# Rule of the Work done by a force

I have been always taught that Work done = force x distance through which the force was applied. but recently in the elasticity I have been taught that W= (1/2) (F X D).

why did we multiply it by a half?, I know that work is the area under the curve and the curve is right-angled triangle like, but why didn't we always multiply it by a half if that's the case? is it a special case?.

• Do you mean 1/2 k$x^2$? – Bob D Jan 28 at 10:19
• no actually, i mean it as it's written – Ziad Al-Qadi Jan 28 at 10:20
• Where did you get this ? Give us a reference – Bob D Jan 28 at 10:25

Work is not generally "force times distance". That is only true when the force is constant. The general formula is (where $$x$$ is the position):

$$W=\int F\,\cdot \mathrm dx$$

An integral is mathematically always the area under the graph, as you also mention.

• For a constant force that graph is a rectangle. Then you can simplify this relation to the rectangle-area formula, width times height, thus "force times distance" (a change in position is a distance): $$W_\text{constant force}=\int F\,\cdot \mathrm dx=F\Delta x$$

• For a linearly growing force the graph is a triangle, as you mention. Then you can simplify this relation to the triangle-area formula, baseline times height times a half, thus "1/2 times final force times distance": $$W_\text{linear force}=\int F\,\cdot \mathrm dx=\frac12 F_\text{final} \Delta x$$

Springs and elastic forces that obey Hooke's law, $$F=kx$$, where $$k$$ is a spring constant, are linear (they grow linearly with position) so that's why you've seen this formula for elastic forces. (Note that Hooke's law is only obeyed by must such elastic materials within certain ranges.)

For other types of non-constant force, if you can find the mathematical pattern (is it linear, quadratic...) then you might be able to simplify the general work formula in each individual case. If not, then you will have to use the general work formula and calculate the integral from scratch every single time.

Often in real-life work you may not even know the exact force expression and it might fluctuate a lot and unpredictively. Then you might instead measure as many force-and-distance data points as possible in order to recreate the graph as closely as possible in that particular case - and then measure/compute the area under this graph. This is the experimental/geometrical way when the force variation is unknown.

• Note that $F$ in your final expression is the final force on the object, a constant, while $F$ in your first expression is a function. Your final expression might be better as $\frac{1}{2}F(\Delta x)\Delta x$. – garyp Jan 28 at 12:38
• @garyo Oh, right, good catch. Edited. – Steeven Jan 28 at 13:26

The factor $$1/2$$ in work done by an elastic force is since elastic force is not constant: $$F(x)=kx$$. This means that the work done when the spring is stretched a small distance $$dx$$ is $$dW=Fdx=kx dx$$. The total work is therefore $$W = \int_{0}^{D} kx dx = \frac{1}{2}kD^2 = \frac{1}{2}kD\cdot D=\frac{1}{2}F(D)\cdot D$$ in agreement with your formula, but note that $$F(x)$$ is not a constant so the formula $$W=\frac{1}{2}F\times D$$ is dependent on the final displacement $$x=D$$.

• In your analysis $D$ is the final displacement of the object. So to say $\frac{1}{2}F\times D$ depends on $x$ makes no sense. $x$ is a dummy variable that gets integrated away. – garyp Jan 28 at 12:31

Instead of thinking of the work as being half the force times the distance, I think it might be better to express the amount of work done against a linearly increasing force as being the average force, i.e. (F₀+F₁)/2, times the distance, where F₀ is the amount of force that had to be applied at the start of motion, and F₁ is the amount of force that had to be applied at the end. In the case where the initial force is zero, that would of course be half of the final force, but in the many real-world scenarios involving a pre-compressed spring, F₀ may be a significant fraction of F₁.