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

by deliciousiron
Tags: 5fx2, spring, work
 P: 2 1. The problem statement, all variables and given/known data "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)??
Mentor
P: 10,813
 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).
 P: 2 yea, that was a typo. but i don't understand why the 40n's weight _isn't_ constant
Mentor
P: 10,813

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

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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,885 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 $$\int F(x)dx= \int kx dx= (k/2)x^2$$.
 Sci Advisor HW Helper Thanks PF Gold P: 4,446 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.
HW Helper
P: 3,337
 Quote by deliciousiron 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).

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