# Can't understand why work done on a spring is given by .5fx^2

• deliciousiron
In summary, a metal wire of diameter .23mm and unstretched length 1.405 is suspended vertically from a fixed point. When a 40N weight is suspended from the lower end of the wire, it stretches by an extension of 10.5mm. The work done on the spring can be calculated using the formula for work done on a spring, which is half of the average force. This is because the 40N force is constant and the work done against the spring is not solely due to this force, but also the force of gravity. Therefore, the work done on the spring is not simply the force multiplied by the distance, but is instead half of the average force multiplied by the distance. This is to account
deliciousiron

## Homework Statement

"a metal wire of diameter .23mm and unstretched length 1.405 is suspended vertically from a fixed point. when a 40n weight was suspended from the lower end of the wire, the wire stretched by an extension of 10.5mm. calculate the work done on the spring."i know the formula for work done on a spring, but i do not understand why it's given by half of the force. the 40n's force is constant, so why do i need to half it?
why isn't it just force*distance/40*(10.5*10^-3)??

why it's given by half of the average force.
It is the full average force.
It is half the maximal force.

the 40n's force is constant, so why do i need to half it.
This is the maximal force (=the force at equilibrium, where it holds the object).

yea, that was a typo.

but i don't understand why the 40n's weight _isn't_ constant

It is assumed that the spring satisfies Hooke's law, so the force (at a specific point) is proportional to the stretch (at that specific point).
The weight is constant, but the force required to extend the spring is not.

For a constant force, F, moving an object over a distance x, the work done is Fx. For a varying force, you can divide it into "infinitesmal" segments, treat the force as if it were a constant over each segment, and sum. In the limit that gives the integral $\int F(x)dx$. If you haven't taken Calculus yet, I'm afraid that won't make much sense but if you have then you know that $$\int F(x)dx= \int kx dx= (k/2)x^2$$.

First you stretch out the spring, then you attach the weight. If you just let the weight drop, it will move up and down, and exhibit simple harmonic motion. At the midway point of the motion, the force of the spring will be kx, and its elastic energy will be 0.5kx2.

deliciousiron said:
i know the formula for work done on a spring, but i do not understand why it's given by half of the force. the 40n's force is constant, so why do i need to half it?
why isn't it just force*distance/40*(10.5*10^-3)??
There are two forces acting on the mass in this situation 1) the force due to gravity 2) the 'contact' force due to the spring. You are totally correct that the work done against gravity is 40N times the distance. But they wanted you to calculate the work done against the spring, not against gravity.

Edit: Actually, you need to decide what you think the question is asking (It is not completely clear). As I think Chestermiller was saying, the question is either asking what is the energy stored in the spring, if someone pulled the spring down far enough that when the mass is attached, it doesn't oscillate. OR the question is saying if you simply attach the mass to the spring, then what is the max amount of energy which will be contained in the spring (since the system is oscillating).

Last edited:

## 1. Why is the work done on a spring given by .5fx^2?

The work done on a spring is given by .5fx^2 because it is a mathematical representation of the energy stored in a spring when it is stretched or compressed. The equation takes into account the force applied to the spring (f) and the displacement of the spring (x^2).

## 2. What do the variables in the equation .5fx^2 represent?

The variable 'f' represents the force applied to the spring, while 'x' represents the displacement of the spring from its equilibrium position. The exponent of 2 indicates that the displacement is squared, which is a common representation of elastic potential energy.

## 3. Can you explain the concept of work done on a spring?

Work done on a spring is the amount of energy required to stretch or compress the spring from its equilibrium position. This energy is stored in the spring as potential energy, and is released when the spring returns to its original position.

## 4. Why is the work done on a spring given by a half of the product of force and displacement?

The work done on a spring is given by a half of the product of force and displacement because only a portion of the force applied to the spring is used to stretch or compress it. The other portion of the force is used to overcome the spring's resistance and maintain equilibrium. Therefore, the equation only takes into account the work done by the effective force on the spring.

## 5. How is the equation .5fx^2 derived?

The equation .5fx^2 is derived from the principle of conservation of energy. When a spring is stretched or compressed, the work done on the spring is equal to the change in potential energy. Using this principle and the formula for elastic potential energy (PE = .5kx^2), we can derive the equation .5fx^2, where k represents the spring constant.

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