Work Done By a Spring's Restoring Force

[rtareen]

Homework Statement

This is not a homework problem but rather a conceptual question.

Relevant Equations

$W_{s}=\frac {1}{2}kx^{2}_{i}-\frac {1}{2}kx ^{2}_{f}$

Hello all.

Right now I am taking physics 1 and were doing the Work-Kinetic Energy Chapter. I was just reading the derivation for the work done by a spring's force. I understand how to get the result but what i don't understand what to make of it. I understand that because a spring's force varies with distance, that we can approximate the work done by taking small intervals of distance,$\Delta x$, from $x_{i}$ to $x_{f}$ and calculating the work the done over these small intervals. Then we can take the sum of all of these works to approximate the total work. Then when we take the limit as $\Delta x$ goes to 0 we get this integral:$-k\int ^{x_{f}}_{x_{i}}xdx = \frac {1}{2}kx^{2}_{i}-\frac {1}{2}kx ^{2}_{f}$

I know this reduces to the less complicated equation $W_{s}=-\frac {1}{2}kx ^{2}_{f}$ when we take the equilibrium position to be $x_{i} = 0$. Why would we ever take a position other than this to be $x_{i}$?

Also, what is the way to think about this, sincethe term with $x_{i}$ comes before the term with $x_{f}$? Any other insight would be appreciated as well.

 

[Orodruin]

Homework Statement:: This is not a homework problem but rather a conceptual question.
Relevant Equations:: $W_{s}=\frac {1}{2}kx^{2}_{i}-\frac {1}{2}kx ^{2}_{f}$

Why would we ever take a position other than this to be xixix_{i}?

Your system may not initially be at the equilibrium position.

The order of the terms is not relevant since addition is commutative. Note that the work done by the spring on the mass is the change in kinetic energy of the mass. If you store potential energy in the spring, the kinetic energy in the mass goes down.

 

[PeroK]

I know this reduces to the less complicated equation $W_{s}=-\frac {1}{2}kx ^{2}_{f}$ when we take the equilibrium position to be $x_{i} = 0$. Why would we ever take a position other than this to be $x_{i}$?

It would make sense also to take $x_i$ to be the starting position, perhaps at the maximum displacement.

Also, what is the way to think about this, sincethe term with $x_{i}$ comes before the term with $x_{f}$? Any other insight would be appreciated as well.

It's often conventional to write $b - a$, rather than $-a + b$; although they are equivalent.

 

[Leo Liu]

I am giving you an intuitive way to comprehend this concept because physics 1 is algebra based.
Please look at the graph below. The reason why $F(x_i)$ is negative when $x_i$ is positive is because $F=-kx$ tells you the sign of $F$ is opposite to the sign of $x$.

If $x_i=0$, the area of the green triangle is 0.
Also notice that the potential energy U is -W. If you want to find U the equation will become:
$$ \Delta U_{spring} = -W_{spring} = \frac{1}{2} kx^2_f - \frac{1}{2}kx^2_i $$

 

[Lnewqban]

... I was just reading the derivation for the work done by a spring's force...

Also, what is the way to think about this, sincethe term with $x_{i}$ comes before the term with $x_{f}$? Any other insight would be appreciated as well.

Think about the work a suspension's spring does to keep a tire pressing over the pavement immediately after it rolls over a bump.
It is possible that the bump effect is strong enough to catapult the tire upward and keep it detached from the pavement for some time, loosing needed grip.
In that case, the job of the spring is to use the energy differential between the maximum compression position $(x_{i})$ and the extension position $(x_{f})$ to push the tire downwards faster that it naturally would fall back down.