Solve the Spring-Block System: Find x1 & x2

[emmy]

Homework Statement

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c07/pict_7_11.gif

In the above figure, we must apply a force of magnitude 82.0 N to hold the block stationary at x=-3.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the block is then again stationary. What are the block's positions, in cm? ((a) positive and (b) negative)

Homework Equations

F_s=-kΔx
-kx_o²-0.5kx² = W

The Attempt at a Solution

I'm sort of lost when it comes to springs, but this is my best shot:
F_s=-kΔx
82N=-k(-3-0cm)
k=2733

W=-kx_o²-0.5kx²
8J=-2733(-3)²- 0.5(2733)(x)²
8J=-24597-0.5(2733)(x)²
24605=-0.5(2733)(x)²
-18.01=x²

And that's where I'm stuck, so I definitely know it must be wrong. The only thing I can think is maybe I got the wrong sign on k, the spring constant. If so,
k=-2733
8J=2733(9)-0.5(-2733)(x)²
-17.99=x²

Either way, it's not a real number, so where did I go wrong...?

 

[ehild]

You used cm units in the expression of energy. Convert cm-s to m-s to get energy in joules.

Write up the change of elastic energy: it is equal to the work of the external force. Your equation is entirely wrong.

ehild

 

[emmy]

You used cm units in the expression of energy. Convert cm-s to m-s to get energy in joules.

Write up the change of elastic energy: it is equal to the work of the external force. Your equation is entirely wrong.

ehild

Right, sorry I did change it when finding the spring constant but forgot otherwise.

Fs=-kΔx
82N=-k(-0.03-0m)
k=2733N/m

To be honest, I'm not sure what elastic energy is, but it sounds like a kind of potential energy--which I don't know an equation for...

External Force, F_s
Work of external force=
W=F•Δx
This is the only equation i could relate from my notes to the work and external force and it can't be used because force isn't constant.

So instead:
W = $$\int_0^{.03m} f(x) \cdot dx$$
I didn't think you could do that for springs... But I'm not sure what else to do

 

[supratim1]

use conservation of energy. recheck your equations once.

 

[emmy]

use conservation of energy. recheck your equations once.

K=K_final-K_initial ?

I'm armed with equations and I'm no longer sure what they do
Though looking back, maybe I messed up on the original equation I used? Unless that one's completely inappropriate for this problem...

Fs=-kΔx
82N=-k(-0.03-0m)
k=2733N/m

W=-0.5kx_o²-0.5kx²
8J=-2733(.5)(-.03)²- 0.5(2733)(x)²
8J=-2.4597-0.5(2733)(x)²
9.23=-0.5(2733)(x)²
-.0068=x²

I'm sorry, i wish I was grasping this... I need to introduce a negative somewhere to get a positive answer so I can take a square root

 

[ehild]

The work of external for is given while compressing the spring. As it is said that the block moves very slowly, the work to accelerate the block to this speed is negligible and the acceleration of the block is zero. The applied force is in equilibrium with the force of the spring, so f(x)=kx. The integral of this force from x=-0.03 to the new position of the block is equal to 8.0 J.

ehild

 

[emmy]

The work of external for is given while compressing the spring. As it is said that the block moves very slowly, the work to accelerate the block to this speed is negligible and the acceleration of the block is zero. The applied force is in equilibrium with the force of the spring, so f(x)=kx. The integral of this force from x=-0.03 to the new position of the block is equal to 8.0 J.

ehild

So in other words,

integral from -.03 to x of kxdx = 8J
(k)(x²/2)|^x_-.03 = 8J
k = 2733N/m

8/2733 = x²/2 - (.03)²/2
0.003377 = x²/2
+ 0.082 = x

and then i'd do the same for the negative but integrate from x to -.03?

 

[ehild]

8/2733 = x²/2 - (.03)²/2
0.003377 = x²/2
+ 0.082 = x

You have two solutions for x, one positive, one negative: x=±0.082 m.

ehild

 

[emmy]

You have two solutions for x, one positive, one negative: x=±0.082 m.

ehild

Thank you very much for your help, I'm not sure I understand completely yet, but I'm going to keep staring at it until I understand!

 

[ehild]

You have got the equation for the position x. It is of second order, like the one x²=4, for example. There are two solutions x1=2 and x2=-2 as the square of both is equal to 4. But you know that, it is simple Maths. And it is valid in Physics, too

ehild