is the energy stored in the spring.

Now you know how to obtain the spring constant and you know how much energy you want to store you can solve for x in the potential energy equation.

 

[clope023]

No, the spring stretches 1.3cm from equilibrium. The units of the spring constant are not joules. The only purpose of the information about the mass is so you can determine the spring constant.




Its the elastic potential energy. That is the energy stored in the spring.

Now you know how to obtain the spring constant and you know how much energy you want to store you can solve for x in the potential energy equation.

ooooooooh, it goes from 12cm to 13.3cm which is 1.3cm which .013m, so

Fx = kx, k = Fx/x = = (3.15kg*9.8m/s^2)/.013m = 30.9N/.013m = 2377N/m

and 10J = 1/2kx^2

20J = kx^2

20J/(2377N/m) = x^2

x = .092m = -9.2cm, 9.2cm?

 

[Kurdt]

Thats good but they want the total length. So that will be 12cm +/- 9.2.

 

[clope023]

Thats good but they want the total length. So that will be 12cm +/- 9.2.

right the initial length when nothing was attached to it, so it would be:

2.8, 21.2? thanks alot!