# Work Done By a Spring's Restoring Force

• rtareen
In summary, the conversation discusses the concept of work done by a spring's force, which can be approximated by taking small intervals of distance and summing them up. The equation for this work includes terms for both initial and final positions, but can be simplified when the initial position is taken to be the equilibrium position. However, it is possible to take a different initial position, such as at the maximum displacement. The order of the terms in the equation is not relevant, as addition is commutative. The conversation also provides some intuitive ways to think about this concept, such as considering the work done by a suspension's spring to keep a tire pressed onto the pavement.
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.

rtareen said:
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.

rtareen said:
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.

rtareen said:
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.

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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:
$$\Delta U_{spring} = -W_{spring} = \frac{1}{2} kx^2_f - \frac{1}{2}kx^2_i$$

Last edited:
rtareen said:
... 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.

## 1. What is the definition of work done by a spring's restoring force?

The work done by a spring's restoring force is the amount of energy transferred to or from an object when the spring is stretched or compressed. This work is calculated by multiplying the force applied to the spring by the distance the spring is stretched or compressed.

## 2. How is the work done by a spring's restoring force related to Hooke's Law?

According to Hooke's Law, the force exerted by a spring is directly proportional to the amount it is stretched or compressed. Therefore, the work done by the spring's restoring force is also directly proportional to the displacement of the spring.

## 3. Can the work done by a spring's restoring force be negative?

Yes, the work done by a spring's restoring force can be negative if the force and displacement act in opposite directions. This means that the spring is either being compressed while the force is pulling in the opposite direction, or stretched while the force is pushing in the opposite direction.

## 4. How is the work done by a spring's restoring force related to the potential energy of the spring?

The work done by a spring's restoring force is equal to the change in potential energy of the spring. When the spring is stretched or compressed, it gains or loses potential energy, respectively. This potential energy can then be converted into kinetic energy when the spring returns to its original position.

## 5. How does the work done by a spring's restoring force affect the motion of an object?

The work done by a spring's restoring force can either increase or decrease the kinetic energy of an object, depending on the direction of the force and displacement. This change in kinetic energy can affect the speed and direction of the object's motion.

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